Calculus: Single Variable Part 3 – Integration Coursera Quiz Answers 2022 [πŸ’―Correct Answer]

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In this third part, the third of five, we talk about how to integrate differential equations, how to integrate, the fundamental theorem of integral calculus, and how to integrate hard problems.

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Calculus: Single Variable Part 3 – Integration Quiz Answers

Week 01: Calculus: Single Variable Part 3 – Integration Coursera Quiz Answers

Main Quiz 01

Q1. \displaystyle \int (4x^3 + 3x^2 + 2x + 1) \, dx =∫(4x3+3x2+2x+1)dx=

  • 4x^4 + 3x^3 + 2x^2 + x + C4x4+3x3+2x2+x+C
  • \displaystyle \frac{1}{4}x^4 + \frac{1}{3}x^3 + \frac{1}{2}x^2 + x + C41​x4+31​x3+21​x2+x+C
  • x^4 + x^3 + x^2 + x + Cx4+x3+x2+x+C
  • 4x^3 + 3x^2 + x + C4x3+3x2+x+C
  • 12x^2 + 6x + 1 + C12x2+6x+1+C
  • 3x^2 + x + C3x2+x+C

Q2. \displaystyle \frac{d}{dx} \int \ln \tan x \, dx =dxdβ€‹βˆ«lntanxdx=

  • \ln \tan x + Clntanx+C
  • \displaystyle \frac{\sec^2 x}{\tan x}tanxsec2x​
  • 00
  • \displaystyle \frac{\sec^2 x}{\tan x} + Ctanxsec2x​+C
  • \ln \tan xlntanx
  • \displaystyle \int \frac{\sec^2 x}{\tan x} \, dx∫tanxsec2x​dx

Q3. \displaystyle \int \left( \frac{d}{dx} e^{-x} \right) \, dx =∫(dxd​eβˆ’x)dx=

  • e^{x} + Cex+C
  • e^{-x}eβˆ’x
  • e^{-x} + Ceβˆ’x+C
  • \displaystyle \frac{d}{dx} \int e^{-x} \, dxdxdβ€‹βˆ«eβˆ’xdx
  • -e^{-x} + Cβˆ’eβˆ’x+C
  • -e^{-x}βˆ’eβˆ’x

Q4. Find the general solution of the differential equation

\frac{dx}{dt} = t^2dtdx​=t2

  • \displaystyle x(t) = \frac{1}{2} t^2 + Cx(t)=21​t2+C
  • x(t) = t^2 + Cx(t)=t2+C
  • \displaystyle x(t) = t^3 + Cx(t)=t3+C
  • \displaystyle x(t) = \frac{1}{3} t^3 + Cx(t)=31​t3+C
  • \displaystyle x(t) = \frac{1}{3} t^3x(t)=31​t3
  • \displaystyle x(t) = t^3x(t)=t3

Q5. Find the general solution of the differential equation

\frac{dx}{dt} = x^2dtdx​=x2

Hint: think in terms of differentials, as we did in the Lecture. This will allow you to bring everything that depends on xx to the left hand side, and everything that depends on tt to the right. Then integrate.

  • \displaystyle x(t) = -\frac{1}{t} + Cx(t)=βˆ’t1​+C
  • \displaystyle x(t) = \frac{1}{-t + C}x(t)=βˆ’t+C1​
  • \displaystyle x(t) = -\frac{1}{t^2} + Cx(t)=βˆ’t21​+C
  • \displaystyle x(t) = -\frac{1}{t}x(t)=βˆ’t1​
  • \displaystyle x(t) = -\frac{1}{t^2}x(t)=βˆ’t21​
  • \displaystyle x(t) = \frac{1}{-t^2 + C}x(t)=βˆ’t2+C1​

Practice Quiz 01

Q1. There is a large class of differential equations β€”the so-called linear onesβ€” for which we can find solutions using the Taylor series method discussed in the Lecture. One such differential equation is

td2xdt2+dxdt+tx=0(βˆ—)

It is a particular case of the more general Bessel differential equation, and one solution of it is given by the Bessel function J_0(t)J0​(t) that we saw in Chapter 1. Notice that (\astβˆ—) involves not only the first derivative \displaystyle \frac{dx}{dt}dtdx​ but also the second derivative \displaystyle \frac{d^2x}{dt^2}dt2d2x​. For this reason, it is said to be a second order differential equation.

In this problem we will content ourselves with finding a relationship (specifically, a recurrence relation) on the coefficients of a Taylor series expansion about t=0t=0 of a solution to our equation. Hence consider the Taylor series

x(t) = \sum_{k=0}^\infty c_k t^kx(t)=k=0βˆ‘βˆžβ€‹ck​tk

Substituting this into (\astβˆ—) will give you two conditions. The first one is c_1 = 0c1​=0. What is the other one?

Note: this problem involves some nontrivial manipulation of indices in summation notation. Do not get discouraged if it feels more difficult than other problems: it is!

  • \displaystyle c_k = \frac{c_{k-2}}{k^2}ck​=k2ckβˆ’2​​
  • \displaystyle c_k = \frac{c_{k-1}}{k^2}ck​=k2ckβˆ’1​​
  • \displaystyle c_k = – \frac{c_{k-1}}{(k-1)^2}ck​=βˆ’(kβˆ’1)2ckβˆ’1​​
  • \displaystyle c_k = – \frac{c_{k-2}}{k^2}ck​=βˆ’k2ckβˆ’2​​
  • \displaystyle c_k = – \frac{c_{k-2}}{(k-2)^2}ck​=βˆ’(kβˆ’2)2ckβˆ’2​​
  • \displaystyle c_k = – \frac{c_{k-1}}{k^2}ck​=βˆ’k2ckβˆ’1​​

Main Quiz 02

Q1. After drinking a cup of coffee, the amount CC of caffeine in a person’s body obeys the differential equation

\frac{dC}{dt}= -\alpha CdtdC​=βˆ’Ξ±C

where the constant \alphaΞ± has an approximate value of 0.14 hours^{-1}βˆ’1.

How many hours will it take a human body to metabolize half of the initial amount of caffeine? Round your answer to the nearest integer.

Q2. The amount II of a radioactive substance in a given sample will decay in time according to the following equation: \frac{dI}{dt} = -\lambda IdtdI​=βˆ’Ξ»I Nuclear engineers and scientists tend to be concerned with the half-life of a substance, that is, the time it takes for the amount of radioactive material to be halved.

Find the half-life of a substance in terms of its decay constant \lambdaΞ».

  • \displaystyle \frac{\ln 2}{\lambda}Ξ»ln2​
  • \displaystyle \frac{2}{\lambda}Ξ»2​
  • \displaystyle \frac{\lambda}{2}2λ​
  • \displaystyle \frac{\lambda}{\ln 2}ln2λ​
  • \displaystyle \frac{\ln \lambda}{2}2lnλ​
  • \displaystyle \frac{2}{\ln \lambda}lnΞ»2​

Q3. In a highly viscous fluid, a falling spherical object of radius rr decelerates right before reaching the bottom of the container. A simple model for this behavior is provided by the equation

\frac{dh}{dt} = – \frac{\alpha}{r}hdtdh​=βˆ’rα​h

where hh is the height of the object measured from the bottom, and \alphaΞ± is a constant that depends on the viscosity of the fluid.

Find the time it would take the object to drop from h = 6rh=6r to h = 2rh=2r in terms of \alphaΞ± and rr.

  • \displaystyle \frac{\alpha}{r} \ln \frac{1}{3}rα​ln31​
  • \displaystyle \frac{\alpha}{r} \ln 12rα​ln12
  • \displaystyle \frac{r}{\alpha} \ln 3Ξ±r​ln3
  • \displaystyle \frac{\alpha}{r} \ln 3rα​ln3
  • \displaystyle \frac{r}{\alpha} \ln \frac{1}{3}Ξ±r​ln31​
  • \displaystyle \frac{r}{\alpha} \ln 12Ξ±r​ln12

Q4. On a cold day you want to brew a nice hot cup of tea. You pour boiling water (at a temperature of 212^\mathrm{o} F212oF) into a mug and drop a tea bag in it. The water cools down in contact with the cold air according to Newton’s law of cooling:

\frac{dT}{dt} = \kappa (A – T)dtdT​=ΞΊ(Aβˆ’T)

where TT is the temperature of the water, A = 32^\mathrm{o} FA=32oF the ambient temperature, and \kappa = 0.36ΞΊ=0.36 min^{-1}βˆ’1.

The threshold for human beings to feel pain when entering in contact with something hot is around 107^\mathrm{o} F107oF. How many seconds do you have to wait until you can safely take a sip? Round your answer to the nearest integer.

Q5. On the night of April 14, 1912, the British passenger liner RMS Titanic collided with an iceberg and sank in the North Atlantic Ocean. The ship lacked enough lifeboats to accommodate all of the passengers, and many of them died from hypothermia in the cold sea waters. Hypothermia is the condition in which the temperature of a human body drops below normal operating levels (around 36^\mathrm{o} C36oC). When the core body temperature drops below 28^\mathrm{o} C28oC, the hypothermia is said to have become severe: major organs shut down and eventually the heart stops.

If the water temperature that night was -2^\mathrm{o} Cβˆ’2oC, how long did it take for passengers of the Titanic to enter severe hypothermia? Recall from lecture that heat transfer is described by Newton’s law of cooling:

\frac{dT}{dt} = \kappa (A – T)dtdT​=ΞΊ(Aβˆ’T)

where TT is the body temperature of a passenger, AA the water temperature, and \kappa = 0.016ΞΊ=0.016 min^{-1}βˆ’1. Give your answer in minutes and round it to the nearest integer.

Practice Quiz 02

Q1. Assume the size PP of a population grows following the differential equation

\frac{dP(t)}{dt} = bP(t)dtdP(t)​=bP(t)

The rate of growth bb is the difference between the birth rate and the mortality rate. The Malthusian model supposes that this rate is constant. Of course we know that does not always happen: factors such as famines, outbreaks of disease or advances in medicine do influence these rates. When modeling some process mathematically it is important to recognize what our assumptions are and when they no longer hold. In this problem and the next we will look at one particular event that would result in a violation of Malthus’ assumption that the growth rate bb is constant over time: the occurrence of famines.

Experimental data suggests that food production FF grows linearly over time:

F(t) = F_0 + stF(t)=F0​+st

We will now make two assumptions:

  • Most food is perishable, so that the amount of food available at any given time is exactly the amount produced at that time. This means that we are not taking into account the effect of the possibility of preserving food for long periods of time.
  • The amount of food that a person in our population eats is, on average, constant and equal to some number \alphaΞ±. That is, the amount of food necessary to keep everybody fed is \alpha P(t)Ξ±P(t).

The so-called Malthusian catastrophe happens when there is not enough food to feed the whole population, that is, when \alpha P(t) = F(t)Ξ±P(t)=F(t).

The Food and Agriculture Organization of the United Nations considers that a person needs around 1800 kcal/day to be considered well-fed. According to a report released in 2002 by the same organization:

  • The world population in 2002 was around 6 billion.
  • The population growth rate was estimated at 1.1% and expected to remain approximately constant for several decades.
  • Total food production in 2002 was determined to be around 6.13 \cdot 10^{15}6.13β‹…1015 kcal, with an expected growth rate of 1.11 \cdot 10^{14}1.11β‹…1014 kcal/year.

[Source: FAO, β€œWorld Agriculture: Towards 2015/2030. Summary Report”, 2002]

Estimate the world population by the year 2030.

  • 8.0 billion
  • 8.8 billion
  • 8.2 billion
  • 8.6 billion
  • 7.8 billion
  • 8.4 billion

Q2. [Continued from previous question] Estimate when the Malthusian catastrophe would happen if our assumptions continue to hold.

  • 39.42 e^{0.011 t} = 61.3 + 1.11 t \Rightarrow t \approx 16639.42e0.011t=61.3+1.11tβ‡’tβ‰ˆ166 years.
  • 39.42 e^{1.1 t} = 61.3 + 1.11 t \Rightarrow t \approx 0.439.42e1.1t=61.3+1.11tβ‡’tβ‰ˆ0.4 years.
  • 0.11 e^{1.1 t} = 61.3 + 1.11 t \Rightarrow t \approx 60.11e1.1t=61.3+1.11tβ‡’tβ‰ˆ6 years.
  • 0.11 e^{0.011 t} = 61.3 + 1.11 t \Rightarrow t \approx 8270.11e0.011t=61.3+1.11tβ‡’tβ‰ˆ827 years.

Main Quiz 03

Q1. Solve the differential equation \displaystyle \frac{dx}{dt} = \frac{x}{t}dtdx​=tx​.

Note: observe that this equation can be rearranged as \displaystyle \frac{dx}{x} = \frac{dt}{t}xdx​=tdt​, which says that the relatives rates of change of xx and tt are equal.

  • x(t) = Ctx(t)=Ct
  • x(t) = Ce^tx(t)=Cet
  • x(t) = \ln t + Cx(t)=lnt+C
  • x(t) = e^t + Cx(t)=et+C
  • x(t) = t + Cx(t)=t+C
  • x(t) = \ln(t + C)x(t)=ln(t+C

Q2. Solve the differential equation \displaystyle \frac{dx}{dt} = \frac{\sqrt{1-x^2}}{\sqrt{1-t^2}}dtdx​=1βˆ’t2​1βˆ’x2​​.

  • x(t) = \arcsin \sin(t + C)x(t)=arcsinsin(t+C)
  • x(t) = \sin ( \arcsin t + C )x(t)=sin(arcsint+C)
  • x(t) = \sin \arcsin(t + C)x(t)=sinarcsin(t+C)
  • x(t) = Ctx(t)=Ct
  • x(t) = t + Cx(t)=t+C
  • x(t) = \arcsin ( \sin t + C)x(t)=arcsin(sint+C)

Q3. Given that x(0)=0x(0)=0 and \displaystyle \frac{dx}{dt} = te^xdtdx​=tex, compute x(1)x(1).

  • x(1) = 0x(1)=0
  • x(1) = \sqrt{2}x(1)=2​
  • x(1) = \ln 2x(1)=ln2
  • x(1) = 2x(1)=2
  • x(1) = \displaystyle \ln \frac{1}{2}x(1)=ln21​
  • x(1) = \displaystyle\frac{1}{2}x(1)=21​

Q4. Which of the following is the integrating factor used to solve the following linear differential equation?

t^2\frac{dx}{dt}=4t-t^5xt2dtdx​=4tβˆ’t5x

  • I(t) = e^{-1/2t^2}I(t)=eβˆ’1/2t2
  • I(t) = e^{-t^6/6}I(t)=eβˆ’t6/6
  • I(t) = e^{t^6/6}I(t)=et6/6
  • I(t) = e^{1/2t^2}I(t)=e1/2t2
  • I(t) = e^{-t^4/4}I(t)=eβˆ’t4/4
  • I(t) = e^{t^4/4}I(t)=et4/4

Q5. Solve the differential equation \displaystyle \frac{dx}{dt} – 5x = 3dtdxβ€‹βˆ’5x=3.

  • \displaystyle x(t) = \frac{5}{3} + C e^{5t}x(t)=35​+Ce5t
  • \displaystyle x(t) = -\frac{3}{5} + C e^{-5t}x(t)=βˆ’53​+Ceβˆ’5t
  • \displaystyle x(t) = \frac{3}{5} + C e^{5t}x(t)=53​+Ce5t
  • \displaystyle x(t) = -\frac{3}{5} + C e^{5t}x(t)=βˆ’53​+Ce5t
  • \displaystyle x(t) = -\frac{5}{3} + C e^{-5t}x(t)=βˆ’35​+Ceβˆ’5t
  • \displaystyle x(t) = -\frac{5}{3} + C e^{5t}x(t)=βˆ’35​+Ce5t

Practice Quiz 03

Q1. Solve the differential equation \displaystyle \frac{dx}{dt} = \frac{x}{1+t} + 2dtdx​=1+tx​+2.

  • \displaystyle x(t) = 1 + t + \frac{C}{1+t}x(t)=1+t+1+tC​
  • x(t) = 2(1+t)\ln(1+t) + C(1+t)x(t)=2(1+t)ln(1+t)+C(1+t)
  • \displaystyle x(t) = C(1 + t) + \frac{1}{1+t}x(t)=C(1+t)+1+t1​
  • x(t) = 2C(1+t)\ln(1+t) + (1+t)x(t)=2C(1+t)ln(1+t)+(1+t)
  • x(t) = 2t(1+t) + C(1+t)x(t)=2t(1+t)+C(1+t)
  • \displaystyle x(t) = \frac{1 + t}{2} + \frac{C}{1+t}x(t)=21+t​+1+tC​

Q2. Suppose that, in order to buy a house, you obtain a mortgage. If the lender advertises an annual interest rate rr, your debt DD will increase exponentially according to the simple O.D.E.

\frac{dD}{dt} = rD.dtdD​=rD.

If you pay your debt at a rate of PP (continuous annual rate), the evolution of your debt will then (under assumptions of continual compounding and payment) obey the linear differential equation

\frac{dD}{dt} = rD – PdtdD​=rDβˆ’P

Using this model, answer the following question: if initial amount of the mortgage is for $400,000, the annual interest rate is 5%, and you pay at a rate of $40,000 every year, how many years will it take you to pay off the debt? Round your answer to the nearest integer

Q3. German physician Ernst Heinrich Weber (1795-1878) is considered one of the fathers of experimental psychology. In his study of perception, he noticed that the perceived difference between two almost-equal stimuli is proportional to the percentual difference between them. In terms of differentials, we can express Weber’s law as

dp = k \frac{dS}{S}dp=kSdS​

where pp is the perceived intensity of a stimulus and SS its actual strength. Observe that \displaystyle \frac{dS}{S}SdS​ is the relative rate of change of SS.

In what way must the magnitude of a stimulus change in time for a human being to perceive a linear growth?

  • Logarithmically.
  • Rationally.
  • Proportional to the square root.
  • Linearly.
  • Exponentially.
  • None of these.

Q4. Some nonlinear differential equations can be reduced to linear ones by a clever change of variables. Bernouilli equations

\frac{dx}{dt} + p(t) x = q(t) x^\alpha, \qquad \alpha \in \mathbb{R}dtdx​+p(t)x=q(t)xΞ±,α∈R

constitute the most important case. Notice that for \alpha=0Ξ±=0 or \alpha=1Ξ±=1 the above equation is already linear. For other values of \alphaΞ±, the substitution u = x^{1-\alpha}u=x1βˆ’Ξ± yields a linear differential equation in the variable uu.

Apply the above change of variables in the case

\frac{dx}{dt} + 2tx = x^3dtdx​+2tx=x3

to find the linear differential equations satisfied by uu.

  • \displaystyle \frac{du}{dt} – 4tu = -2dtduβ€‹βˆ’4tu=βˆ’2
  • \displaystyle \frac{du}{dt} + 2tu = 1dtdu​+2tu=1
  • \displaystyle \frac{du}{dt} + 2tu = -2dtdu​+2tu=βˆ’2
  • \displaystyle \frac{du}{dt} – 4tu = 1dtduβ€‹βˆ’4tu=1
  • \displaystyle \frac{du}{dt} = -2-2tdtdu​=βˆ’2βˆ’2t
  • \displaystyle \frac{du}{dt} = 1-2tdtdu​=1βˆ’2t

Main Quiz 04

Q1. For a differential equation

\frac{dx}{dt} = f(x)dtdx​=f(x)

with an equilibrium at x=0x=0, the linearization of the differential equation at the equilibrium is, as per our example in class,

\frac{dx}{dt} = f'(0)x ,dtdx​=fβ€²(0)x,

and one hopes that the solution to the linearized equation provides a good approximation to the original, nonlinear equation. (It does, so long as f'(0)\neq 0fβ€²(0)​=0).

The differential equation

\frac{dx}{dt} = (e^x-1)(x-1)dtdx​=(exβˆ’1)(xβˆ’1)

has an equilibrium at x=0x=0. Which of the following is the linearized equation at this equilibrium?

  • \displaystyle \frac{dx}{dt} = -exdtdx​=βˆ’ex
  • \displaystyle \frac{dx}{dt} = 0dtdx​=0
  • \displaystyle \frac{dx}{dt} = e^xdtdx​=ex
  • \displaystyle \frac{dx}{dt} = xdtdx​=x
  • \displaystyle \frac{dx}{dt} = -xdtdx​=βˆ’x
  • \displaystyle \frac{dx}{dt} = exdtdx​=ex

Q2. The differential equation in Question 1 has another equilibrium at x=1x=1. Let’s look at that. Recall, for a differential equation

\frac{dx}{dt} = f(x)dtdx​=f(x)

with an equilibrium at x=ax=a, the linearization of the differential equation at the equilibrium is best expressed as a differential equation on a new variable, h=x-ah=xβˆ’a. You should think of hh as a small perturbation to the equilibrium. This perturbation changes at a rate given by the linearization at aa:

\frac{dh}{dt} = f'(a)h ,dtdh​=fβ€²(a)h,

and one hopes that the solution to the linearized equation provides a good approximation for the original, nonlinear equation. (It does, so long as f'(a)\neq 0fβ€²(a)​=0).

The differential equation

\frac{dx}{dt} = (e^x-1)(x-1)dtdx​=(exβˆ’1)(xβˆ’1)

has an equilibrium at x=1x=1. What is the linearized equation at this equilibrium?

  • \displaystyle\frac{dh}{dt} = -hdtdh​=βˆ’h
  • \displaystyle\frac{dh}{dt} = (e-1)hdtdh​=(eβˆ’1)h
  • \displaystyle\frac{dh}{dt} = -ehdtdh​=βˆ’eh
  • \displaystyle\frac{dh}{dt} = ehdtdh​=eh
  • \displaystyle\frac{dh}{dt} = hdtdh​=h
  • \displaystyle\frac{dh}{dt} = 0dtdh​=0

Q3. It is easy to determine stability or instability of an equilibrium. Recall that an equilibrium x=ax=a for \displaystyle\frac{dx}{dt}=f(x)dtdx​=f(x) is stable if f'(a)\lt 0fβ€²(a)<0 and is unstable if f'(a)\gt 0fβ€²(a)>0.

Recall from Lecture 18, Newton’s Law of Heat Transfer, which states that

\frac{dT}{dt} = \kappa ( A – T ),dtdT​=ΞΊ(Aβˆ’T),

where \kappa>0ΞΊ>0 is a thermal conductivity constant and AA is the (constant) ambient temperature. Find and classify the equilibria in this system.

  • T=-AT=βˆ’A is the unique equilibrium. It is unstable.
  • There are no equilibria, which is why the temperature never quite stops changing.
  • T=\kappaT=ΞΊ is the unique equilibrium. It is unstable.
  • T=\kappaT=ΞΊ is the unique equilibrium. It is stable.
  • T=AT=A is the unique equilibrium. It is stable.
  • There are equilibria at T=0T=0 (unstable) and T=AT=A (stable).

Q4. Consider the differential equation

\frac{dx}{dt} = f(x)dtdx​=f(x)

and assume that f(a)=0f(a)=0 for some constant aa.

Assume also that the function ff has a Taylor expansion at x = ax=a,

f(x) = b(x-a) + O\big((x-a)^2\big)f(x)=b(xβˆ’a)+O((xβˆ’a)2)

for some constant bb.

Which of the following statements about the differential equation are true? Select all that apply.

  • bb is a stable equilibrium if a \gt 0a>0.
  • bb is a stable equilibrium if a \lt 0a<0.
  • aa is a stable equilibrium if b \lt 0b<0.
  • bb is an unstable equilibrium if a \gt 0a>0.
  • aa is a stable equilibrium if b \gt 0b>0.
  • Neither aa nor bb is an equilibrium point.
  • aa is an unstable equilibrium if b \gt 0b>0.
  • aa is an unstable equilibrium if b \lt 0b<0.

Q5. Consider the differential equation

\frac{dy}{dt} = -2y + y^2 + y^3dtdy​=βˆ’2y+y2+y3

Find and classify all the equilibria.

  • y=0y=0: unstable. y=1y=1 and y=-2y=βˆ’2: unstable.
  • y=1y=1: stable. y=0y=0 and y=-2y=βˆ’2: unstable.
  • y=0y=0: stable. y=-1y=βˆ’1 and y=2y=2: unstable.
  • y=0y=0: stable. y=1y=1 and y=-2y=βˆ’2: unstable.
  • y=2y=2: unstable. y=0y=0 and y=-1y=βˆ’1: stable.
  • y=0y=0, y=1y=1, and y=-2y=βˆ’2: stable.

Q6. Recall from Lecture 19 how we computed the terminal velocity of a falling body with linear drag given by

m\frac{dv}{dt} = mg – \kappa v ,mdtdv​=mgβˆ’ΞΊv,

where, of course, mm is mass, gg is gravitation, vv is velocity, and \kappa>0ΞΊ>0 is the drag coefficient. Can you see how easily one can solve for the equilibrium v_\infty = mg/\kappavβˆžβ€‹=mg/ΞΊ?

Very good. Now, let’s use a more realistic model of drag that is quadratic as opposed to linear:

m\frac{dv}{dt} = mg – \lambda v^2 ,mdtdv​=mgβˆ’Ξ»v2,

where \lambda>0Ξ»>0 is a constant drag coefficient. This differential equation is not as easy to solve (but soon you will learn how). Is there is terminal velocity? What is it?

  • \displaystyle v_\infty = \sqrt{\frac{mg}{\lambda}}vβˆžβ€‹=Ξ»mg​​ is an unstable equilibrium; there is no terminal velocity.
  • \displaystyle v_\infty = 0vβˆžβ€‹=0 is an unstable equilibrium; there is no terminal velocity.
  • \displaystyle v_\infty = 0vβˆžβ€‹=0 is an unstable equilibrium; this is the terminal velocity.
  • \displaystyle v_\infty = \frac{mg}{\lambda}vβˆžβ€‹=Ξ»mg​ is an unstable equilibrium; there is no terminal velocity.
  • \displaystyle v_\infty = \sqrt{\frac{mg}{\lambda}}vβˆžβ€‹=Ξ»mg​​ is a stable equilibrium; this is the terminal velocity.
  • \displaystyle v_\infty = \frac{mg}{\lambda}vβˆžβ€‹=Ξ»mg​ is a stable equilibrium; this is the terminal velocity.

Q7. Recall that with continuous compounding at an interest rate of r>0r>0, an investment I(t)I(t) with initial investment I_0=I(0)I0​=I(0) is I(t)=I_0e^{rt}I(t)=I0​ert. What happens if you wish to withdraw funds from the investment at a rate of spending SS, where S>0S>0 is constant? The differential equation is:

\frac{dI}{dt} = rI – SdtdI​=rIβˆ’S

Your goals are as follow. You have an initial investment I_0I0​, and you cannot change it or the rate rr. You want to be able to spend as much as possible but you also don’t want to ever spend all your money. What amount of spending rate SS can you bear?

Hint: If you’re not sure what to do, find and classify the equilibria in this model and think about which initial conditions lead to which long-term behaviors.

  • Make sure S \le I_0S≀I0​.
  • Make sure S \le rS≀r.
  • Any S \gt 0S>0 will lead to eventual loss of all funds.
  • Make sure S/r \le 1S/r≀1.
  • Make sure S \ge r I_0Sβ‰₯rI0​
  • Make sure S \le r I_0S≀rI0​.

Practice Quiz 04

Q1. In our lesson, we looked at two oscillators with β€œsinusoidal” coupling. Other types of coupling are possible as well. Consider the system of two oscillators modeled by

\frac{d\theta_1}{dt} = 2 + \epsilon(e^{\theta_1-\theta_2}-1) \quad ; \quad \frac{d\theta_2}{dt} = 2 + \epsilon(1-e^{\theta_1-\theta_2})dtdΞΈ1​​=2+Ο΅(eΞΈ1β€‹βˆ’ΞΈ2β€‹βˆ’1);dtdΞΈ2​​=2+Ο΅(1βˆ’eΞΈ1β€‹βˆ’ΞΈ2​)

Consider the phase difference \varphi = \theta_2 – \theta_1Ο†=ΞΈ2β€‹βˆ’ΞΈ1​. Note that \varphi=0Ο†=0 (where the oscillators are coupled) is an equilibrium. What is the linearized equation for \varphiΟ† about 00?

This looks intimidating, but is very straightforward. If you’re not sure how to start, compute \displaystyle \frac{d\varphi}{dt}dtdφ​. Then linearize this about \varphi=0Ο†=0.

  • \displaystyle \frac{d\varphi}{dt}=\epsilon\varphidtdφ​=ϡφ
  • \displaystyle \frac{d\varphi}{dt}=-\epsilon\varphidtdφ​=βˆ’Ο΅Ο†
  • \displaystyle \frac{d\varphi}{dt}=2\epsilon\varphidtdφ​=2ϡφ
  • \displaystyle \frac{d\varphi}{dt}=\epsilondtdφ​=Ο΅
  • \displaystyle \frac{d\varphi}{dt}=2\varphidtdφ​=2Ο†
  • \displaystyle \frac{d\varphi}{dt}=-2\epsilon\varphidtdφ​=βˆ’2ϡφ

Q2. Biology is full of oscillators. Consider the following biomechanical example. When you walk with your arms free, you may find that they swing back and forth in a periodic manner. Let’s model each arm as a simple oscillator obeying

\frac{d\theta_i}{dt} = adtdΞΈi​​=a

where \theta_1ΞΈ1​ and \theta_2ΞΈ2​ encode the state of your arm. These are not the angle at which your arms reside; rather, these are abstract measurements of how far along each periodic swing your arm is. Your two arms are connected by your torso, which acts as a torsional spring. When you swing one arm out it imparts an impulse through your shoulders to the other arm (try walking around!).

Which of the following is a reasonable model for the coupling function FF between your two arms? Here, \epsilon>0Ο΅>0 is a very small positive constant and the model is:

\frac{d\theta_1}{dt} = a + \epsilon F(\theta_2-\theta_1), \qquad \frac{d\theta_2}{dt} = a – \epsilon F(\theta_2-\theta_1)dtdΞΈ1​​=a+Ο΅F(ΞΈ2β€‹βˆ’ΞΈ1​),dtdΞΈ2​​=aβˆ’Ο΅F(ΞΈ2β€‹βˆ’ΞΈ1​)

Hint: I suggest you first derive the equation for phase difference \varphi=\theta_2-\theta_1Ο†=ΞΈ2β€‹βˆ’ΞΈ1​ in terms of F(\varphi)F(Ο†). Then, consider what happens when you walk around. Do your arms become phase-locked somehow? What stable equilibrium for \varphiΟ† do you note?

  • F(\varphi)=-\sin(\varphi)F(Ο†)=βˆ’sin(Ο†)
  • F(\varphi)=\sin(\varphi)F(Ο†)=sin(Ο†)
  • F(\varphi)=\tan(\varphi)F(Ο†)=tan(Ο†)
  • F(\varphi)=-\cos(\varphi)F(Ο†)=βˆ’cos(Ο†)
  • F(\varphi)=\cos(\varphi)F(Ο†)=cos(Ο†)

Q3. For the brave: The following model of a biochemical β€œswitch” comes from the (very pleasant) text of Strogatz, Nonlinear Dynamics and Chaos. Let G(t)G(t) denote a concentration of a certain gene product, where the gene is switched β€œon” or β€œoff” depending on the presence of a chemical signal. The production of GG is influenced both by the chemical signal and by autocatalysis (a feedback mechanism) and decay (a natural degradation). After simplification, the behavior of GG follows:

\frac{dG}{dt} = a – bG – \frac{G}{1+G^2}dtdG​=aβˆ’bGβˆ’1+G2G​

where aa and bb are positive constants and G(t)\geq 0G(t)β‰₯0.

The number and types of equilibria depend on the values of aa and bb. Unfortunately, solving for them analytically (that is, finding exact values) is… challenging. Without determining exactly where the equilibria are, what can you say about them?

  • For some values of aa and bb, there are no equilibria.
  • All equilibria are unstable.
  • There are at most three equilibria. When there are three, then two are stable and one is unstable.
  • There is always at least one equilibrium.
  • All equilibria are stable.

Week 02 :Calculus: Single Variable Part 3 – Integration Coursera Quiz Answers

Main Quiz 01

Q1. \displaystyle \int 3\cos x \, dx =∫3cosxdx=

  • -3\cos x + Cβˆ’3cosx+C
  • \sin 3x + Csin3x+C
  • 3\sin x + C3sinx+C
  • \cos 3x + Ccos3x+C
  • -3\sin x + Cβˆ’3sinx+C
  • 3\cos x + C3cosx+C

Q2. \displaystyle \int x \sec^2 x^2 \, dx =∫xsec2x2dx=

  • \displaystyle \frac{1}{2}x \sec^2 x^2 + C21​xsec2x2+C
  • \displaystyle \frac{1}{2} \tan x^2 + C21​tanx2+C
  • x^2 \tan x^2 + Cx2tanx2+C
  • 2 \sec x^2 + C2secx2+C
  • \displaystyle \frac{1}{2}x \tan^2 x^2 + C21​xtan2x2+C
  • \tan x^2 + Ctanx2+C

Q3. \displaystyle \int \frac{4x}{(x^2 – 1)^3} \, dx =∫(x2βˆ’1)34x​dx=

  • \displaystyle \frac{4}{3(x^2 – 1)^2} + C3(x2βˆ’1)24​+C
  • \displaystyle \frac{1}{(x^2 – 1)^2} + C(x2βˆ’1)21​+C
  • \displaystyle -\frac{1}{(x^2 – 1)^2} + Cβˆ’(x2βˆ’1)21​+C
  • \displaystyle -\frac{2x}{(x^2 – 1)^2} + Cβˆ’(x2βˆ’1)22x​+C
  • \displaystyle \frac{2}{3(x^2 – 1)^2} + C3(x2βˆ’1)22​+C
  • \displaystyle -\frac{2}{3(x^2 – 1)^2} + Cβˆ’3(x2βˆ’1)22​+C

Q4. \displaystyle \int \frac{e^{\sqrt{x}}}{\sqrt{x}} \, dx =∫x​ex​​dx=

  • e^{\sqrt{x} / 2} + Cex​/2+C
  • 2\sqrt{x} \, e^{\sqrt{x}} + C2x​ex​+C
  • \displaystyle \frac{1}{2}\sqrt{x} \, e^{\sqrt{x}} + C21​x​ex​+C
  • e^{2\sqrt{x}} + Ce2x​+C
  • \displaystyle e^{\sqrt{x}} + Cex​+C
  • 2e^{\sqrt{x}} + C2ex​+C

Q5. \displaystyle \int \frac{ \ln(15x^5)}{x} \, dx =∫xln(15x5)​dx=

  • \displaystyle \frac{5}{2} \ln(15x) + C25​ln(15x)+C
  • \displaystyle \frac{5}{2}x^6 +C25​x6+C
  • \displaystyle \frac{1}{60} \ln^2(15x^5) + C601​ln2(15x5)+C
  • 10\ln^2(15x) + C10ln2(15x)+C
  • \displaystyle \frac{1}{10} \ln^2(15x^5) + C101​ln2(15x5)+C
  • \displaystyle \frac{1}{10} \ln(15x) + C101​ln(15x)+C

Q6. \displaystyle \int \frac{x\, dx}{\sqrt{x+3}} =∫x+3​xdx​=

  • \displaystyle \frac{2}{3} (x+3)^{3/2} + C32​(x+3)3/2+C
  • \displaystyle \frac{1}{3} (x-3) \sqrt{x+3} + C31​(xβˆ’3)x+3​+C
  • \displaystyle \frac{1}{3} \left[ (x+3)^{3/2} – 2(x+3)^{1/2} \right] + C31​[(x+3)3/2βˆ’2(x+3)1/2]+C
  • \displaystyle \frac{2}{3}(x-6)\sqrt{x+3} + C32​(xβˆ’6)x+3​+C
  • \displaystyle \frac{1}{3} (x+6)\sqrt{x+3} + C31​(x+6)x+3​+C
  • \displaystyle \frac{2}{3} \left[ 2(x+3)^{3/2} + (x+3)^{1/2} \right] + C32​[2(x+3)3/2+(x+3)1/2]+C

Practice Quiz 01

Q1. Apply the substitution \displaystyle u = \frac{1}{x}u=x1​ to calculate the integral

I(x) = \int \frac{dx}{x\sqrt{x^2-1}}I(x)=∫xx2βˆ’1​dx​

assuming x > 0x>0.

  • \displaystyle I(x) = \arcsin \frac{1}{x} + CI(x)=arcsinx1​+C
  • \displaystyle I(x) = 2\sqrt{x^2-1} + CI(x)=2x2βˆ’1​+C
  • \displaystyle I(x) = -\arctan \frac{1}{x} + CI(x)=βˆ’arctanx1​+C
  • \displaystyle I(x) = \arctan \frac{1}{x} + CI(x)=arctanx1​+C
  • \displaystyle I(x) = -\arcsin \frac{1}{x} + CI(x)=βˆ’arcsinx1​+C
  • \displaystyle I(x) = -2\sqrt{x^2-1} + CI(x)=βˆ’2x2βˆ’1​+C

Q2. Apply the substitution u = \sqrt{x^2-1}u=x2βˆ’1​ to calculate the integral

I(x) = \int \frac{dx}{x\sqrt{x^2-1}}I(x)=∫xx2βˆ’1​dx​

Note: this is exactly the same integral as that of the previous problem! In Lecture 23, we will learn of yet another substitution that allows us to solve this integral.

  • \displaystyle I(x) = \arcsin\frac{1}{\sqrt{x^2-1}} + CI(x)=arcsinx2βˆ’1​1​+C
  • I(x) = 2\sqrt{x^2-1} + CI(x)=2x2βˆ’1​+C
  • I(x) = -\arctan\sqrt{x^2-1} + CI(x)=βˆ’arctanx2βˆ’1​+C
  • I(x) = \arctan\sqrt{x^2-1} + CI(x)=arctanx2βˆ’1​+C
  • I(x) = -2\sqrt{x^2-1} + CI(x)=βˆ’2x2βˆ’1​+C
  • \displaystyle I(x) = -\arcsin\frac{1}{\sqrt{x^2-1}} + CI(x)=βˆ’arcsinx2βˆ’1​1​+C

Main Quiz 02

Q1. \displaystyle \int x e^{x/2} \, dx =∫xex/2dx=

  • \displaystyle \frac{1}{4} x^2 e^{x/2} + C41​x2ex/2+C
  • 2e^{x/2} (x-1) + C2ex/2(xβˆ’1)+C
  • \displaystyle \frac{1}{2} e^{x/2} (x-2) + C21​ex/2(xβˆ’2)+C
  • \displaystyle \frac{1}{4} e^{x/2} (2x-1) + C41​ex/2(2xβˆ’1)+C
  • 2e^{x/2} (x-2) + C2ex/2(xβˆ’2)+C
  • e^{x^2/4} + Cex2/4+C

Q2. \displaystyle \int x^2e^{x/2} \, dx =∫x2ex/2dx=

  • e^{x/2} (x^2 – 2x + 4) + Cex/2(x2βˆ’2x+4)+C
  • e^{x/2} (x^2 – 4x – 8) + Cex/2(x2βˆ’4xβˆ’8)+C
  • 2e^{x/2} (x – 2) + C2ex/2(xβˆ’2)+C
  • 2e^{x/2} (x^2 – 4x – 8) + C2ex/2(x2βˆ’4xβˆ’8)+C
  • 2e^{x/2} (x^2 – 4x + 8) + C2ex/2(x2βˆ’4x+8)+C
  • \displaystyle e^{x/2} \left( \frac{1}{2}x^2 – \frac{1}{8}x + \frac{1}{16} \right) + Cex/2(21​x2βˆ’81​x+161​)+C

Q3. \displaystyle \int 3x \ln x \, dx =∫3xlnxdx=

  • \displaystyle \frac{3x^2}{2} \left( \ln x – 1 \right) + C23x2​(lnxβˆ’1)+C
  • \displaystyle \frac{x^2}{4} \left( 2\ln x – 1 \right) + C4x2​(2lnxβˆ’1)+C
  • \displaystyle \frac{3x^2}{4} \left( \ln x – 1 \right) + C43x2​(lnxβˆ’1)+C
  • \displaystyle \frac{x^2}{2} \left( 3 \ln x – x \right) + C2x2​(3lnxβˆ’x)+C
  • \displaystyle \frac{3x}{2} \left( 4 \ln x – 1 \right) + C23x​(4lnxβˆ’1)+C
  • \displaystyle \frac{3x^2}{4} \left( 2\ln x – 1 \right) + C43x2​(2lnxβˆ’1)+C

Q4. \displaystyle \int 3x^2 \ln x \, dx =∫3x2lnxdx=

  • \displaystyle x^3 \left( \ln x – \frac{1}{3} \right) + Cx3(lnxβˆ’31​)+C
  • \displaystyle \frac{1}{3}x^3( \ln x – 1) + C31​x3(lnxβˆ’1)+C
  • \displaystyle x ( \ln x – 1 ) + Cx(lnxβˆ’1)+C
  • \displaystyle x^2\left( \ln x – \frac{1}{2} \right) + Cx2(lnxβˆ’21​)+C
  • \displaystyle 3x^2\ln x + \frac{1}{4}x^4 + C3x2lnx+41​x4+C
  • \displaystyle 3x^2\ln x – \frac{1}{4}x^4 + C3x2lnxβˆ’41​x4+C

Q5. \displaystyle \int x^2 \cos\frac{x}{2} \, dx =∫x2cos2x​dx=

Hint: no mathematician ever remembers those reduction formulas in the Lecture, but we all remember that we get to them by repeated integration by parts.

  • \displaystyle (2x^2+16) \sin\frac{x}{2} – 8x\cos\frac{x}{2} + C(2x2+16)sin2xβ€‹βˆ’8xcos2x​+C
  • \displaystyle (16-2x^2) \sin\frac{x}{2} + 8x\cos\frac{x}{2} + C(16βˆ’2x2)sin2x​+8xcos2x​+C
  • \displaystyle (2x^2-16) \sin\frac{x}{2} – 8x\cos\frac{x}{2} + C(2x2βˆ’16)sin2xβ€‹βˆ’8xcos2x​+C
  • \displaystyle (2x^2+16) \sin\frac{x}{2} + 8x\cos\frac{x}{2} + C(2x2+16)sin2x​+8xcos2x​+C
  • \displaystyle (16-2x^2) \sin\frac{x}{2} – 8x\cos\frac{x}{2} + C(16βˆ’2x2)sin2xβ€‹βˆ’8xcos2x​+C
  • \displaystyle (2x^2-16) \sin\frac{x}{2} + 8x\cos\frac{x}{2} + C(2x2βˆ’16)sin2x​+8xcos2x​+C

Q6. \displaystyle \int e^{2x}\sin 3x \, dx =∫e2xsin3xdx=

  • \displaystyle \frac{2}{3}e^{2x}\sin 3x + \frac{1}{3}e^{2x}\cos 3x + C32​e2xsin3x+31​e2xcos3x+C
  • \displaystyle \frac{2}{3}e^{2x}\cos 3x + \frac{3}{13}e^{2x}\sin 3x + C32​e2xcos3x+133​e2xsin3x+C
  • \displaystyle \frac{2}{13}e^{2x}\sin 3x – \frac{3}{13}e^{2x}\cos 3x + C132​e2xsin3xβˆ’133​e2xcos3x+C
  • \displaystyle \frac{1}{2}e^{2x}\sin 3x – \frac{3}{4}e^{2x}\cos 3x + C21​e2xsin3xβˆ’43​e2xcos3x+C
  • \displaystyle \frac{2}{3}e^{2x}\sin 3x – \frac{4}{3}e^{2x}\cos 3x + C32​e2xsin3xβˆ’34​e2xcos3x+C
  • \displaystyle \frac{2}{13}e^{2x}\cos 3x – \frac{3}{13}e^{2x}\sin 3x + C132​e2xcos3xβˆ’133​e2xsin3x+C

Q7. Apply integration by parts with u = \ln xu=lnx and dv = dxdv=dx to find an antiderivative for the logarithm β€”that is, to solve the following integral:

I(x) = \int \ln x \, dxI(x)=∫lnxdx

  • I(x) = x \ln x + CI(x)=xlnx+C
  • I(x) = x \left( \ln x – 1 \right) + CI(x)=x(lnxβˆ’1)+C
  • \displaystyle I(x) = \frac{\ln^2 x}{2} + CI(x)=2ln2x​+C
  • \displaystyle I(x) = \frac{\ln^2 x}{x^2} + CI(x)=x2ln2x​+C
  • I(x) = x \left( \ln x – x \right) + CI(x)=x(lnxβˆ’x)+C
  • \displaystyle I(x) = \frac{1}{x} + CI(x)=x1​+C

Practice Quiz 02

Q1. \displaystyle \int \ln^2 x \, dx =∫ln2xdx=

  • \displaystyle \frac{1}{2}\ln^2 x – x\ln x + C21​ln2xβˆ’xlnx+C
  • \ln^2 x + x\ln x – 2x + Cln2x+xlnxβˆ’2x+C
  • \ln^2 x – 2x + Cln2xβˆ’2x+C
  • x \ln^2 x – \ln x + 2x + Cxln2xβˆ’lnx+2x+C
  • x \left( \ln^2 x – 2\ln x + 2 \right) + Cx(ln2xβˆ’2lnx+2)+C
  • x \left( \ln^2 x + \ln x – 2 \right) + Cx(ln2x+lnxβˆ’2)+C

Q2. \displaystyle \int \sin(\ln x) \, dx =∫sin(lnx)dx=

  • \displaystyle \frac{x}{2} \left[ \sin(\ln x) + \cos(\ln x)\right] + C2x​[sin(lnx)+cos(lnx)]+C
  • x \left[ \sin(\ln x) + \cos(\ln x)\right] + Cx[sin(lnx)+cos(lnx)]+C
  • 2x \sin(\ln x) + C2xsin(lnx)+C
  • x \left[ \sin(\ln x) – \cos(\ln x)\right] + Cx[sin(lnx)βˆ’cos(lnx)]+C
  • \displaystyle \frac{x}{2} \left[ \sin(\ln x) – \cos(\ln x)\right] + C2x​[sin(lnx)βˆ’cos(lnx)]+C
  • 2x \cos(\ln x) + C2xcos(lnx)+C

Q3. \displaystyle \int \arcsin 2x \, dx =∫arcsin2xdx=

  • \displaystyle x\arcsin 2x + \frac{1}{2} \sqrt{1 – x^2} + Cxarcsin2x+21​1βˆ’x2​+C
  • \displaystyle x\arcsin 2x + \frac{1}{2} \sqrt{1 – 4x^2} + Cxarcsin2x+21​1βˆ’4x2​+C
  • \displaystyle \arcsin 2x + \frac{1}{4} \sqrt{1 – 2x^2} + Carcsin2x+41​1βˆ’2x2​+C
  • 2x\arcsin 2x + 8 \sqrt{1 – x^2} + C2xarcsin2x+81βˆ’x2​+C
  • x\arcsin 2x + 2 \sqrt{1 – x^2} + Cxarcsin2x+21βˆ’x2​+C
  • \arcsin 2x + 2 \sqrt{1 – 4x^2} + Carcsin2x+21βˆ’4x2​+C

Main Quiz 03

Q1. Which of the following integrals do you get when performing the substitution x = \sin\thetax=sinΞΈ in \displaystyle \int x^5 \sqrt{1-x^2} \, dx∫x51βˆ’x2​dx ?

Note: in Lecture 28 we will see how to attack integrals involving powers of trigonometric functions.

  • \displaystyle \int \sin^2\theta \cos^5 \theta \, d\theta∫sin2ΞΈcos5ΞΈdΞΈ
  • \displaystyle – \int \sin^2\theta \cos^5 \theta \, d\thetaβˆ’βˆ«sin2ΞΈcos5ΞΈdΞΈ
  • \displaystyle – \int \sin^5\theta \cos\theta \, d\thetaβˆ’βˆ«sin5ΞΈcosΞΈdΞΈ
  • \displaystyle \int \sin^5\theta \cos^2\theta \, d\theta∫sin5ΞΈcos2ΞΈdΞΈ
  • \displaystyle – \int \sin^5\theta \cos^2\theta \, d\thetaβˆ’βˆ«sin5ΞΈcos2ΞΈdΞΈ
  • \displaystyle \int \sin^5\theta \cos\theta \, d\theta∫sin5ΞΈcosΞΈdΞΈ

Q2. \displaystyle \int \frac{x^2}{\sqrt{4-x^2}} \, dx =∫4βˆ’x2​x2​dx=

  • \displaystyle -2 \arccos\frac{x}{2} – \frac{1}{2}x \sqrt{4-x^2} + Cβˆ’2arccos2xβ€‹βˆ’21​x4βˆ’x2​+C
  • \displaystyle 2 \arcsin\frac{x}{2} – \frac{1}{2}x \sqrt{4-x^2} + C2arcsin2xβ€‹βˆ’21​x4βˆ’x2​+C
  • \displaystyle x – \frac{1}{4}x \sqrt{4 – x^2} + Cxβˆ’41​x4βˆ’x2​+C
  • \displaystyle -\frac{1}{2} \arccos\frac{x}{2} – \frac{1}{8}x \sqrt{4-x^2} + Cβˆ’21​arccos2xβ€‹βˆ’81​x4βˆ’x2​+C
  • \displaystyle \frac{x}{2} – \frac{1}{4} \cos 2x + C2xβ€‹βˆ’41​cos2x+C
  • \displaystyle \frac{1}{2} \arcsin\frac{x}{2} – \frac{1}{8}x \sqrt{4-x^2} + C21​arcsin2xβ€‹βˆ’81​x4βˆ’x2​+C

Q3. In Lecture we saw that we can use the substitution \displaystyle x = \frac{b}{a}\sin\thetax=ab​sinΞΈ to (hopefully) calculate integrals involving \sqrt{b^2 – a^2x^2}b2βˆ’a2x2​. Another equally suitable subsitution in this case is \displaystyle x = \frac{b}{a}\cos\thetax=ab​cosΞΈ. Use the latter to compute

\displaystyle \int \frac {\sqrt{1-x^2}} {x^2} \, dx =∫x21βˆ’x2​​dx=

  • \displaystyle -\frac{ \sqrt{1-x^2}}{2x^2} – \frac{1}{2} \arccos x + Cβˆ’2x21βˆ’x2β€‹β€‹βˆ’21​arccosx+C
  • \displaystyle -\frac{ \sqrt{1-x^2} }{x} + \arccos x + Cβˆ’x1βˆ’x2​​+arccosx+C
  • \displaystyle -\frac{ \sqrt{1-x^2} }{x} – 2\arccos x + Cβˆ’x1βˆ’x2β€‹β€‹βˆ’2arccosx+C
  • \displaystyle \frac{ \sqrt{1-x^2} }{x^2} + \arccos x + Cx21βˆ’x2​​+arccosx+C
  • x \sqrt{1-x^2} – \arccos x + Cx1βˆ’x2β€‹βˆ’arccosx+C
  • \displaystyle \frac{ \sqrt{1-x^2} }{x^2} + 2\arccos x + Cx21βˆ’x2​​+2arccosx+C

Q4. \displaystyle \int (1-x^2)^{-3/2} \, dx =∫(1βˆ’x2)βˆ’3/2dx=

  • \displaystyle \arcsin x + \frac{x}{\sqrt{1-x^2}} + Carcsinx+1βˆ’x2​x​+C
  • \displaystyle \frac{x}{\sqrt{1-x^2}} + C1βˆ’x2​x​+C
  • \displaystyle -2\sqrt{1-x^2} + Cβˆ’21βˆ’x2​+C
  • \displaystyle \frac{1}{2}\arcsin x + \sqrt{1-x^2} + C21​arcsinx+1βˆ’x2​+C
  • \displaystyle \arccos x + \frac{1}{\sqrt{1-x^2}} + Carccosx+1βˆ’x2​1​+C
  • \displaystyle \frac{1}{\sqrt{1-x^2}} + C1βˆ’x2​1​+C

Q5. \displaystyle \int \frac{x}{\sqrt{1+x^2}} \, dx =∫1+x2​x​dx=

  • \displaystyle \ln |x + 1| + \frac{1}{ \sqrt{1 + x^2}} + Cln∣x+1∣+1+x2​1​+C
  • \displaystyle \frac{x}{1 + x^2} + C1+x2x​+C
  • \sqrt{1+x^2} + C1+x2​+C
  • \displaystyle \frac{1}{\sqrt{1+x^2}} + C1+x2​1​+C
  • \ln \big| x + \sqrt{1+x^2} \big| + Clnβˆ£βˆ£βˆ£β€‹x+1+x2β€‹βˆ£βˆ£βˆ£β€‹+C
  • \displaystyle \frac{1}{2} \left( \mathrm{arcsinh}\, x – x \sqrt{1+x^2}\right) + C21​(arcsinhxβˆ’x1+x2​)+C

Q6. For the appropriate value of aa, the substitution x = a \sec\thetax=asecΞΈ helps in reducing the integral \displaystyle \int \frac{x^4 \, dx}{\sqrt{x^2 – 4}}∫x2βˆ’4​x4dx​ to an integral involving trigonometric functions β€”which we will see how to deal with in Lecture 28. What value should you take for aa, and what integral do you get after performing the substitution?

  • For a = 2a=2, we obtain the integral \displaystyle \int 2^{10} \sec^5\theta \, d\theta∫210sec5ΞΈdΞΈ
  • For a = 2a=2, we obtain the integral \displaystyle \int 2^4 \sec^5\theta \, d\theta∫24sec5ΞΈdΞΈ
  • For a = 2a=2, we obtain the integral \displaystyle \int 2^5 \sec^5\theta \, d\theta∫25sec5ΞΈdΞΈ
  • For a = 4a=4, we obtain the integral \displaystyle \int 2^8 \sec^4\theta \, d\theta∫28sec4ΞΈdΞΈ
  • For a = 2a=2, we obtain the integral \displaystyle \int 2^4 \sec^4\theta \tan\theta \, d\theta∫24sec4ΞΈtanΞΈdΞΈ
  • For a = 4a=4, we obtain the integral \displaystyle \int 2^8 \sec^5\theta \, d\theta∫28sec5ΞΈdΞΈ

Practice Quiz 03

Q1. In Homework 21 (challenge), we use the substitutions \displaystyle u = \frac{1}{x}u=x1​ and u = \sqrt{x^2-1}u=x2βˆ’1​ to compute the integral

I(x) = \int \frac{dx}{x\sqrt{x^2-1}}I(x)=∫xx2βˆ’1​dx​

Another possible substitution is x = \sec\thetax=secΞΈ. Using it, compute I(x)I(x).

  • \displaystyle \frac{\sqrt{x^2-1}}{2x^2} + C2x2x2βˆ’1​​+C
  • \displaystyle \frac{1}{2}\sqrt{x^2-1} – \mathrm{arcsec}\, x + C21​x2βˆ’1β€‹βˆ’arcsecx+C
  • x \arccos \sqrt{x^2-1} + Cxarccosx2βˆ’1​+C
  • \displaystyle \sqrt{x^2-1}\arccos x + Cx2βˆ’1​arccosx+C
  • \displaystyle \mathrm{arcsec}\, \frac{1}{x} + Carcsecx1​+C
  • \displaystyle \arccos \frac{1}{x} + Carccosx1​+C

Q2. \displaystyle \int \frac{dx}{\sqrt{x^2-6x+10}} =∫x2βˆ’6x+10​dx​=

(there may be multiple correct answers)

  • \mathrm{arcsinh}\, (x-3) + Carcsinh(xβˆ’3)+C
  • \displaystyle \frac{1}{2} \mathrm{arcsinh}\,(x-3) + \frac{1}{4} x \sqrt{x^2-6x+10} + C21​arcsinh(xβˆ’3)+41​xx2βˆ’6x+10​+C
  • \displaystyle \frac{\sqrt{x^2-6x+10}}{2x} + C2xx2βˆ’6x+10​​+C
  • \mathrm{arcsinh}\,\sqrt{x^2-6x+10} + Carcsinhx2βˆ’6x+10​+C
  • \ln \big| x – 3 + \sqrt{x^2 – 6x + 10}\big| + Clnβˆ£βˆ£βˆ£β€‹xβˆ’3+x2βˆ’6x+10β€‹βˆ£βˆ£βˆ£β€‹+C
  • \displaystyle \frac{1}{2}\mathrm{arccosh}\, (x-3) – \frac{\sqrt{x^2-6x+10}}{4x} + C21​arccosh(xβˆ’3)βˆ’4xx2βˆ’6x+10​​+C

Q3. \displaystyle \int \frac{dx}{\sqrt{x^2-2x-8}} =∫x2βˆ’2xβˆ’8​dx​=

(there may be multiple correct answers)

  • \displaystyle \frac{1}{2}\sqrt{x^2-2x-8} + \mathrm{arccosh}\frac{x-1}{3} + C21​x2βˆ’2xβˆ’8​+arccosh3xβˆ’1​+C
  • \displaystyle -\sqrt{x^2-2x-8} + Cβˆ’x2βˆ’2xβˆ’8​+C
  • \displaystyle \ln \big| x-1 + \sqrt{x^2-2x-8} \big| + Clnβˆ£βˆ£βˆ£β€‹xβˆ’1+x2βˆ’2xβˆ’8β€‹βˆ£βˆ£βˆ£β€‹+C
  • \displaystyle \mathrm{arccosh}\frac{x-1}{3} + Carccosh3xβˆ’1​+C
  • \displaystyle \frac{1}{3}\ln \left| x – 1 + \sqrt{x^2-2x-8} \right| + C31​lnβˆ£βˆ£βˆ£βˆ£β€‹xβˆ’1+x2βˆ’2xβˆ’8β€‹βˆ£βˆ£βˆ£βˆ£β€‹+C
  • \displaystyle \frac{1}{3}\sqrt{x^2-2x-8} – \mathrm{arccosh}\frac{x-1}{3} + C31​x2βˆ’2xβˆ’8β€‹βˆ’arccosh3xβˆ’1​+C

Main Quiz 04

Q1. \displaystyle \int \frac{5+x}{x^2+x-6} \, dx =∫x2+xβˆ’65+x​dx=

  • \displaystyle \frac{1}{5} \ln \left| \frac{x +3}{x – 2} \right| + C51​lnβˆ£βˆ£βˆ£βˆ£βˆ£β€‹xβˆ’2x+3β€‹βˆ£βˆ£βˆ£βˆ£βˆ£β€‹+C
  • \ln |x – 2| – \ln |x+3| + Cln∣xβˆ’2βˆ£βˆ’ln∣x+3∣+C
  • \displaystyle \frac{7}{5} \ln \left| \frac{x – 2}{x+3} \right| + C57​lnβˆ£βˆ£βˆ£βˆ£βˆ£β€‹x+3xβˆ’2β€‹βˆ£βˆ£βˆ£βˆ£βˆ£β€‹+C
  • \displaystyle \ln |x^2 – x + 6| + \frac{11}{\sqrt{23}} \arctan\frac{2x – 1}{\sqrt{23}} + Cln∣x2βˆ’x+6∣+23​11​arctan23​2xβˆ’1​+C
  • \displaystyle \frac{2}{5} \ln |x – 2| – \frac{7}{5} \ln |x+3| + C52​ln∣xβˆ’2βˆ£βˆ’57​ln∣x+3∣+C
  • \displaystyle \frac{7}{5} \ln |x – 2| – \frac{2}{5} \ln |x+3| + C57​ln∣xβˆ’2βˆ£βˆ’52​ln∣x+3∣+C

Q2. \displaystyle \int \frac{2x + 3}{6x^2 + 5x + 1} \, dx =∫6x2+5x+12x+3​dx=

  • \displaystyle \frac{7}{3}\ln |3x + 1| – \frac{2}{3}\ln |2x + 1| + C37​ln∣3x+1βˆ£βˆ’32​ln∣2x+1∣+C
  • \displaystyle \frac{7}{3}\ln |2x + 1| – 4\ln |3x + 1| + C37​ln∣2x+1βˆ£βˆ’4ln∣3x+1∣+C
  • \displaystyle 7\ln |3x + 1| – \frac{2}{3}\ln |2x + 1| + C7ln∣3x+1βˆ£βˆ’32​ln∣2x+1∣+C
  • \displaystyle \frac{7}{3}\ln |3x + 1| – 2\ln |2x + 1| + C37​ln∣3x+1βˆ£βˆ’2ln∣2x+1∣+C
  • \displaystyle \frac{7}{3}\ln |2x + 1| – 2\ln |3x + 1| + C37​ln∣2x+1βˆ£βˆ’2ln∣3x+1∣+C
  • 7\ln |3x + 1| – 2\ln |2x + 1| + C7ln∣3x+1βˆ£βˆ’2ln∣2x+1∣+C

Q3. \displaystyle \int \frac{x^2-x+5}{(x-2)(x-1)(x+3)} \, dx =∫(xβˆ’2)(xβˆ’1)(x+3)x2βˆ’x+5​dx=

  • \displaystyle \frac{2}{5} \ln |x – 2| – \frac{3}{4} \ln |x-1| + C52​ln∣xβˆ’2βˆ£βˆ’43​ln∣xβˆ’1∣+C
  • \displaystyle -\frac{1}{4} \ln \left| \frac{x^2 + x – 6}{x-1} \right| + Cβˆ’41​lnβˆ£βˆ£βˆ£βˆ£βˆ£β€‹xβˆ’1x2+xβˆ’6β€‹βˆ£βˆ£βˆ£βˆ£βˆ£β€‹+C
  • \displaystyle \frac{7}{5} \ln |x – 2| – \ln |x – 1| + \frac{1}{20} \ln |x + 3| + C57​ln∣xβˆ’2βˆ£βˆ’ln∣xβˆ’1∣+201​ln∣x+3∣+C
  • \displaystyle \frac{7}{5} \ln |x – 2| – \frac{5}{4} \ln |x – 1| + \frac{17}{20} \ln |x + 3| + C57​ln∣xβˆ’2βˆ£βˆ’45​ln∣xβˆ’1∣+2017​ln∣x+3∣+C
  • \displaystyle \frac{1}{4} \ln \left| \frac{x^2 + x – 6}{x-1} \right| + C41​lnβˆ£βˆ£βˆ£βˆ£βˆ£β€‹xβˆ’1x2+xβˆ’6β€‹βˆ£βˆ£βˆ£βˆ£βˆ£β€‹+C
  • \displaystyle \frac{37}{43} \ln |(x – 2)(x – 1)(x + 3)| + C4337​ln∣(xβˆ’2)(xβˆ’1)(x+3)∣+C

Q4. \displaystyle \int \frac{2x-1}{x^3-x} \, dx =∫x3βˆ’x2xβˆ’1​dx=

  • \displaystyle \ln \left| \frac{x(x-1)}{x+1} \right| + Clnβˆ£βˆ£βˆ£βˆ£βˆ£β€‹x+1x(xβˆ’1)β€‹βˆ£βˆ£βˆ£βˆ£βˆ£β€‹+C
  • \displaystyle \ln |x| – \frac{1}{2} \ln |x-2| – \frac{1}{2} \ln |x+1| + Cln∣xβˆ£βˆ’21​ln∣xβˆ’2βˆ£βˆ’21​ln∣x+1∣+C
  • \displaystyle \ln \left| \frac{x^2(x-1)}{(x+1)^3} \right| + Clnβˆ£βˆ£βˆ£βˆ£βˆ£β€‹(x+1)3x2(xβˆ’1)β€‹βˆ£βˆ£βˆ£βˆ£βˆ£β€‹+C
  • \displaystyle \ln \left| \frac{x+1}{x(x-1)} \right| + Clnβˆ£βˆ£βˆ£βˆ£βˆ£β€‹x(xβˆ’1)x+1β€‹βˆ£βˆ£βˆ£βˆ£βˆ£β€‹+C
  • \displaystyle \ln |x| + \frac{1}{2}\ln|x-1| – \frac{3}{2}\ln |x+1| + Cln∣x∣+21​ln∣xβˆ’1βˆ£βˆ’23​ln∣x+1∣+C
  • \displaystyle \ln \left| \frac{(x+1)^3}{x^2(x-1)} \right| + Clnβˆ£βˆ£βˆ£βˆ£βˆ£β€‹x2(xβˆ’1)(x+1)3β€‹βˆ£βˆ£βˆ£βˆ£βˆ£β€‹+C

Q5. \displaystyle{\int} \frac{x^2-3}{x^2-4} \, dx =∫x2βˆ’4x2βˆ’3​dx=

Hint: start by performing long division of the numerator by the denominator.

  • \displaystyle x+\frac{1}{4} \ln|x^2-4|+Cx+41​ln∣x2βˆ’4∣+C
  • \displaystyle x + \frac{1}{4}\ln\left|\frac{x-2}{x+2}\right| + Cx+41​lnβˆ£βˆ£βˆ£βˆ£βˆ£β€‹x+2xβˆ’2β€‹βˆ£βˆ£βˆ£βˆ£βˆ£β€‹+C
  • \displaystyle \frac{3}{4} \ln\left|\frac{x+2}{x-2}\right|+C43​lnβˆ£βˆ£βˆ£βˆ£βˆ£β€‹xβˆ’2x+2β€‹βˆ£βˆ£βˆ£βˆ£βˆ£β€‹+C
  • \displaystyle x-\frac{1}{x}-4x+Cxβˆ’x1β€‹βˆ’4x+C
  • \displaystyle x-\frac{3}{4} \ln\left|\frac{x-2}{x+2}\right|+Cxβˆ’43​lnβˆ£βˆ£βˆ£βˆ£βˆ£β€‹x+2xβˆ’2β€‹βˆ£βˆ£βˆ£βˆ£βˆ£β€‹+C
  • \displaystyle x + \frac{1}{2}\ln|x-2| – \frac{1}{4}\ln|x+2| + Cx+21​ln∣xβˆ’2βˆ£βˆ’41​ln∣x+2∣+C

Q6. \displaystyle \int \frac{dx}{x^2 – 4x + 8} =∫x2βˆ’4x+8dx​=

Hint: complete the square in the denominator and perform a judicious substitution to get an integral of the form

\displaystyle \int \frac{du}{1+u^2}∫1+u2du​.

  • \displaystyle \frac{1}{4} \arctan \frac{x+2}{3} + C41​arctan3x+2​+C
  • \displaystyle \frac{1}{2} \arctan \frac{x-2}{2} + C21​arctan2xβˆ’2​+C
  • \displaystyle \frac{1}{2} \arctan \left(1 + \frac{x}{2}\right) + C21​arctan(1+2x​)+C
  • \displaystyle \frac{1}{4} \arctan \left(1 + \frac{x}{2}\right) + C41​arctan(1+2x​)+C
  • \displaystyle \frac{1}{4} \arctan \frac{x-2}{2} + C41​arctan2xβˆ’2​+C
  • \displaystyle \frac{1}{2} \arctan \frac{x+2}{3} + C21​arctan3x+2​+C

Practice Quiz 04

Q1. \displaystyle \int \frac{x^3+10x^2+33x+36}{x^2+4x+3} \, dx =∫x2+4x+3x3+10x2+33x+36​dx=

  • \ln |x + 1| – \ln |x+3| + Cln∣x+1βˆ£βˆ’ln∣x+3∣+C
  • \displaystyle \frac{8}{9} \ln |x + 1| + \frac{4}{9} \ln |x+3| + C98​ln∣x+1∣+94​ln∣x+3∣+C
  • \displaystyle \frac{1}{2}x^2 + 6x + 6 \ln |x+1| + C21​x2+6x+6ln∣x+1∣+C
  • \displaystyle \frac{1}{2}x^2 + \ln |x^2 + 4x + 3| + C21​x2+ln∣x2+4x+3∣+C
  • x^2 + 6 \ln |x+1| + Cx2+6ln∣x+1∣+C
  • \displaystyle \frac{1}{2}x^2 + 6x + 3 \ln |x+1| + 3 \ln|x+3|+ C21​x2+6x+3ln∣x+1∣+3ln∣x+3∣+C

Q2. \displaystyle \int \frac{x+2}{(x-1)^2} \, dx =∫(xβˆ’1)2x+2​dx=

Hint: remember from the Lecture that we can deal with multiple roots in the denominator by using a partial fraction decomposition of the form

\frac{P(x)}{(x-r)^n} = \frac{A_1}{x-r} + \frac{A_2}{(x-r)^2} + \cdots + \frac{A_n}{(x-r)^n}(xβˆ’r)nP(x)​=xβˆ’rA1​​+(xβˆ’r)2A2​​+β‹―+(xβˆ’r)nAn​​

Clearing denominators results in the equation:

P(x) = A_1 (x-r)^{n-1} + A_2 (x-r)^{n-2} + \cdots + A_n \tag{$\ast$}P(x)=A1​(xβˆ’r)nβˆ’1+A2​(xβˆ’r)nβˆ’2+β‹―+An​(βˆ—)

Notice that, in this case, the direct technique for finding the coefficients A_1, \ldots, A_nA1​,…,An​ β€”substituting x=rx=r in (\astβˆ—)β€” does not quite work: it only gives you the last coefficient, A_nAn​. In order to find the other coefficients, A_1, \ldots, A_{n-1}A1​,…,Anβˆ’1​, you can revert to equating the coefficients of each power of xx on both sides of the equation (\astβˆ—).

  • \displaystyle \ln |x-1| + \frac{3}{x-1} + Cln∣xβˆ’1∣+xβˆ’13​+C
  • \displaystyle \ln |x-1| – \frac{3}{x-1} + Cln∣xβˆ’1βˆ£βˆ’xβˆ’13​+C
  • 3\ln|x+1| + C3ln∣x+1∣+C
  • \displaystyle \ln |x-1| + \frac{2}{x-1} + Cln∣xβˆ’1∣+xβˆ’12​+C
  • \ln|x+1| + 3\ln|x+1|^2 + Cln∣x+1∣+3ln∣x+1∣2+C
  • \ln|x+1| – 3\ln|x+1|^2 + Cln∣x+1βˆ£βˆ’3ln∣x+1∣2+C

Q3. Consider the integral

I(x) = \int \frac{dx}{x^4 – 6x^3 + 12x^2}I(x)=∫x4βˆ’6x3+12x2dx​

Factoring the denominator will get you a repeated linear factor β€”which we have just seen how to deal withβ€” and an irreducible quadratic polynomial β€”which we know leads to an arctangent (and maybe something more). Now just put everything together to solve this integral!

Hint: when trying to solve for the coefficients of the partial fraction decomposition (there should be four of them), you might feel a little bit overwhelmed. Don’t worry! Start by plugging in x=0x=0 β€”that should give you one of the coefficients. To find the rest of them, remember that two polynomials are equal if and only if the coefficients of each of the terms are equal. That is,

1 = a x^3 + b x^2 + c x + d1=ax3+bx2+cx+d

if and only if a=0a=0, b=0b=0, c=0c=0 and d=1d=1.

  • \displaystyle I(x) = \frac{1}{36x} \left[ \sqrt{3} x \arctan\frac{x-3}{\sqrt{3}} +3\right] + CI(x)=36x1​[3​xarctan3​xβˆ’3​+3]+C
  • \displaystyle I(x) = -\frac{1}{36x} \left[ \sqrt{3} x \arctan\frac{x-3}{\sqrt{3}} +3\right] + CI(x)=βˆ’36x1​[3​xarctan3​xβˆ’3​+3]+C
  • \displaystyle I(x) = \frac{1}{24} \left[ \ln |x| – \frac{2}{x} – \frac{1}{2}\ln\left(x^2-6x+12\right) – \frac{1}{\sqrt{3}} \arctan\frac{x-3}{\sqrt{3}} \right] + CI(x)=241​[ln∣xβˆ£βˆ’x2β€‹βˆ’21​ln(x2βˆ’6x+12)βˆ’3​1​arctan3​xβˆ’3​]+C
  • \displaystyle I(x) = \frac{1}{24} \left[ \ln |x| – \frac{2}{x} – \frac{1}{2}\ln\left(x^2-6x+12\right) + \frac{1}{\sqrt{3}} \arctan\frac{x-3}{\sqrt{3}} \right] + CI(x)=241​[ln∣xβˆ£βˆ’x2β€‹βˆ’21​ln(x2βˆ’6x+12)+3​1​arctan3​xβˆ’3​]+C
  • \displaystyle I(x) = \frac{1}{48} \left[2 \ln |x| – \frac{1}{x} – \frac{1}{2}\ln\left(x^2-6x+12\right) – \frac{2}{3} \arctan{(x-3)} \right] + CI(x)=481​[2ln∣xβˆ£βˆ’x1β€‹βˆ’21​ln(x2βˆ’6x+12)βˆ’32​arctan(xβˆ’3)]+C
  • \displaystyle I(x) = \frac{1}{24} \left[ \ln |x| – \frac{2}{x} + \frac{1}{2}\ln\left(x^2-6x+12\right) -\frac{1}{\sqrt{3}} \arctan\frac{x-3}{\sqrt{3}} \right] + CI(x)=241​[ln∣xβˆ£βˆ’x2​+21​ln(x2βˆ’6x+12)βˆ’3​1​arctan3​xβˆ’3​]+C

Week 03:Calculus: Single Variable Part 3 – Integration Coursera Quiz Answers

Main Quiz 01

Q1. Recall the definition of definite integrals through Riemann sums:

\int_{x=a}^b f(x)\, dx = \lim_{P: \Delta x \to 0} \sum_{i = 1}^n f(x_i) (\Delta x)_i∫x=ab​f(x)dx=P:Ξ”xβ†’0lim​i=1βˆ‘n​f(xi​)(Ξ”x)i​

Here PP is a partition of the interval [a, b][a,b] into nn intervals P_iPi​, each of width (\Delta x)_i(Ξ”x)i​. The point x_ixi​ is a sampling of P_iPi​, that is, a point in the interval P_iPi​. The limit is taken over all partitions as the width of the subdivisions gets smaller and smaller.

One particular choice of partition and sampling that can be used to numerically evaluate definite integrals is the following. With nn fixed, divide the interval [a, b][a,b] into nn subintervals P_iPi​ of common length (\Delta x)_i = (b-a)/n(Ξ”x)i​=(bβˆ’a)/n. For the sampling, choose the right endpoint of each P_iPi​; this gives you the formula

x_i = a + i \frac{b-a}{n}xi​=a+inbβˆ’a​

With these choices of partition and sampling, compute the Riemann sums for the integral

\int_{x=1}^2 \frac{dx}{x}∫x=12​xdx​

for n=1n=1, n=2n=2 and n=3n=3 subdivisions.

Note: in the next Lecture we will learn that

\int_{x=1}^2 \frac{dx}{x} = \ln 2 \simeq 0.693∫x=12​xdx​=ln2≃0.693

Compare this value to the ones you obtain from the Riemann sums.

  • \displaystyle 1, \frac{3}{2} = 1.5, \frac{11}{6} \simeq 1.8331,23​=1.5,611​≃1.833
  • \displaystyle \frac{1}{2} = 0.5, \frac{5}{6} \simeq 0.833, \frac{13}{12} \simeq 1.08321​=0.5,65​≃0.833,1213​≃1.083
  • \displaystyle 1, \frac{1}{2} = 0.5, \frac{5}{6} \simeq 0.8331,21​=0.5,65​≃0.833
  • \displaystyle \frac{1}{2} = 0.5, \frac{7}{12} \simeq 0.583, \frac{37}{60} \simeq 0.61721​=0.5,127​≃0.583,6037​≃0.617
  • \displaystyle \frac{1}{2} = 0.5, \frac{5}{6} \simeq 0.833, \frac{5}{12} \simeq 0.41721​=0.5,65​≃0.833,125​≃0.417
  • \displaystyle 1, \frac{5}{6} \simeq 0.833, \frac{47}{60} \simeq 0.7831,65​≃0.833,6047​≃0.783

Q2. With the same choices of partition and sampling as in the previous problem, evaluate the Riemann sum for the integral

\int_{x=0}^3 x^2 \, dx∫x=03​x2dx

for an arbitrary number nn of subdivisions.

You might need to use any of the following formulas:

\sum_{i=1}^n i = \frac{n(n+1)}{2}, \quad \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}, \quad \sum_{i=1}^n i^3 = \frac{n^2(n+1)^2}{4}i=1βˆ‘n​i=2n(n+1)​,i=1βˆ‘n​i2=6n(n+1)(2n+1)​,i=1βˆ‘n​i3=4n2(n+1)2​

  • \displaystyle 9 + \frac{9}{n} + \frac{9}{n^2}9+n9​+n29​
  • \displaystyle 9 + \frac{9}{2n} + \frac{3}{2n^2}9+2n9​+2n23​
  • \displaystyle 9 + \frac{2n}{9} + \frac{2n^2}{3}9+92n​+32n2​
  • \displaystyle 9 + \frac{27}{2n} + \frac{9}{2n^2}9+2n27​+2n29​
  • \displaystyle 9 + \frac{2n}{27} + \frac{2n^2}{9}9+272n​+92n2​
  • 99

Q3. The line y=xy=x, the xx-axis and the vertical line x=2x=2 bound a triangle of area 22. Thus,

I = \int_{x=0}^2 x \, dx = 2I=∫x=02​xdx=2

Evaluating the Riemann sum for nn subdivisions for the above integral with the same choices of partition and sampling as in the previous problem yields an approximation RS(n)RS(n) for its value II. The error E(n)E(n) we commit by using this approximation is defined to be the difference

E(n) = RS(n) – IE(n)=RS(n)βˆ’I

Some of the following statements are true, while others are false. Among the correct ones, which one is optimal (assuming that you want to know how bad the error can be)?

  • E(n)E(n) is in O(1/n^3)O(1/n3) as n \to +\inftynβ†’+∞
  • E(n)E(n) is in O(n^2)O(n2) as n \to +\inftynβ†’+∞
  • E(n)E(n) is in O(1/n)O(1/n) as n \to +\inftynβ†’+∞
  • E(n)E(n) is in O(1/n^2)O(1/n2) as n \to +\inftynβ†’+∞
  • E(n)E(n) is in O(n)O(n) as n \to +\inftynβ†’+∞
  • E(n)E(n) is in O(1)O(1) as n \to +\inftynβ†’+∞

Q4. Evaluate \displaystyle \int_{x=-\pi/4}^{\pi/4} \left( x^2 + \ln|\cos x|\right) \sin \frac{x}{2} \, dx∫x=βˆ’Ο€/4Ο€/4​(x2+ln∣cosx∣)sin2x​dx. Provide a numeric answer.

Hint: before you try to compute anything, think

Q5. Every function can be expressed as the sum of an even function β€”called its even partβ€” and an odd function β€”its odd part. They are given by the formulas

f^{\mathrm{even}}(x) = \frac{f(x) + f(-x)}{2}, \qquad f^{\mathrm{odd}}(x) = \frac{f(x) – f(-x)}{2}feven(x)=2f(x)+f(βˆ’x)​,fodd(x)=2f(x)βˆ’f(βˆ’x)​

Notice that, indeed,

f^{\mathrm{even}}(-x) = \frac{f(-x) + f(x)}{2} = f^{\mathrm{even}}(x), \qquad f^{\mathrm{odd}}(-x) = \frac{f(-x) – f(x)}{2} = -f^{\mathrm{odd}}(x)feven(βˆ’x)=2f(βˆ’x)+f(x)​=feven(x),fodd(βˆ’x)=2f(βˆ’x)βˆ’f(x)​=βˆ’fodd(x)

and

f^{\mathrm{even}}(x) + f^{\mathrm{odd}}(x) = \frac{f(x) + f(-x)}{2} + \frac{f(x) – f(-x)}{2} = f(x)feven(x)+fodd(x)=2f(x)+f(βˆ’x)​+2f(x)βˆ’f(βˆ’x)​=f(x)

If f(x) = e^xf(x)=ex, which of the following statements are true? Select all that apply.

  • \displaystyle f^{\mathrm{odd}}(x) = 0fodd(x)=0
  • \displaystyle f^{\mathrm{even}}(x) = \sinh xfeven(x)=sinhx
  • \displaystyle f^{\mathrm{odd}}(x) = \sinh xfodd(x)=sinhx
  • \displaystyle f^{\mathrm{even}}(x) = 0feven(x)=0
  • \displaystyle f^{\mathrm{even}}(x) = \cosh xfeven(x)=coshx
  • \displaystyle f^{\mathrm{odd}}(x) = \cosh xfodd(x)=coshx

Practice Quiz 01

Q1. \displaystyle \frac{d}{dx} \int_{t=0}^{\arcsin x} \ln \left| \sin t + \cos t \right| \, dt =dxdβ€‹βˆ«t=0arcsinx​ln∣sint+cost∣dt=

  • \displaystyle \ln \left| x + \sqrt{1-x^2} \right| \arcsin x lnβˆ£βˆ£βˆ£βˆ£β€‹x+1βˆ’x2β€‹βˆ£βˆ£βˆ£βˆ£β€‹arcsinx
  • \ln \left| x + \sqrt{1-x^2} \right|lnβˆ£βˆ£βˆ£β€‹x+1βˆ’x2β€‹βˆ£βˆ£βˆ£β€‹
  • \displaystyle \frac{1}{\sqrt{1-x^2}}\ln \left| x + \sqrt{1-x^2} \right|1βˆ’x2​1​lnβˆ£βˆ£βˆ£βˆ£β€‹x+1βˆ’x2β€‹βˆ£βˆ£βˆ£βˆ£β€‹
  • \ln \arcsin xlnarcsinx
  • \arcsin\left( \ln \left| x + \sqrt{1-x^2} \right| \right)arcsin(lnβˆ£βˆ£βˆ£β€‹x+1βˆ’x2β€‹βˆ£βˆ£βˆ£β€‹)
  • 0

Q2. \displaystyle \frac{d}{dx} \int_{t=\sin x}^{\tan x} e^{-t^2} \, dt =dxdβ€‹βˆ«t=sinxtanx​eβˆ’t2dt=

  • \displaystyle \int_{2\sin x\, cos x}^{2 \tan x\, \sec^2 x} e^{-t^2} \, dt∫2sinxcosx2tanxsec2x​eβˆ’t2dt
  • e^{-4\tan^2 x \sec^4 x} – e^{-4\sin^2 x \cos^2 x}eβˆ’4tan2xsec4xβˆ’eβˆ’4sin2xcos2x
  • 00
  • e^{-\tan^2 x}\sec^2 x – e^{-\sin^2 x}\cos xeβˆ’tan2xsec2xβˆ’eβˆ’sin2xcosx
  • 2e^{-\tan^2 x}\tan x \sec^2 x – 2e^{-\sin^2 x}\sin x \cos x2eβˆ’tan2xtanxsec2xβˆ’2eβˆ’sin2xsinxcosx
  • 2(\tan x \sec^2 x – \sin x \cos x)e^{-t^2}2(tanxsec2xβˆ’sinxcosx)eβˆ’t2

Q3. Which of the following is the leading order term in the Taylor series about x=0x=0 of

f(x) = \int_{0}^x\ln(\cosh(t)) dtf(x)=∫0x​ln(cosh(t))dt

Hint: yes, there’s more than one way to do this problem… Try using the F.T.I.C. to compute the derivatives.

  • \displaystyle \frac{x^2}{6}6x2​
  • \displaystyle \frac{x^2}{2}2x2​
  • \displaystyle -\frac{x^5}{60}βˆ’60x5​
  • \displaystyle \frac{x^3}{2}2x3​
  • \displaystyle \frac{x^3}{6}6x3​
  • \displaystyle -\frac{x^3}{18}βˆ’18x3​

Q4. We usually use Riemann sums to approximate integrals, but we can go the other way, too, using an antiderivative to approximate a sum. Using only your head (no paper, no calculator), tell me which of the following is the best estimate for

\sum_{n=0}^{100} n^3n=0βˆ‘100​n3

  • 2.5\times 10^72.5Γ—107
  • 2.5\times 10^82.5Γ—108
  • 6.6\times 10^76.6Γ—107
  • 1.0\times 10^71.0Γ—107
  • 3.3\times 10^73.3Γ—107
  • 1.0\times 10^81.0Γ—108

Main Quiz 02

Q1. \displaystyle \int_{x=-1}^1 \frac{dx}{1+x^2} =∫x=βˆ’11​1+x2dx​=

  • \displaystyle \frac{\pi}{4}4π​
  • \piΟ€
  • 2\pi2Ο€
  • \displaystyle \frac{\pi}{2}2π​
  • 00
  • \displaystyle \frac{\pi}{3}3π​

Q2. \displaystyle \int_{x=0}^{3} 5x \sqrt{x+1} \, dx =∫x=03​5xx+1​dx=

  • \displaystyle \frac{116}{3}3116​
  • 3030
  • 12\sqrt{3}123​
  • 00
  • -12\sqrt{3}βˆ’123​
  • \displaystyle -\frac{116}{3}βˆ’3116​

Q3. \displaystyle \int_{x=-\pi}^\pi \frac{d}{dx} ( x\cos x ) \, dx =∫x=βˆ’Ο€Ο€β€‹dxd​(xcosx)dx=

  • x\cos x + Cxcosx+C
  • 00
  • 2\pi2Ο€
  • \cos x – x\sin x + Ccosxβˆ’xsinx+C
  • x\cos xxcosx
  • -2\piβˆ’2Ο€

Q4. \displaystyle \frac{d}{dx} \int_{x=-\pi}^\pi x\cos x \, dx =dxdβ€‹βˆ«x=βˆ’Ο€Ο€β€‹xcosxdx=

  • \cos x – x\sin x + Ccosxβˆ’xsinx+C
  • 2\pi2Ο€
  • x\cos x + Cxcosx+C
  • x\cos xxcosx
  • 00
  • -2\piβˆ’2Ο€

Q5. \displaystyle \frac{d}{dx} \int_{t=0}^x \cos t \, dt =dxdβ€‹βˆ«t=0x​costdt=

  • \sin xsinx
  • \sin x – 1sinxβˆ’1
  • 1 – \sin x1βˆ’sinx
  • 1 – \cos x1βˆ’cosx
  • \cos x – 1cosxβˆ’1
  • \cos xcosx

Week 04:Calculus: Single Variable Part 3 – Integration Coursera Quiz Answers

Main Quiz 01

Q1. Recall that we say that an integral is improper if:

  • the integrand blows up somewhere inside or on the boundary of the domain of integration,
  • the domain of integration is not bounded, or
  • both of the above.

You could also say that an integral that is not improper is proper (although this term is not used much).

Among the integrals below, select all of the improper ones and none of the proper ones.

  • \displaystyle \int_{\theta=-\pi/4}^{\pi/4} \sec \theta \, d\theta∫θ=βˆ’Ο€/4Ο€/4​secΞΈdΞΈ
  • \displaystyle \int_{x=0}^1 \frac{dx}{\sqrt{x} – 10}∫x=01​xβ€‹βˆ’10dx​
  • \displaystyle \int_{x=0}^1 \frac{dx}{x-1}∫x=01​xβˆ’1dx​
  • \displaystyle \int_{x=-1}^1 \ln(1+x^2) \, dx∫x=βˆ’11​ln(1+x2)dx
  • \displaystyle \int_{t=0}^{+\infty} t^{x-1} e^{-t} \, dt∫t=0+βˆžβ€‹txβˆ’1eβˆ’tdt
  • \displaystyle \int_{x=0}^1 \frac{dx}{x^2}∫x=01​x2dx​
  • \displaystyle \int_{x=1}^{2} \sqrt{2-x} \, dx∫x=12​2βˆ’x​dx
  • \displaystyle \int_{\theta=\pi/4}^{3\pi/4} \sec\theta \, d\theta∫θ=Ο€/43Ο€/4​secΞΈdΞΈ
  • \displaystyle \int_{x=-\infty}^{+\infty} e^{-x^2/2} \, dx∫x=βˆ’βˆž+βˆžβ€‹eβˆ’x2/2dx
  • \displaystyle \int_{x=-1,000,000}^{1,000,000} e^{-x^2/2} \, dx∫x=βˆ’1,000,0001,000,000​eβˆ’x2/2dx

Q2. Consider the integral

\int_{x=0}^1 \frac{e^{-x}}{x} \, dx∫x=01​xeβˆ’x​dx

It is improper because its integrand blows up at x=0x=0. Using your knowledge of Taylor series, determine the leading order behavior of the integrand near x=0x=0 and decide whether the integral converges or diverges.

Note: observe that you do not need to find an antiderivative in order to determine whether an improper integral converges or diverges!

  • \displaystyle \frac{e^{-x}}{x} = 1 + O(x)xeβˆ’x​=1+O(x), so the integral converges.
  • \displaystyle \frac{e^{-x}}{x} = 1 + O(x)xeβˆ’x​=1+O(x), so the integral diverges.
  • \displaystyle \frac{e^{-x}}{x} = e^{-x} + O(xe^{-x})xeβˆ’x​=eβˆ’x+O(xeβˆ’x), so the integral converges.
  • \displaystyle \frac{e^{-x}}{x} = \frac{1}{x} + O(1)xeβˆ’x​=x1​+O(1), so the integral diverges.
  • \displaystyle \frac{e^{-x}}{x} = \frac{1}{x} + O(1)xeβˆ’x​=x1​+O(1), so the integral converges.
  • \displaystyle \frac{e^{-x}}{x} = e^{-x} + O(xe^{-x})xeβˆ’x​=eβˆ’x+O(xeβˆ’x), so the integral diverges.

Q3. By analyzing the leading order behavior of the integrand near x=0x=0, decide whether the following integral converges or diverges.

\int_{x=0}^1 \frac{\cos^2 x}{\sqrt{x}} \, dx∫x=01​x​cos2x​dx

  • \displaystyle \frac{\cos^2 x}{\sqrt{x}} = 1 + O(x^4)x​cos2x​=1+O(x4), so the integral diverges.
  • \displaystyle \frac{\cos^2 x}{\sqrt{x}} = \sqrt{x} + O(x^{3/2})x​cos2x​=x​+O(x3/2), so the integral converges.
  • \displaystyle \frac{\cos^2 x}{\sqrt{x}} = 1 + O(x^{4})x​cos2x​=1+O(x4), so the integral converges.
  • \displaystyle \frac{\cos^2 x}{\sqrt{x}} = \frac{1}{\sqrt{x}} + O(x^{3/2})x​cos2x​=x​1​+O(x3/2), so the integral converges.
  • \displaystyle \frac{\cos^2 x}{\sqrt{x}} = \frac{1}{\sqrt{x}} + O(x^{3/2})x​cos2x​=x​1​+O(x3/2), so the integral diverges.
  • \displaystyle \frac{\cos^2 x}{\sqrt{x}} = \sqrt{x} + O(x^{3/2})x​cos2x​=x​+O(x3/2), so the integral diverges.

Q4. Calculate

I = \int_{x=1}^2 \frac{dx}{\sqrt{x – 1}}I=∫x=12​xβˆ’1​dx​

Hint: remember from the Lecture that improper integrals are calculated as limits. In this case,

I = \lim_{T \to 1^+} \int_{x=T}^2 \frac{dx}{\sqrt{x – 1}}I=Tβ†’1+limβ€‹βˆ«x=T2​xβˆ’1​dx​

  • I = 1I=1
  • \displaystyle I = \frac{\sqrt{2}}{2} – 1I=22β€‹β€‹βˆ’1
  • I = -2I=βˆ’2
  • The integral diverges.
  • I = 2I=2
  • I = \sqrt{2} – 1I=2β€‹βˆ’1

Q5. Calculate

I = \int_{x=0}^4 \frac{2 \, dx}{\sqrt{16 – x^2}}I=∫x=04​16βˆ’x2​2dx​

  • I = 0I=0
  • \displaystyle I = -\frac{\pi}{2}I=βˆ’2π​
  • I = -\piI=βˆ’Ο€
  • I = \piI=Ο€
  • \displaystyle I = \frac{\pi}{2}I=2π​
  • The integral diverges

Q6. Determine the leading order term of the integrand as x \to +\inftyxβ†’+∞ to decide whether the following integral converges or diverges.

\int_{x=1}^{+\infty} \frac{\sqrt[3]{x + 3}}{x^3} \, dx∫x=1+βˆžβ€‹x33x+3​​dx

  • \displaystyle \frac{\sqrt[3]{x + 3}}{x^3} = \frac{1}{x^3} + O\left(\frac{1}{x^4}\right)x33x+3​​=x31​+O(x41​), so the integral diverges.
  • \displaystyle \frac{\sqrt[3]{x + 3}}{x^3} = \frac{1}{x^{10/3}} + O\left(\frac{1}{x^{13/3}}\right)x33x+3​​=x10/31​+O(x13/31​), so the integral diverges.
  • \displaystyle \frac{\sqrt[3]{x + 3}}{x^3} = \frac{1}{x^{8/3}} + O\left(\frac{1}{x^{11/3}}\right)x33x+3​​=x8/31​+O(x11/31​), so the integral diverges.
  • \displaystyle \frac{\sqrt[3]{x + 3}}{x^3} = \frac{1}{x^3} + O\left(\frac{1}{x^4}\right)x33x+3​​=x31​+O(x41​), so the integral converges.
  • \displaystyle \frac{\sqrt[3]{x + 3}}{x^3} = \frac{1}{x^{8/3}} + O\left(\frac{1}{x^{11/3}}\right)x33x+3​​=x8/31​+O(x11/31​), so the integral converges.
  • \displaystyle \frac{\sqrt[3]{x + 3}}{x^3} = \frac{1}{x^{10/3}} + O\left(\frac{1}{x^{13/3}}\right)x33x+3​​=x10/31​+O(x13/31​), so the integral converges.

Q7. Does the integral converge or diverge? \int_{x=1}^{+\infty} \frac{1-5^{-x}}{x} \, dx∫x=1+βˆžβ€‹x1βˆ’5βˆ’x​dx

  • The leading order term in the integrand as x \to +\inftyxβ†’+∞ is 5^{-x}5βˆ’x, so the integral converges.
  • The leading order term in the integrand as x \to +\inftyxβ†’+∞ is 11, so the integral diverges.
  • The leading order term in the integrand as x \to +\inftyxβ†’+∞ is 5^{-x}5βˆ’x, so the integral diverges.
  • The leading order term in the integrand as x \to +\inftyxβ†’+∞ is \displaystyle \frac{1}{x}x1​, so the integral converges.
  • The leading order term in the integrand as x \to +\inftyxβ†’+∞ is 11, so the integral converges.
  • The leading order term in the integrand as x \to +\inftyxβ†’+∞ is \displaystyle \frac{1}{x}x1​, so the integral diverges.

Q8. Calculate

I = \int_{x=0}^{+\infty} e^{-x} \sin x \, dxI=∫x=0+βˆžβ€‹eβˆ’xsinxdx

Hint: again, remember that this integral is defined as the limit

I = \lim_{T \to +\infty} \int_{x=0}^T e^{-x} \sin x \, dxI=Tβ†’+∞limβ€‹βˆ«x=0T​eβˆ’xsinxdx

  • I = 0I=0
  • \displaystyle I = -\frac{1}{2}I=βˆ’21​
  • I = -1I=βˆ’1
  • \displaystyle I = \frac{1}{2}I=21​
  • I = 1I=1
  • The integral diverges

Practice Quiz 01

Q1`. Look at the following integral:

\int_{x=1}^{+\infty} \frac{1}{\sqrt{x^3 -1} } \, dx∫x=1+βˆžβ€‹x3βˆ’1​1​dx

The integrand blows up at x=1x=1, and so it is an improper integral. But it is also improper because the domain of integration is unbounded!

Analyze the behavior of the integrand both near x=1x=1 and as x \to +\inftyxβ†’+∞ to decide whether the integral converges or diverges. From the statements below, choose all the correct ones and none of the incorrect ones.

  • The integral diverges.
  • The integral converges.
  • Near x=1x=1, the integrand is \displaystyle \frac{1}{\sqrt{3(x-1)}} + O\left(\sqrt{x-1}\right)3(xβˆ’1)​1​+O(xβˆ’1​).
  • As x \to +\inftyxβ†’+∞, the integrand is \displaystyle \frac{1}{x^{3/2}} + O\left(\frac{1}{x^{9/2}}\right)x3/21​+O(x9/21​).
  • As x \to +\inftyxβ†’+∞, the integrand is \displaystyle \frac{1}{x^{1/3}} + O(x^{2/3})x1/31​+O(x2/3).
  • Near x=1x=1, the integrand is \displaystyle \frac{1}{(x-1)^{3/2}} + O\left(\frac{1}{(x-1)^{1/2}}\right)(xβˆ’1)3/21​+O((xβˆ’1)1/21​).

Q2. Consider the following two integrals:

I_1 = \int_{x=2}^{+\infty} \frac{dx}{\sqrt{x^3 – 8}}, \qquad I_2 = \int_{x=2}^{+\infty} \frac{1}{\sqrt{(x-2)^3}} \, dxI1​=∫x=2+βˆžβ€‹x3βˆ’8​dx​,I2​=∫x=2+βˆžβ€‹(xβˆ’2)3​1​dx

They look very much alike. In fact, both of their integrands blow up at x=2x=2, and the domain of integration β€”the same one in both casesβ€” is unbounded. But which of the following statements is true?

  • Both I_1I1​ and I_2I2​ diverge.
  • Both I_1I1​ and I_2I2​ converge.
  • I_2I2​ converges, but I_1I1​ diverges.
  • I_1I1​ converges, but I_2I2​ diverges.

Q3. Until now we have used asymptotic analysis to relate an improper integral to a pp-integral. But sometimes the leading order term is not a power of xx…

Identify the leading order term as x \to +\inftyxβ†’+∞ of the integrand of

\int_{x=1}^{+\infty} \frac{1}{\sinh x} \, dx∫x=1+βˆžβ€‹sinhx1​dx

Use this information to decide whether it converges or diverges.

  • \displaystyle \frac{1}{\sinh x} = \frac{e^{-x}}{x} + O(xe^{-x})sinhx1​=xeβˆ’x​+O(xeβˆ’x), so the integral diverges.
  • \displaystyle \frac{1}{\sinh x} = \frac{e^{-x}}{x} + O(xe^{-x})sinhx1​=xeβˆ’x​+O(xeβˆ’x), so the integral converges.
  • \displaystyle \frac{1}{\sinh x} = e^x + O(e^{3x})sinhx1​=ex+O(e3x), so the integral converges.
  • \displaystyle \frac{1}{\sinh x} = e^x + O(e^{3x})sinhx1​=ex+O(e3x), so the integral diverges.
  • \displaystyle \frac{1}{\sinh x} = 2e^{-x} + O(e^{-3x})sinhx1​=2eβˆ’x+O(eβˆ’3x), so the integral converges.
  • \displaystyle \frac{1}{\sinh x} = 2e^{-x} + O(e^{-3x})sinhx1​=2eβˆ’x+O(eβˆ’3x), so the integral diverges.

Q4. For p \geq 0pβ‰₯0 an integer, consider the following integral:I_p = \int_{x=1}^{+\infty} \frac{dx}{\ln^p x}Ip​=∫x=1+βˆžβ€‹lnpxdx​ Which of the following statements is true?

Two hints:

  1. Think about the growth of \ln^p xlnpx as x \to +\inftyxβ†’+∞. Can you compare it to the growth of x^qxq for some value of qq ?
  2. Recall from Lecture 25 that if g(x) \geq f(x)g(x)β‰₯f(x) for every x \in [a,b]x∈[a,b], then\int_{x=a}^b g(x) \,dx \geq \int_{x=a}^b f(x) \,dx∫x=ab​g(x)dxβ‰₯∫x=ab​f(x)dx This is also true if the domain of integration is unbounded! Now, from the comparison between the orders of growth above, what can you deduce?
  • I_pIp​ converges for any value of pp.
  • I_pIp​ converges for p \gt 1p>1 and diverges for p \leq 1p≀1.
  • I_pIp​ converges for p \leq 1p≀1 and diverges for p \gt 1p>1.
  • I_pIp​ diverges for any value of pp.
  • I_pIp​ converges for p \geq 1pβ‰₯1 and diverges for p=0p=0.

Q5. We have alluded to the fact that the integral of e^{-t^2}eβˆ’t2 is difficult. Let’s say I told you that

\int_{0}^\infty e^{-t^2}dt = \frac{\sqrt{\pi}}{2} .∫0βˆžβ€‹eβˆ’t2dt=2π​​.

Use this fact to compute \left(-\frac{1}{2}\right)!(βˆ’21​)! (that is, negative one-half, factorial).

Hint: use the \GammaΞ“-function and a substitution.

  • \displaystyle\frac{\pi}{2}2π​
  • \pi^2Ο€2
  • \displaystyle\frac{\pi^2}{2}2Ο€2​
  • \piΟ€
  • \sqrt{\pi}π​
  • \displaystyle\frac{\sqrt{\pi}}{2}2π​​

Main Quiz 02

Q1. \displaystyle \int \sin ^2 x \cos ^2 x \, dx =∫sin2xcos2xdx=

Hint: you may (or may not) need to use any of the following reduction formulae:

\int \cos^n x \, dx = \frac{\cos^{n-1}x \sin x}{n} + \frac{n-1}{n} \int \cos^{n-2}x \, dx∫cosnxdx=ncosnβˆ’1xsinx​+nnβˆ’1β€‹βˆ«cosnβˆ’2xdx

\int \sin^n x \, dx = -\frac{\sin^{n-1}x \cos x}{n} + \frac{n-1}{n} \int \sin^{n-2}x \, dx∫sinnxdx=βˆ’nsinnβˆ’1xcosx​+nnβˆ’1β€‹βˆ«sinnβˆ’2xdx

More than one answer may be correct.

  • \displaystyle \frac{1}{8} x + \frac{1}{16} \sin 2x – \frac{1}{4} \sin x \cos^3 x + C81​x+161​sin2xβˆ’41​sinxcos3x+C
  • \displaystyle \frac{1}{8} x + \frac{1}{32} \cos 4x + C81​x+321​cos4x+C
  • \displaystyle \frac{1}{8} x + \frac{1}{16} \cos 2x – \frac{1}{4} \sin x \cos^3 x + C81​x+161​cos2xβˆ’41​sinxcos3x+C
  • \displaystyle \frac{1}{8} x – \frac{1}{16} \cos 2x + \frac{1}{4} \sin^3 x \cos x + C81​xβˆ’161​cos2x+41​sin3xcosx+C
  • \displaystyle \frac{1}{8} x – \frac{1}{16} \sin 2x + \frac{1}{4} \sin^3 x \cos x + C81​xβˆ’161​sin2x+41​sin3xcosx+C
  • \displaystyle \frac{1}{8} x – \frac{1}{32} \sin 4x + C81​xβˆ’321​sin4x+C

Q2. \displaystyle \int \sin ^3 \frac{x}{2} \cos ^3 \frac{x}{2} \, dx =∫sin32x​cos32x​dx=

(More than one answer may be correct.)

  • \displaystyle -\frac{1}{3} \sin^6 \frac{x}{2} + \frac{1}{2} \sin^4 \frac{x}{2} + Cβˆ’31​sin62x​+21​sin42x​+C
  • \displaystyle \frac{1}{3} \cos^6 \frac{x}{2} -\frac{1}{2} \cos^4 \frac{x}{2} + C31​cos62xβ€‹βˆ’21​cos42x​+C
  • \displaystyle \frac{1}{24} \cos^3 x -\frac{1}{8} \cos x + C241​cos3xβˆ’81​cosx+C
  • \displaystyle \frac{1}{6} \sin^6 \frac{x}{2} – \frac{1}{4} \sin^4 \frac{x}{2} + C61​sin62xβ€‹βˆ’41​sin42x​+C
  • \displaystyle \frac{1}{96} \big( \cos x – 9\cos 3x \big) + C961​(cosxβˆ’9cos3x)+C
  • \displaystyle -2\sin^5 \frac{x}{2} + 2 \sin^3 \frac{x}{2} + Cβˆ’2sin52x​+2sin32x​+C

Q3. \displaystyle \int \frac{x^3\, dx}{\sqrt{9-x^2}} =∫9βˆ’x2​x3dx​=

Hint: use a trigonometric substitution to obtain a trigonometric integral.

  • \displaystyle – \bigg( 6 + \frac{x^2}{3} \bigg) \sqrt{9-x^2} + Cβˆ’(6+3x2​)9βˆ’x2​+C
  • \displaystyle -3\sqrt{9-x^2} – \frac{1}{81} (9-x^2)^{3/2} + Cβˆ’39βˆ’x2β€‹βˆ’811​(9βˆ’x2)3/2+C
  • \displaystyle \frac{1}{6} (9-x^2)^{1/2} + C61​(9βˆ’x2)1/2+C
  • 3 (9-x^2)^{3/2} + C3(9βˆ’x2)3/2+C
  • \displaystyle \bigg( \frac{1}{3} + 9x^2 \bigg) \sqrt{9-x^2} + C(31​+9x2)9βˆ’x2​+C
  • \displaystyle 9\sqrt{9-x^2} – \frac{1}{3} (9-x^2)^{3/2} + C99βˆ’x2β€‹βˆ’31​(9βˆ’x2)3/2+C

Q4. \displaystyle \int 5\tan ^5 x \sec ^3 x \, dx =∫5tan5xsec3xdx=

  • \displaystyle\frac{5}{6} \tan^6 x + \frac{5}{8} \tan^8 x + C65​tan6x+85​tan8x+C
  • \displaystyle\frac{5}{7} \sec^{14} x – 2\sec^{10} x + \frac{5}{3} \sec^6 x + C75​sec14xβˆ’2sec10x+35​sec6x+C
  • \displaystyle \sec^5 x – \frac{10}{3} \sec^3 x + 5\sec x + Csec5xβˆ’310​sec3x+5secx+C
  • \displaystyle\frac{4}{5} \sec^5 x \tan^5 x – \frac{4}{3} \sec^3 x \tan^3 x + C54​sec5xtan5xβˆ’34​sec3xtan3x+C
  • \displaystyle \frac{5}{7} \tan^7 x + \tan^5 x + C75​tan7x+tan5x+C
  • \displaystyle \frac{5}{7} \sec^7 x – 2 \sec^5 x + \frac{5}{3} \sec^3 x + C75​sec7xβˆ’2sec5x+35​sec3x+C

Q5. \displaystyle \int 7\tan ^4 x \sec ^4 x \, dx =∫7tan4xsec4xdx=

  • \displaystyle \frac{1}{7} \tan^7 x + \frac{1}{5} \tan^5 x + C71​tan7x+51​tan5x+C
  • \displaystyle \frac{7}{5} \tan^5 x + \frac{7}{3} \tan^3 x + C57​tan5x+37​tan3x+C
  • \displaystyle \sec^{7} x + \frac{7}{5} \sec^{5} x + Csec7x+57​sec5x+C
  • \displaystyle \tan^7 x + \frac{7}{5} \tan^5 x + Ctan7x+57​tan5x+C
  • \displaystyle \frac{1}{2} \sec^{14} x + \frac{7}{10} \sec^{10} x + C21​sec14x+107​sec10x+C
  • \displaystyle \tan^7 x – \frac{7}{5} \tan^5 x + Ctan7xβˆ’57​tan5x+C

Q6. Assuming 0 \lt x \lt \pi0<x<Ο€, compute

\displaystyle \int \cot x \sqrt{1 – \cos 2x} \, dx =∫cotx1βˆ’cos2x​dx=

Hint: remember the double-angle formula

\cos 2x = \cos^2 x – \sin^2 xcos2x=cos2xβˆ’sin2x

  • \cos^2 x + Ccos2x+C
  • \sqrt{2} \cos x + C2​cosx+C
  • \displaystyle \frac{x}{2} + \frac{ \cos 2x}{2} + C2x​+2cos2x​+C
  • \sqrt{2} \sin x + C2​sinx+C
  • \cos x + Ccosx+C
  • \sin x + Csinx+C

Main Quiz 03

Q1. \displaystyle \int \frac{(\arcsin x)^2}{\sqrt{1-x^2}}\, dx =∫1βˆ’x2​(arcsinx)2​dx=

  • \displaystyle \frac{1}{3} \sin^3 x + C31​sin3x+C
  • \displaystyle \frac{1}{3} \left( \arcsin\sqrt{1-x^2} \right)^3 + C31​(arcsin1βˆ’x2​)3+C
  • \displaystyle \frac{1}{3} \sqrt{1-(\arcsin x)^3} + C31​1βˆ’(arcsinx)3​+C
  • \displaystyle \frac{2}{3} \sqrt{1-\sin^3 x} + C32​1βˆ’sin3x​+C
  • \displaystyle \sqrt{\frac{1}{3}(\arcsin x)^3} + C31​(arcsinx)3​+C
  • \displaystyle \frac{(\arcsin x)^3}{3\sqrt{1-x^2}} + C31βˆ’x2​(arcsinx)3​+C
  • \displaystyle \frac{1}{3} (\arcsin x)^3 + C31​(arcsinx)3+C
  • \displaystyle \frac{x^3}{3} + C3x3​+C

Q2. \displaystyle \int \frac{dx}{x^2-x-6} =∫x2βˆ’xβˆ’6dx​=

  • \displaystyle \frac{1}{3} \ln\left| \frac{x-3}{x+2} \right| + C31​lnβˆ£βˆ£βˆ£βˆ£βˆ£β€‹x+2xβˆ’3β€‹βˆ£βˆ£βˆ£βˆ£βˆ£β€‹+C
  • \displaystyle -\frac{1}{3} \ln\left| \frac{x-3}{x+2} \right| + Cβˆ’31​lnβˆ£βˆ£βˆ£βˆ£βˆ£β€‹x+2xβˆ’3β€‹βˆ£βˆ£βˆ£βˆ£βˆ£β€‹+C
  • \displaystyle -\frac{1}{3} \ln\left| \frac{x+3}{x-2} \right| + Cβˆ’31​lnβˆ£βˆ£βˆ£βˆ£βˆ£β€‹xβˆ’2x+3β€‹βˆ£βˆ£βˆ£βˆ£βˆ£β€‹+C
  • \displaystyle \frac{1}{5} \ln\left| \frac{x-3}{x+2} \right| + C51​lnβˆ£βˆ£βˆ£βˆ£βˆ£β€‹x+2xβˆ’3β€‹βˆ£βˆ£βˆ£βˆ£βˆ£β€‹+C
  • \displaystyle -\frac{1}{5} \ln\left| \frac{x-3}{x+2} \right| + Cβˆ’51​lnβˆ£βˆ£βˆ£βˆ£βˆ£β€‹x+2xβˆ’3β€‹βˆ£βˆ£βˆ£βˆ£βˆ£β€‹+C
  • \displaystyle -\frac{1}{5} \ln\left| \frac{x+3}{x-2} \right| + Cβˆ’51​lnβˆ£βˆ£βˆ£βˆ£βˆ£β€‹xβˆ’2x+3β€‹βˆ£βˆ£βˆ£βˆ£βˆ£β€‹+C
  • \displaystyle \frac{1}{3} \ln\left| \frac{x+3}{x-2} \right| + C31​lnβˆ£βˆ£βˆ£βˆ£βˆ£β€‹xβˆ’2x+3β€‹βˆ£βˆ£βˆ£βˆ£βˆ£β€‹+C
  • \displaystyle \frac{1}{5} \ln\left| \frac{x+3}{x-2} \right| + C51​lnβˆ£βˆ£βˆ£βˆ£βˆ£β€‹xβˆ’2x+3β€‹βˆ£βˆ£βˆ£βˆ£βˆ£β€‹+C

Q3. Exactly half of the following integrals converge: which are they? In order to receive full credit for this problem, you must select all the integrals that converge and none of those that diverge.

  • \displaystyle \int_{x=0}^1 \frac{\cos x}{x^2}dx∫x=01​x2cosx​dx
  • \displaystyle \int_{x=0}^3 \frac{dx}{x-1}∫x=03​xβˆ’1dx​
  • \displaystyle \int_{x=0}^1 \frac{\sin(\pi x)}{x^{3/2}} dx∫x=01​x3/2sin(Ο€x)​dx
  • \displaystyle \int_{x=1}^2 \frac{2}{\sqrt[3]{x-1}} dx∫x=12​3xβˆ’1​2​dx
  • \displaystyle \int_{x=0}^{+\infty} \frac{2x + 3}{x^2 + 3x + 6}\, dx∫x=0+βˆžβ€‹x2+3x+62x+3​dx
  • \displaystyle \int_{x=2}^{+\infty} \frac{e^{-x}}{x}dx∫x=2+βˆžβ€‹xeβˆ’x​dx
  • \displaystyle \int_{x=2}^{+\infty} \frac{dx}{x^{1/2}(x-1)^{3/2}}∫x=2+βˆžβ€‹x1/2(xβˆ’1)3/2dx​
  • \displaystyle \int_{x=2}^{+\infty} \frac{x}{1+x^2}dx∫x=2+βˆžβ€‹1+x2x​dx

Q4. \displaystyle \int_{x=0}^{\pi/2} x \sin 2x \,dx =∫x=0Ο€/2​xsin2xdx=

  • \displaystyle \frac{1}{4}41​
  • \displaystyle -\frac{\pi}{4}βˆ’4π​
  • \displaystyle -\frac{1}{4}βˆ’41​
  • \displaystyle \frac{\pi}{2}2π​
  • \displaystyle -\frac{1}{2}βˆ’21​
  • \displaystyle -\frac{\pi}{2}βˆ’2π​
  • \displaystyle \frac{\pi}{4}4π​
  • \displaystyle \frac{1}{2}21​

Q5. Compute \displaystyle \int \frac{dx}{x^2-2x+10}∫x2βˆ’2x+10dx​. Hint: partial fractions doesn’t seem to work on this problem, since you don’t have real roots for the denominator…is there another technique you could use?

  • \displaystyle \frac{1}{3} \arctan\frac{x+1}{5} + C31​arctan5x+1​+C
  • \displaystyle \frac{1}{2} \arctan\frac{x+1}{2} + C21​arctan2x+1​+C
  • \displaystyle \frac{1}{3} \arctan\frac{x+1}{3} + C31​arctan3x+1​+C
  • \displaystyle \frac{1}{3} \arctan\frac{x-1}{5} + C31​arctan5xβˆ’1​+C
  • \displaystyle \frac{1}{2} \arctan\frac{x-1}{4} + C21​arctan4xβˆ’1​+C
  • \displaystyle \frac{1}{2} \arctan\frac{x+1}{4} + C21​arctan4x+1​+C
  • \displaystyle \frac{1}{2} \arctan\frac{x-1}{2} + C21​arctan2xβˆ’1​+C
  • \displaystyle \frac{1}{3} \arctan\frac{x-1}{3} + C31​arctan3xβˆ’1​+C

Q6. Solve the following separable differential equation \displaystyle \frac{dx}{dt} = e^{t-x}dtdx​=etβˆ’x.

  • \displaystyle x = -\ln\left(C-e^t\right)x=βˆ’ln(Cβˆ’et)
  • \displaystyle x = \left( \frac{t^2}{2} + C \right) e^{-2t}x=(2t2​+C)eβˆ’2t
  • \displaystyle x = \ln\left(Ce^{-2t}\right)x=ln(Ceβˆ’2t)
  • \displaystyle x = \frac{t}{2} – \frac{1}{4} + Cx=2tβ€‹βˆ’41​+C
  • \displaystyle x = \frac{t^2}{4} + \frac{1}{2} + Ce^{-2t}x=4t2​+21​+Ceβˆ’2t
  • \displaystyle x = -\frac{t}{2} – \frac{1}{4} + Ce^{2t}x=βˆ’2tβ€‹βˆ’41​+Ce2t
  • \displaystyle x = \ln\left(e^t+C\right)x=ln(et+C)
  • \displaystyle x =t+ \ln Cx=t+lnC

Q7. Compute \displaystyle \int \cos^3 4x \,dx∫cos34xdx .

  • \displaystyle \frac{1}{16} \cos^4 4x + C161​cos44x+C
  • \displaystyle \cos 4x + \frac{1}{4}\cos^4 4x + Ccos4x+41​cos44x+C
  • \displaystyle \frac{1}{4} \cos^4 4x + C41​cos44x+C
  • \displaystyle \frac{1}{4}\cos 4x + \frac{1}{12}\cos^4 4x + C41​cos4x+121​cos44x+C
  • \displaystyle \sin 4x – \frac{1}{3}\sin^3 4x + Csin4xβˆ’31​sin34x+C
  • \displaystyle \frac{1}{4}\sin 4x – \frac{1}{12}\sin^3 4x + C41​sin4xβˆ’121​sin34x+C
  • 12\cos^2 4x + C12cos24x+C
  • \displaystyle \frac{1}{4}\sin 4x + \frac{1}{12}\sin^3 4x + C41​sin4x+121​sin34x+C

Q8. \displaystyle \left.\frac{d}{dx}\right|_{x=\pi} \int_{t=0}^{x} \frac{\cos 3t}{\sqrt{1+t}} \,dt =dxdβ€‹βˆ£βˆ£βˆ£βˆ£βˆ£β€‹x=Ο€β€‹βˆ«t=0x​1+t​cos3t​dt=

  • \displaystyle \frac{\pi}{\sqrt{1+\pi}}1+π​π​
  • 00
  • -\piβˆ’Ο€
  • \displaystyle -\frac{1}{\sqrt{1+\pi}}βˆ’1+π​1​
  • \displaystyle \int_0^{\pi} \frac{d}{dt}\!\left( \frac{\cos 3t}{\sqrt{1+t}} \right) dt∫0π​dtd​(1+t​cos3t​)dt
  • \displaystyle -1βˆ’1
  • \displaystyle \frac{\cos(3\ln \pi)}{\pi\sqrt{1+\ln \pi}}Ο€1+lnπ​cos(3lnΟ€)​
  • \displaystyle \frac{1}{\sqrt{1+\pi}}1+π​1​

Q9. Which of the following is the integrating factor, II, used to solve the linear differential equation

e^{3t}\frac{dx}{dt}=2-e^txe3tdtdx​=2βˆ’etx

  • I = e^{e^{t}}I=eet, that is, I= {\rm exp}\left( e^{t}\right)I=exp(et)
  • I = 2e^{3t}I=2e3t, that is, I= 2\,{\rm exp }\left( 3t\right)I=2exp(3t)
  • I = e^{\frac{1}{2}e^{2t}}I=e21​e2t, that is, I= {\rm exp}\left(\frac{1}{2}e^{2t}\right)I=exp(21​e2t)
  • I = e^{-e^{-t}}I=eβˆ’eβˆ’t, that is, I= {\rm exp}\left( -e^{-t}\right)I=exp(βˆ’eβˆ’t)
  • I = 2e^{-3t}I=2eβˆ’3t, that is, I= 2\,{\rm exp }\left( -3t\right)I=2exp(βˆ’3t)
  • I = e^{\frac{1}{3}e^{3t}}I=e31​e3t, that is, I= {\rm exp}\left( \frac{1}{3}e^{3t}\right)I=exp(31​e3t)
  • I = e^{-\frac{1}{3}e^{-3t}}I=eβˆ’31​eβˆ’3t, that is, I= {\rm exp}\left( -\frac{1}{3}e^{-3t}\right)I=exp(βˆ’31​eβˆ’3t)
  • I = e^{-\frac{1}{2}e^{-2t}}I=eβˆ’21​eβˆ’2t, that is, I= {\rm exp}\left( -\frac{1}{2}e^{-2t}\right)I=exp(βˆ’21​eβˆ’2t)

Q10. The size z(t)z(t) of a hailstone evolves according to the differential equation

\frac{dz}{dt} = A \sqrt{z} – B\sqrt{z^3}dtdz​=Azβ€‹βˆ’Bz3​

where AA and BB are positive constants. Without solving the differential equation, determine the limiting size \displaystyle \lim_{t\to+\infty}z(t)tβ†’+∞lim​z(t) in the case where z(0)=1z(0)=1. Hint: think in terms of equilibria…

  • \displaystyle\frac{\sqrt{A}}{B}BA​​
  • \displaystyle-\frac{A}{B}βˆ’BA​
  • \displaystyle\sqrt{\frac{A}{B}}BA​​
  • \displaystyle\frac{A}{B}BA​
  • \displaystyle-\frac{B}{A}βˆ’AB​
  • \displaystyle\frac{B}{A}AB​
  • \displaystyle\frac{A}{\sqrt{B}}B​A​
  • \displaystyle\sqrt{\frac{B}{A}}AB

More About This Course

Calculus is one of the greatest things that people have thought of. It helps us understand everything from the orbits of planets to the best size for a city to how often a heart beats. This quick course covers the main ideas of Calculus with one variable, with a focus on understanding the ideas and how to use them. This course is perfect for students who are just starting out in engineering, the physical sciences, or the social sciences.

The course is different because:

1) Taylor series and approximations are introduced and used from the start;
2) a new way of combining discrete and continuous forms of calculus is used;
3) the emphasis is on the ideas rather than the calculations; and
4) the course is taught in a clear, dynamic, and unified way.

SKILLS YOU WILL GAIN

  • Differential Equations
  • Integration By Parts
  • Improper Integral
  • Integration By Substitution

Conclusion

Hopefully, this article will be useful for you to find all theΒ Week, final assessment, and Peer Graded Assessment Answers of Calculus: Single Variable Part 3 – Integration Quiz of CourseraΒ and grab some premium knowledge with less effort. If this article really helped you in any way then make sure to share it with your friends on social media and let them also know about this amazing training. You can also check out our other courseΒ Answers.Β So, be with us guys we will share a lot more free courses and their exam/quiz solutions also, and follow ourΒ Techno-RJΒ BlogΒ for more updates.

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  20. γ€Š539ε½©εˆΈοΌšε°η£ηš„ε°η’ΊεΉΈγ€‹

    ε“Žε‘€οΌŒθͺͺεˆ°ε°η£ηš„ε½©εˆΈιŠζˆ²οΌŒδ½ ζ€ŽιΊΌε―θƒ½δΈηŸ₯道539ε½©εˆΈε‘’οΌŸζ―ζ¬‘”539ι–‹ηŽ”οΌŒιƒ½ζœ‰ι‚£ιΊΌε€šδΊΊη·ŠεΌ΅εœ°η›―θ‘—θž’εΉ•οΌŒεΏƒζƒ³οΌšγ€Œι€™ζ¬‘ζœƒδΈζœƒθΌͺεˆ°ζˆ‘οΌŸγ€γ€‚

    ### 539ε½©εˆΈοΌŒι‚£ζ˜―δ»€ιΊΌδΎ†ι ­οΌŸ

    嘿,539ε½©εˆΈε―δΈζ˜―ζ˜¨ε€©ζ‰ζœ‰ηš„ζ–°ιδΊ‹οΌŒεƒεœ¨ε°η£ε·²ηΆ“ι™ͺδΌ΄δΊ†ζˆ‘ε€‘ε₯½ε€šεΉ΄δΊ†γ€‚η°‘ε–ηš„ηŽ©ζ³•οΌŒε°ε°ηš„ζŠ•ζ³¨οΌŒε»ζœ‰θ‘—δΈε°ηš„ζœŸεΎ…οΌŒι›£ζ€ͺεƒι€™ιΊΌε—ζ­‘θΏŽγ€‚

    ### 539ι–‹ηŽοΌŒζ˜―ε ΄θ¦–θ¦Ίη››ε΄οΌ

    每欑”539ι–‹ηŽ”οΌŒιƒ½εƒζ˜―δΈ€ε ΄ε°εž‹ηš„η―€η›γ€‚ε°ˆζ₯­ηš„δΈ»ζŒδΊΊγ€ζ˜ŽδΊηš„η‡ˆε…‰οΌŒι‚„ζœ‰ι‚£ε°ε°ˆζ₯­ηš„ζŠ½ηŽζ©Ÿε™¨οΌŒζ―ζ¬‘ιƒ½εΈΆη΅¦ζˆ‘ε€‘δΈε°ηš„εˆΊζΏ€γ€‚

    ### θ·Ÿζˆ‘δΈ€θ΅·ηŽ©539?

    ζƒ³ηŽ©539οΌŸθΆ…η°‘ε–οΌθ΅°εˆ°θ‘—δΈŠοΌŒζ‰Ύε€‹ε½©εˆΈθ‘ŒοΌŒιΈδΊ”ε€‹δ½ ε–œζ­‘ηš„θ™Ÿη’ΌοΌŒθ²·δΈ‹δΎ†ε°±ε°δΊ†γ€‚η•Άη„ΆοΌŒηΎεœ¨η§‘ζŠ€ι€™ιΊΌη™Όι”οΌŒεεœ¨εΆθ£‘δΉŸθƒ½θ²·οΌŒε€šζ–ΉδΎΏοΌ

    ### 539ι–‹ηŽοΌŒι‚£εˆΊζΏ€ηš„ζ„Ÿθ¦ΊοΌ

    每欑”539ι–‹ηŽ”οΌŒηœŸηš„ζ˜―θ“δΊΊζ—’ζœŸεΎ…εˆη·ŠεΌ΅γ€‚ζƒ³εƒδΈ€δΈ‹οΌŒε¦‚ζžœι€™ζ¬‘δΈ­δΊ†οΌŒζ˜―δΈζ˜―ε―δ»₯εŽ»εƒι‚£εΆδΈ€η›΄ζƒ³εŽ»δ½†εˆθ¦ΊεΎ—ε€ͺθ²΄ηš„ι€ε»³οΌŸ

    ### ζœ€εΎŒθͺͺε…©ε₯

    539ε½©εˆΈοΌŒηœŸηš„ζ˜―ε€‹ε°η’ΊεΉΈγ€‚δ½†ε˜ΏοΌŒηŽ©ε½©εˆΈδΉŸθ¦ζœ‰εΊ¦οΌŒεˆ₯ε€ͺζ²‰θΏ·ε“¦οΌεΈŒζœ›ζ―ζ¬‘”539ι–‹ηŽ”οΌŒιƒ½θƒ½εΈΆη΅¦δ½ δΈ€ι»žι»žηš„ι©šε–œε’ŒεΏ«ζ¨‚γ€‚

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