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- Calculus: Single Variable Part 3 – Integration Quiz Answers
- Week 01: Calculus: Single Variable Part 3 β Integration Coursera Quiz Answers
- Practice Quiz 01
- Week 02 :Calculus: Single Variable Part 3 β Integration Coursera Quiz Answers
- Main Quiz 01
- Week 03:Calculus: Single Variable Part 3 β Integration Coursera Quiz Answers
- Practice Quiz 01
- Week 04:Calculus: Single Variable Part 3 β Integration Coursera Quiz Answers
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About Calculus: Single Variable Part 3 – Integration Course
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 β Integration Coursera Quiz Answers
Main Quiz 01
Q1. \displaystyle \int (4x^3 + 3x^2 + 2x + 1) \, dx =β«(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 + C41βx4+31βx3+21βx2+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 =dxdββ«lntanxdx=
- \ln \tan x + Clntanx+C
- \displaystyle \frac{\sec^2 x}{\tan x}tanxsec2xβ
- 00
- \displaystyle \frac{\sec^2 x}{\tan x} + Ctanxsec2xβ+C
- \ln \tan xlntanx
- \displaystyle \int \frac{\sec^2 x}{\tan x} \, dxβ«tanxsec2xβdx
Q3. \displaystyle \int \left( \frac{d}{dx} e^{-x} \right) \, dx =β«(dxdβeβx)dx=
- e^{x} + Cex+C
- e^{-x}eβx
- e^{-x} + Ceβx+C
- \displaystyle \frac{d}{dx} \int e^{-x} \, dxdxdββ«eβxdx
- -e^{-x} + Cβeβx+C
- -e^{-x}βeβx
Q4. Find the general solution of the differential equation
\frac{dx}{dt} = t^2dtdxβ=t2
- \displaystyle x(t) = \frac{1}{2} t^2 + Cx(t)=21βt2+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)=31βt3+C
- \displaystyle x(t) = \frac{1}{3} t^3x(t)=31βt3
- \displaystyle x(t) = t^3x(t)=t3
Q5. Find the general solution of the differential equation
\frac{dx}{dt} = x^2dtdxβ=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)=βt1β+C
- \displaystyle x(t) = \frac{1}{-t + C}x(t)=βt+C1β
- \displaystyle x(t) = -\frac{1}{t^2} + Cx(t)=βt21β+C
- \displaystyle x(t) = -\frac{1}{t}x(t)=βt1β
- \displaystyle x(t) = -\frac{1}{t^2}x(t)=βt21β
- \displaystyle x(t) = \frac{1}{-t^2 + C}x(t)=βt2+C1β
Practice Quiz 01
Q1. There is a large class of differential equations βthe 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(β)
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)J0β(t) that we saw in Chapter 1. Notice that (\astβ) involves not only the first derivative \displaystyle \frac{dx}{dt}dtdxβ but also the second derivative \displaystyle \frac{d^2x}{dt^2}dt2d2xβ. 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=0βββckβtk
Substituting this into (\astβ) will give you two conditions. The first one is c_1 = 0c1β=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}ckβ=k2ckβ2ββ
- \displaystyle c_k = \frac{c_{k-1}}{k^2}ckβ=k2ckβ1ββ
- \displaystyle c_k = β \frac{c_{k-1}}{(k-1)^2}ckβ=β(kβ1)2ckβ1ββ
- \displaystyle c_k = β \frac{c_{k-2}}{k^2}ckβ=βk2ckβ2ββ
- \displaystyle c_k = β \frac{c_{k-2}}{(k-2)^2}ckβ=β(kβ2)2ckβ2ββ
- \displaystyle c_k = β \frac{c_{k-1}}{k^2}ckβ=βk2ckβ1ββ
Main Quiz 02
Q1. After drinking a cup of coffee, the amount CC of caffeine in a personβs body obeys the differential equation
\frac{dC}{dt}= -\alpha CdtdCβ=βΞ±C
where the constant \alphaΞ± has an approximate value of 0.14 hours^{-1}β1.
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 IdtdIβ=βΞ»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»ln2β
- \displaystyle \frac{2}{\lambda}Ξ»2β
- \displaystyle \frac{\lambda}{2}2Ξ»β
- \displaystyle \frac{\lambda}{\ln 2}ln2Ξ»β
- \displaystyle \frac{\ln \lambda}{2}2lnΞ»β
- \displaystyle \frac{2}{\ln \lambda}lnΞ»2β
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} = β \frac{\alpha}{r}hdtdhβ=βrΞ±βh
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}rΞ±βln31β
- \displaystyle \frac{\alpha}{r} \ln 12rΞ±βln12
- \displaystyle \frac{r}{\alpha} \ln 3Ξ±rβln3
- \displaystyle \frac{\alpha}{r} \ln 3rΞ±βln3
- \displaystyle \frac{r}{\alpha} \ln \frac{1}{3}Ξ±rβln31β
- \displaystyle \frac{r}{\alpha} \ln 12Ξ±rβln12
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 Newtonβs law of cooling:
\frac{dT}{dt} = \kappa (A β T)dtdTβ=ΞΊ(AβT)
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}β1.
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} Cβ2oC, how long did it take for passengers of the Titanic to enter severe hypothermia? Recall from lecture that heat transfer is described by Newtonβs law of cooling:
\frac{dT}{dt} = \kappa (A β T)dtdTβ=ΞΊ(AβT)
where TT is the body temperature of a passenger, AA the water temperature, and \kappa = 0.016ΞΊ=0.016 min^{-1}β1. 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)β=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 Malthusβ 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)=F0β+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, βWorld Agriculture: Towards 2015/2030. Summary Reportβ, 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+1.11tβtβ166 years.
- 39.42 e^{1.1 t} = 61.3 + 1.11 t \Rightarrow t \approx 0.439.42e1.1t=61.3+1.11tβtβ0.4 years.
- 0.11 e^{1.1 t} = 61.3 + 1.11 t \Rightarrow t \approx 60.11e1.1t=61.3+1.11tβtβ6 years.
- 0.11 e^{0.011 t} = 61.3 + 1.11 t \Rightarrow t \approx 8270.11e0.011t=61.3+1.11tβtβ827 years.
Main Quiz 03
Q1. Solve the differential equation \displaystyle \frac{dx}{dt} = \frac{x}{t}dtdxβ=txβ.
Note: observe that this equation can be rearranged as \displaystyle \frac{dx}{x} = \frac{dt}{t}xdxβ=tdtβ, 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}}dtdxβ=1βt2β1βx2ββ.
- 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^xdtdxβ=tex, compute x(1)x(1).
- x(1) = 0x(1)=0
- x(1) = \sqrt{2}x(1)=2β
- x(1) = \ln 2x(1)=ln2
- x(1) = 2x(1)=2
- x(1) = \displaystyle \ln \frac{1}{2}x(1)=ln21β
- x(1) = \displaystyle\frac{1}{2}x(1)=21β
Q4. Which of the following is the integrating factor used to solve the following linear differential equation?
t^2\frac{dx}{dt}=4t-t^5xt2dtdxβ=4tβt5x
- I(t) = e^{-1/2t^2}I(t)=eβ1/2t2
- I(t) = e^{-t^6/6}I(t)=eβt6/6
- 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)=eβt4/4
- I(t) = e^{t^4/4}I(t)=et4/4
Q5. Solve the differential equation \displaystyle \frac{dx}{dt} β 5x = 3dtdxββ5x=3.
- \displaystyle x(t) = \frac{5}{3} + C e^{5t}x(t)=35β+Ce5t
- \displaystyle x(t) = -\frac{3}{5} + C e^{-5t}x(t)=β53β+Ceβ5t
- \displaystyle x(t) = \frac{3}{5} + C e^{5t}x(t)=53β+Ce5t
- \displaystyle x(t) = -\frac{3}{5} + C e^{5t}x(t)=β53β+Ce5t
- \displaystyle x(t) = -\frac{5}{3} + C e^{-5t}x(t)=β35β+Ceβ5t
- \displaystyle x(t) = -\frac{5}{3} + C e^{5t}x(t)=β35β+Ce5t
Practice Quiz 03
Q1. Solve the differential equation \displaystyle \frac{dx}{dt} = \frac{x}{1+t} + 2dtdxβ=1+txβ+2.
- \displaystyle x(t) = 1 + t + \frac{C}{1+t}x(t)=1+t+1+tCβ
- 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+t1β
- 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+tβ+1+tCβ
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.dtdDβ=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 β PdtdDβ=rDβP
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 Weberβs law as
dp = k \frac{dS}{S}dp=kSdSβ
where pp is the perceived intensity of a stimulus and SS its actual strength. Observe that \displaystyle \frac{dS}{S}SdSβ 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}dtdxβ+p(t)x=q(t)xΞ±,Ξ±βR
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=x1βΞ± yields a linear differential equation in the variable uu.
Apply the above change of variables in the case
\frac{dx}{dt} + 2tx = x^3dtdxβ+2tx=x3
to find the linear differential equations satisfied by uu.
- \displaystyle \frac{du}{dt} β 4tu = -2dtduββ4tu=β2
- \displaystyle \frac{du}{dt} + 2tu = 1dtduβ+2tu=1
- \displaystyle \frac{du}{dt} + 2tu = -2dtduβ+2tu=β2
- \displaystyle \frac{du}{dt} β 4tu = 1dtduββ4tu=1
- \displaystyle \frac{du}{dt} = -2-2tdtduβ=β2β2t
- \displaystyle \frac{du}{dt} = 1-2tdtduβ=1β2t
Main Quiz 04
Q1. For a differential equation
\frac{dx}{dt} = f(x)dtdxβ=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 ,dtdxβ=fβ²(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 0fβ²(0)ξ β=0).
The differential equation
\frac{dx}{dt} = (e^x-1)(x-1)dtdxβ=(exβ1)(xβ1)
has an equilibrium at x=0x=0. Which of the following is the linearized equation at this equilibrium?
- \displaystyle \frac{dx}{dt} = -exdtdxβ=βex
- \displaystyle \frac{dx}{dt} = 0dtdxβ=0
- \displaystyle \frac{dx}{dt} = e^xdtdxβ=ex
- \displaystyle \frac{dx}{dt} = xdtdxβ=x
- \displaystyle \frac{dx}{dt} = -xdtdxβ=βx
- \displaystyle \frac{dx}{dt} = exdtdxβ=ex
Q2. The differential equation in Question 1 has another equilibrium at x=1x=1. Letβs look at that. Recall, for a differential equation
\frac{dx}{dt} = f(x)dtdxβ=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=xβa. 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 ,dtdhβ=fβ²(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 0fβ²(a)ξ β=0).
The differential equation
\frac{dx}{dt} = (e^x-1)(x-1)dtdxβ=(exβ1)(xβ1)
has an equilibrium at x=1x=1. What is the linearized equation at this equilibrium?
- \displaystyle\frac{dh}{dt} = -hdtdhβ=βh
- \displaystyle\frac{dh}{dt} = (e-1)hdtdhβ=(eβ1)h
- \displaystyle\frac{dh}{dt} = -ehdtdhβ=βeh
- \displaystyle\frac{dh}{dt} = ehdtdhβ=eh
- \displaystyle\frac{dh}{dt} = hdtdhβ=h
- \displaystyle\frac{dh}{dt} = 0dtdhβ=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)dtdxβ=f(x) is stable if f'(a)\lt 0fβ²(a)<0 and is unstable if f'(a)\gt 0fβ²(a)>0.
Recall from Lecture 18, Newtonβs Law of Heat Transfer, which states that
\frac{dT}{dt} = \kappa ( A β T ),dtdTβ=ΞΊ(AβT),
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=βA 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)dtdxβ=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(xβa)+O((xβa)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^3dtdyβ=β2y+y2+y3
Find and classify all the equilibria.
- y=0y=0: unstable. y=1y=1 and y=-2y=β2: unstable.
- y=1y=1: stable. y=0y=0 and y=-2y=β2: unstable.
- y=0y=0: stable. y=-1y=β1 and y=2y=2: unstable.
- y=0y=0: stable. y=1y=1 and y=-2y=β2: unstable.
- y=2y=2: unstable. y=0y=0 and y=-1y=β1: stable.
- y=0y=0, y=1y=1, and y=-2y=β2: 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 β \kappa v ,mdtdvβ=mgβΞΊv,
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/\kappavββ=mg/ΞΊ?
Very good. Now, letβs use a more realistic model of drag that is quadratic as opposed to linear:
m\frac{dv}{dt} = mg β \lambda v^2 ,mdtdvβ=mgβΞ»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}}vββ=Ξ»mgββ is an unstable equilibrium; there is no terminal velocity.
- \displaystyle v_\infty = 0vββ=0 is an unstable equilibrium; there is no terminal velocity.
- \displaystyle v_\infty = 0vββ=0 is an unstable equilibrium; this is the terminal velocity.
- \displaystyle v_\infty = \frac{mg}{\lambda}vββ=Ξ»mgβ is an unstable equilibrium; there is no terminal velocity.
- \displaystyle v_\infty = \sqrt{\frac{mg}{\lambda}}vββ=Ξ»mgββ is a stable equilibrium; this is the terminal velocity.
- \displaystyle v_\infty = \frac{mg}{\lambda}vββ=Ξ»mgβ 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)I0β=I(0) is I(t)=I_0e^{rt}I(t)=I0βert. 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 β SdtdIβ=rIβS
Your goals are as follow. You have an initial investment I_0I0β, and you cannot change it or the rate rr. You want to be able to spend as much as possible but you also donβt want to ever spend all your money. What amount of spending rate SS can you bear?
Hint: If youβre 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_0Sβ€I0β.
- Make sure S \le rSβ€r.
- Any S \gt 0S>0 will lead to eventual loss of all funds.
- Make sure S/r \le 1S/rβ€1.
- Make sure S \ge r I_0Sβ₯rI0β
- Make sure S \le r I_0Sβ€rI0β.
Practice Quiz 04
Q1. In our lesson, we looked at two oscillators with βsinusoidalβ 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})dtdΞΈ1ββ=2+Ο΅(eΞΈ1ββΞΈ2ββ1);dtdΞΈ2ββ=2+Ο΅(1βeΞΈ1ββΞΈ2β)
Consider the phase difference \varphi = \theta_2 β \theta_1Ο=ΞΈ2ββΞΈ1β. 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 youβre not sure how to start, compute \displaystyle \frac{d\varphi}{dt}dtdΟβ. Then linearize this about \varphi=0Ο=0.
- \displaystyle \frac{d\varphi}{dt}=\epsilon\varphidtdΟβ=Ο΅Ο
- \displaystyle \frac{d\varphi}{dt}=-\epsilon\varphidtdΟβ=βΟ΅Ο
- \displaystyle \frac{d\varphi}{dt}=2\epsilon\varphidtdΟβ=2Ο΅Ο
- \displaystyle \frac{d\varphi}{dt}=\epsilondtdΟβ=Ο΅
- \displaystyle \frac{d\varphi}{dt}=2\varphidtdΟβ=2Ο
- \displaystyle \frac{d\varphi}{dt}=-2\epsilon\varphidtdΟβ=β2Ο΅Ο
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. Letβs model each arm as a simple oscillator obeying
\frac{d\theta_i}{dt} = adtdΞΈiββ=a
where \theta_1ΞΈ1β and \theta_2ΞΈ2β 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 β \epsilon F(\theta_2-\theta_1)dtdΞΈ1ββ=a+Ο΅F(ΞΈ2ββΞΈ1β),dtdΞΈ2ββ=aβΟ΅F(ΞΈ2ββΞΈ1β)
Hint: I suggest you first derive the equation for phase difference \varphi=\theta_2-\theta_1Ο=ΞΈ2ββΞΈ1β 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)=βsin(Ο)
- F(\varphi)=\sin(\varphi)F(Ο)=sin(Ο)
- F(\varphi)=\tan(\varphi)F(Ο)=tan(Ο)
- F(\varphi)=-\cos(\varphi)F(Ο)=βcos(Ο)
- F(\varphi)=\cos(\varphi)F(Ο)=cos(Ο)
Q3. For the brave: The following model of a biochemical βswitchβ 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 βonβ or βoffβ 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 β bG β \frac{G}{1+G^2}dtdGβ=aβbGβ1+G2Gβ
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 β Integration Coursera Quiz Answers
Main Quiz 01
Q1. \displaystyle \int 3\cos x \, dx =β«3cosxdx=
- -3\cos x + Cβ3cosx+C
- \sin 3x + Csin3x+C
- 3\sin x + C3sinx+C
- \cos 3x + Ccos3x+C
- -3\sin x + Cβ3sinx+C
- 3\cos x + C3cosx+C
Q2. \displaystyle \int x \sec^2 x^2 \, dx =β«xsec2x2dx=
- \displaystyle \frac{1}{2}x \sec^2 x^2 + C21βxsec2x2+C
- \displaystyle \frac{1}{2} \tan x^2 + C21βtanx2+C
- x^2 \tan x^2 + Cx2tanx2+C
- 2 \sec x^2 + C2secx2+C
- \displaystyle \frac{1}{2}x \tan^2 x^2 + C21βxtan2x2+C
- \tan x^2 + Ctanx2+C
Q3. \displaystyle \int \frac{4x}{(x^2 β 1)^3} \, dx =β«(x2β1)34xβdx=
- \displaystyle \frac{4}{3(x^2 β 1)^2} + C3(x2β1)24β+C
- \displaystyle \frac{1}{(x^2 β 1)^2} + C(x2β1)21β+C
- \displaystyle -\frac{1}{(x^2 β 1)^2} + Cβ(x2β1)21β+C
- \displaystyle -\frac{2x}{(x^2 β 1)^2} + Cβ(x2β1)22xβ+C
- \displaystyle \frac{2}{3(x^2 β 1)^2} + C3(x2β1)22β+C
- \displaystyle -\frac{2}{3(x^2 β 1)^2} + Cβ3(x2β1)22β+C
Q4. \displaystyle \int \frac{e^{\sqrt{x}}}{\sqrt{x}} \, dx =β«xβexββdx=
- e^{\sqrt{x} / 2} + Cexβ/2+C
- 2\sqrt{x} \, e^{\sqrt{x}} + C2xβexβ+C
- \displaystyle \frac{1}{2}\sqrt{x} \, e^{\sqrt{x}} + C21βxβexβ+C
- e^{2\sqrt{x}} + Ce2xβ+C
- \displaystyle e^{\sqrt{x}} + Cexβ+C
- 2e^{\sqrt{x}} + C2exβ+C
Q5. \displaystyle \int \frac{ \ln(15x^5)}{x} \, dx =β«xln(15x5)βdx=
- \displaystyle \frac{5}{2} \ln(15x) + C25βln(15x)+C
- \displaystyle \frac{5}{2}x^6 +C25βx6+C
- \displaystyle \frac{1}{60} \ln^2(15x^5) + C601βln2(15x5)+C
- 10\ln^2(15x) + C10ln2(15x)+C
- \displaystyle \frac{1}{10} \ln^2(15x^5) + C101βln2(15x5)+C
- \displaystyle \frac{1}{10} \ln(15x) + C101βln(15x)+C
Q6. \displaystyle \int \frac{x\, dx}{\sqrt{x+3}} =β«x+3βxdxβ=
- \displaystyle \frac{2}{3} (x+3)^{3/2} + C32β(x+3)3/2+C
- \displaystyle \frac{1}{3} (x-3) \sqrt{x+3} + C31β(xβ3)x+3β+C
- \displaystyle \frac{1}{3} \left[ (x+3)^{3/2} β 2(x+3)^{1/2} \right] + C31β[(x+3)3/2β2(x+3)1/2]+C
- \displaystyle \frac{2}{3}(x-6)\sqrt{x+3} + C32β(xβ6)x+3β+C
- \displaystyle \frac{1}{3} (x+6)\sqrt{x+3} + C31β(x+6)x+3β+C
- \displaystyle \frac{2}{3} \left[ 2(x+3)^{3/2} + (x+3)^{1/2} \right] + C32β[2(x+3)3/2+(x+3)1/2]+C
Practice Quiz 01
Q1. Apply the substitution \displaystyle u = \frac{1}{x}u=x1β to calculate the integral
I(x) = \int \frac{dx}{x\sqrt{x^2-1}}I(x)=β«xx2β1βdxβ
assuming x > 0x>0.
- \displaystyle I(x) = \arcsin \frac{1}{x} + CI(x)=arcsinx1β+C
- \displaystyle I(x) = 2\sqrt{x^2-1} + CI(x)=2x2β1β+C
- \displaystyle I(x) = -\arctan \frac{1}{x} + CI(x)=βarctanx1β+C
- \displaystyle I(x) = \arctan \frac{1}{x} + CI(x)=arctanx1β+C
- \displaystyle I(x) = -\arcsin \frac{1}{x} + CI(x)=βarcsinx1β+C
- \displaystyle I(x) = -2\sqrt{x^2-1} + CI(x)=β2x2β1β+C
Q2. Apply the substitution u = \sqrt{x^2-1}u=x2β1β to calculate the integral
I(x) = \int \frac{dx}{x\sqrt{x^2-1}}I(x)=β«xx2β1βdxβ
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)=arcsinx2β1β1β+C
- I(x) = 2\sqrt{x^2-1} + CI(x)=2x2β1β+C
- I(x) = -\arctan\sqrt{x^2-1} + CI(x)=βarctanx2β1β+C
- I(x) = \arctan\sqrt{x^2-1} + CI(x)=arctanx2β1β+C
- I(x) = -2\sqrt{x^2-1} + CI(x)=β2x2β1β+C
- \displaystyle I(x) = -\arcsin\frac{1}{\sqrt{x^2-1}} + CI(x)=βarcsinx2β1β1β+C
Main Quiz 02
Q1. \displaystyle \int x e^{x/2} \, dx =β«xex/2dx=
- \displaystyle \frac{1}{4} x^2 e^{x/2} + C41βx2ex/2+C
- 2e^{x/2} (x-1) + C2ex/2(xβ1)+C
- \displaystyle \frac{1}{2} e^{x/2} (x-2) + C21βex/2(xβ2)+C
- \displaystyle \frac{1}{4} e^{x/2} (2x-1) + C41βex/2(2xβ1)+C
- 2e^{x/2} (x-2) + C2ex/2(xβ2)+C
- e^{x^2/4} + Cex2/4+C
Q2. \displaystyle \int x^2e^{x/2} \, dx =β«x2ex/2dx=
- e^{x/2} (x^2 β 2x + 4) + Cex/2(x2β2x+4)+C
- e^{x/2} (x^2 β 4x β 8) + Cex/2(x2β4xβ8)+C
- 2e^{x/2} (x β 2) + C2ex/2(xβ2)+C
- 2e^{x/2} (x^2 β 4x β 8) + C2ex/2(x2β4xβ8)+C
- 2e^{x/2} (x^2 β 4x + 8) + C2ex/2(x2β4x+8)+C
- \displaystyle e^{x/2} \left( \frac{1}{2}x^2 β \frac{1}{8}x + \frac{1}{16} \right) + Cex/2(21βx2β81βx+161β)+C
Q3. \displaystyle \int 3x \ln x \, dx =β«3xlnxdx=
- \displaystyle \frac{3x^2}{2} \left( \ln x β 1 \right) + C23x2β(lnxβ1)+C
- \displaystyle \frac{x^2}{4} \left( 2\ln x β 1 \right) + C4x2β(2lnxβ1)+C
- \displaystyle \frac{3x^2}{4} \left( \ln x β 1 \right) + C43x2β(lnxβ1)+C
- \displaystyle \frac{x^2}{2} \left( 3 \ln x β x \right) + C2x2β(3lnxβx)+C
- \displaystyle \frac{3x}{2} \left( 4 \ln x β 1 \right) + C23xβ(4lnxβ1)+C
- \displaystyle \frac{3x^2}{4} \left( 2\ln x β 1 \right) + C43x2β(2lnxβ1)+C
Q4. \displaystyle \int 3x^2 \ln x \, dx =β«3x2lnxdx=
- \displaystyle x^3 \left( \ln x β \frac{1}{3} \right) + Cx3(lnxβ31β)+C
- \displaystyle \frac{1}{3}x^3( \ln x β 1) + C31βx3(lnxβ1)+C
- \displaystyle x ( \ln x β 1 ) + Cx(lnxβ1)+C
- \displaystyle x^2\left( \ln x β \frac{1}{2} \right) + Cx2(lnxβ21β)+C
- \displaystyle 3x^2\ln x + \frac{1}{4}x^4 + C3x2lnx+41βx4+C
- \displaystyle 3x^2\ln x β \frac{1}{4}x^4 + C3x2lnxβ41βx4+C
Q5. \displaystyle \int x^2 \cos\frac{x}{2} \, dx =β«x2cos2xβdx=
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} β 8x\cos\frac{x}{2} + C(2x2+16)sin2xββ8xcos2xβ+C
- \displaystyle (16-2x^2) \sin\frac{x}{2} + 8x\cos\frac{x}{2} + C(16β2x2)sin2xβ+8xcos2xβ+C
- \displaystyle (2x^2-16) \sin\frac{x}{2} β 8x\cos\frac{x}{2} + C(2x2β16)sin2xββ8xcos2xβ+C
- \displaystyle (2x^2+16) \sin\frac{x}{2} + 8x\cos\frac{x}{2} + C(2x2+16)sin2xβ+8xcos2xβ+C
- \displaystyle (16-2x^2) \sin\frac{x}{2} β 8x\cos\frac{x}{2} + C(16β2x2)sin2xββ8xcos2xβ+C
- \displaystyle (2x^2-16) \sin\frac{x}{2} + 8x\cos\frac{x}{2} + C(2x2β16)sin2xβ+8xcos2xβ+C
Q6. \displaystyle \int e^{2x}\sin 3x \, dx =β«e2xsin3xdx=
- \displaystyle \frac{2}{3}e^{2x}\sin 3x + \frac{1}{3}e^{2x}\cos 3x + C32βe2xsin3x+31βe2xcos3x+C
- \displaystyle \frac{2}{3}e^{2x}\cos 3x + \frac{3}{13}e^{2x}\sin 3x + C32βe2xcos3x+133βe2xsin3x+C
- \displaystyle \frac{2}{13}e^{2x}\sin 3x β \frac{3}{13}e^{2x}\cos 3x + C132βe2xsin3xβ133βe2xcos3x+C
- \displaystyle \frac{1}{2}e^{2x}\sin 3x β \frac{3}{4}e^{2x}\cos 3x + C21βe2xsin3xβ43βe2xcos3x+C
- \displaystyle \frac{2}{3}e^{2x}\sin 3x β \frac{4}{3}e^{2x}\cos 3x + C32βe2xsin3xβ34βe2xcos3x+C
- \displaystyle \frac{2}{13}e^{2x}\cos 3x β \frac{3}{13}e^{2x}\sin 3x + C132βe2xcos3xβ133βe2xsin3x+C
Q7. Apply integration by parts with u = \ln xu=lnx and dv = dxdv=dx to find an antiderivative for the logarithm βthat is, to solve the following integral:
I(x) = \int \ln x \, dxI(x)=β«lnxdx
- I(x) = x \ln x + CI(x)=xlnx+C
- I(x) = x \left( \ln x β 1 \right) + CI(x)=x(lnxβ1)+C
- \displaystyle I(x) = \frac{\ln^2 x}{2} + CI(x)=2ln2xβ+C
- \displaystyle I(x) = \frac{\ln^2 x}{x^2} + CI(x)=x2ln2xβ+C
- I(x) = x \left( \ln x β x \right) + CI(x)=x(lnxβx)+C
- \displaystyle I(x) = \frac{1}{x} + CI(x)=x1β+C
Practice Quiz 02
Q1. \displaystyle \int \ln^2 x \, dx =β«ln2xdx=
- \displaystyle \frac{1}{2}\ln^2 x β x\ln x + C21βln2xβxlnx+C
- \ln^2 x + x\ln x β 2x + Cln2x+xlnxβ2x+C
- \ln^2 x β 2x + Cln2xβ2x+C
- x \ln^2 x β \ln x + 2x + Cxln2xβlnx+2x+C
- x \left( \ln^2 x β 2\ln x + 2 \right) + Cx(ln2xβ2lnx+2)+C
- x \left( \ln^2 x + \ln x β 2 \right) + Cx(ln2x+lnxβ2)+C
Q2. \displaystyle \int \sin(\ln x) \, dx =β«sin(lnx)dx=
- \displaystyle \frac{x}{2} \left[ \sin(\ln x) + \cos(\ln x)\right] + C2xβ[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) β \cos(\ln x)\right] + Cx[sin(lnx)βcos(lnx)]+C
- \displaystyle \frac{x}{2} \left[ \sin(\ln x) β \cos(\ln x)\right] + C2xβ[sin(lnx)βcos(lnx)]+C
- 2x \cos(\ln x) + C2xcos(lnx)+C
Q3. \displaystyle \int \arcsin 2x \, dx =β«arcsin2xdx=
- \displaystyle x\arcsin 2x + \frac{1}{2} \sqrt{1 β x^2} + Cxarcsin2x+21β1βx2β+C
- \displaystyle x\arcsin 2x + \frac{1}{2} \sqrt{1 β 4x^2} + Cxarcsin2x+21β1β4x2β+C
- \displaystyle \arcsin 2x + \frac{1}{4} \sqrt{1 β 2x^2} + Carcsin2x+41β1β2x2β+C
- 2x\arcsin 2x + 8 \sqrt{1 β x^2} + C2xarcsin2x+81βx2β+C
- x\arcsin 2x + 2 \sqrt{1 β x^2} + Cxarcsin2x+21βx2β+C
- \arcsin 2x + 2 \sqrt{1 β 4x^2} + Carcsin2x+21β4x2β+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} \, dxβ«x51βx2βdx ?
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\thetaβ«sin2ΞΈcos5ΞΈdΞΈ
- \displaystyle β \int \sin^2\theta \cos^5 \theta \, d\thetaββ«sin2ΞΈcos5ΞΈdΞΈ
- \displaystyle β \int \sin^5\theta \cos\theta \, d\thetaββ«sin5ΞΈcosΞΈdΞΈ
- \displaystyle \int \sin^5\theta \cos^2\theta \, d\thetaβ«sin5ΞΈcos2ΞΈdΞΈ
- \displaystyle β \int \sin^5\theta \cos^2\theta \, d\thetaββ«sin5ΞΈcos2ΞΈdΞΈ
- \displaystyle \int \sin^5\theta \cos\theta \, d\thetaβ«sin5ΞΈcosΞΈdΞΈ
Q2. \displaystyle \int \frac{x^2}{\sqrt{4-x^2}} \, dx =β«4βx2βx2βdx=
- \displaystyle -2 \arccos\frac{x}{2} β \frac{1}{2}x \sqrt{4-x^2} + Cβ2arccos2xββ21βx4βx2β+C
- \displaystyle 2 \arcsin\frac{x}{2} β \frac{1}{2}x \sqrt{4-x^2} + C2arcsin2xββ21βx4βx2β+C
- \displaystyle x β \frac{1}{4}x \sqrt{4 β x^2} + Cxβ41βx4βx2β+C
- \displaystyle -\frac{1}{2} \arccos\frac{x}{2} β \frac{1}{8}x \sqrt{4-x^2} + Cβ21βarccos2xββ81βx4βx2β+C
- \displaystyle \frac{x}{2} β \frac{1}{4} \cos 2x + C2xββ41βcos2x+C
- \displaystyle \frac{1}{2} \arcsin\frac{x}{2} β \frac{1}{8}x \sqrt{4-x^2} + C21βarcsin2xββ81βx4βx2β+C
Q3. In Lecture we saw that we can use the substitution \displaystyle x = \frac{b}{a}\sin\thetax=abβsinΞΈ to (hopefully) calculate integrals involving \sqrt{b^2 β a^2x^2}b2βa2x2β. Another equally suitable subsitution in this case is \displaystyle x = \frac{b}{a}\cos\thetax=abβcosΞΈ. Use the latter to compute
\displaystyle \int \frac {\sqrt{1-x^2}} {x^2} \, dx =β«x21βx2ββdx=
- \displaystyle -\frac{ \sqrt{1-x^2}}{2x^2} β \frac{1}{2} \arccos x + Cβ2x21βx2βββ21βarccosx+C
- \displaystyle -\frac{ \sqrt{1-x^2} }{x} + \arccos x + Cβx1βx2ββ+arccosx+C
- \displaystyle -\frac{ \sqrt{1-x^2} }{x} β 2\arccos x + Cβx1βx2βββ2arccosx+C
- \displaystyle \frac{ \sqrt{1-x^2} }{x^2} + \arccos x + Cx21βx2ββ+arccosx+C
- x \sqrt{1-x^2} β \arccos x + Cx1βx2ββarccosx+C
- \displaystyle \frac{ \sqrt{1-x^2} }{x^2} + 2\arccos x + Cx21βx2ββ+2arccosx+C
Q4. \displaystyle \int (1-x^2)^{-3/2} \, dx =β«(1βx2)β3/2dx=
- \displaystyle \arcsin x + \frac{x}{\sqrt{1-x^2}} + Carcsinx+1βx2βxβ+C
- \displaystyle \frac{x}{\sqrt{1-x^2}} + C1βx2βxβ+C
- \displaystyle -2\sqrt{1-x^2} + Cβ21βx2β+C
- \displaystyle \frac{1}{2}\arcsin x + \sqrt{1-x^2} + C21βarcsinx+1βx2β+C
- \displaystyle \arccos x + \frac{1}{\sqrt{1-x^2}} + Carccosx+1βx2β1β+C
- \displaystyle \frac{1}{\sqrt{1-x^2}} + C1βx2β1β+C
Q5. \displaystyle \int \frac{x}{\sqrt{1+x^2}} \, dx =β«1+x2βxβdx=
- \displaystyle \ln |x + 1| + \frac{1}{ \sqrt{1 + x^2}} + Clnβ£x+1β£+1+x2β1β+C
- \displaystyle \frac{x}{1 + x^2} + C1+x2xβ+C
- \sqrt{1+x^2} + C1+x2β+C
- \displaystyle \frac{1}{\sqrt{1+x^2}} + C1+x2β1β+C
- \ln \big| x + \sqrt{1+x^2} \big| + Clnβ£β£β£βx+1+x2ββ£β£β£β+C
- \displaystyle \frac{1}{2} \left( \mathrm{arcsinh}\, x β x \sqrt{1+x^2}\right) + C21β(arcsinhxβx1+x2β)+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 β 4}}β«x2β4βx4dxβ to an integral involving trigonometric functions βwhich 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\thetaβ«210sec5ΞΈdΞΈ
- For a = 2a=2, we obtain the integral \displaystyle \int 2^4 \sec^5\theta \, d\thetaβ«24sec5ΞΈdΞΈ
- For a = 2a=2, we obtain the integral \displaystyle \int 2^5 \sec^5\theta \, d\thetaβ«25sec5ΞΈdΞΈ
- For a = 4a=4, we obtain the integral \displaystyle \int 2^8 \sec^4\theta \, d\thetaβ«28sec4ΞΈdΞΈ
- For a = 2a=2, we obtain the integral \displaystyle \int 2^4 \sec^4\theta \tan\theta \, d\thetaβ«24sec4ΞΈtanΞΈdΞΈ
- For a = 4a=4, we obtain the integral \displaystyle \int 2^8 \sec^5\theta \, d\thetaβ«28sec5ΞΈdΞΈ
Practice Quiz 03
Q1. In Homework 21 (challenge), we use the substitutions \displaystyle u = \frac{1}{x}u=x1β and u = \sqrt{x^2-1}u=x2β1β to compute the integral
I(x) = \int \frac{dx}{x\sqrt{x^2-1}}I(x)=β«xx2β1βdxβ
Another possible substitution is x = \sec\thetax=secΞΈ. Using it, compute I(x)I(x).
- \displaystyle \frac{\sqrt{x^2-1}}{2x^2} + C2x2x2β1ββ+C
- \displaystyle \frac{1}{2}\sqrt{x^2-1} β \mathrm{arcsec}\, x + C21βx2β1ββarcsecx+C
- x \arccos \sqrt{x^2-1} + Cxarccosx2β1β+C
- \displaystyle \sqrt{x^2-1}\arccos x + Cx2β1βarccosx+C
- \displaystyle \mathrm{arcsec}\, \frac{1}{x} + Carcsecx1β+C
- \displaystyle \arccos \frac{1}{x} + Carccosx1β+C
Q2. \displaystyle \int \frac{dx}{\sqrt{x^2-6x+10}} =β«x2β6x+10βdxβ=
(there may be multiple correct answers)
- \mathrm{arcsinh}\, (x-3) + Carcsinh(xβ3)+C
- \displaystyle \frac{1}{2} \mathrm{arcsinh}\,(x-3) + \frac{1}{4} x \sqrt{x^2-6x+10} + C21βarcsinh(xβ3)+41βxx2β6x+10β+C
- \displaystyle \frac{\sqrt{x^2-6x+10}}{2x} + C2xx2β6x+10ββ+C
- \mathrm{arcsinh}\,\sqrt{x^2-6x+10} + Carcsinhx2β6x+10β+C
- \ln \big| x β 3 + \sqrt{x^2 β 6x + 10}\big| + Clnβ£β£β£βxβ3+x2β6x+10ββ£β£β£β+C
- \displaystyle \frac{1}{2}\mathrm{arccosh}\, (x-3) β \frac{\sqrt{x^2-6x+10}}{4x} + C21βarccosh(xβ3)β4xx2β6x+10ββ+C
Q3. \displaystyle \int \frac{dx}{\sqrt{x^2-2x-8}} =β«x2β2xβ8βdxβ=
(there may be multiple correct answers)
- \displaystyle \frac{1}{2}\sqrt{x^2-2x-8} + \mathrm{arccosh}\frac{x-1}{3} + C21βx2β2xβ8β+arccosh3xβ1β+C
- \displaystyle -\sqrt{x^2-2x-8} + Cβx2β2xβ8β+C
- \displaystyle \ln \big| x-1 + \sqrt{x^2-2x-8} \big| + Clnβ£β£β£βxβ1+x2β2xβ8ββ£β£β£β+C
- \displaystyle \mathrm{arccosh}\frac{x-1}{3} + Carccosh3xβ1β+C
- \displaystyle \frac{1}{3}\ln \left| x β 1 + \sqrt{x^2-2x-8} \right| + C31βlnβ£β£β£β£βxβ1+x2β2xβ8ββ£β£β£β£β+C
- \displaystyle \frac{1}{3}\sqrt{x^2-2x-8} β \mathrm{arccosh}\frac{x-1}{3} + C31βx2β2xβ8ββarccosh3xβ1β+C
Main Quiz 04
Q1. \displaystyle \int \frac{5+x}{x^2+x-6} \, dx =β«x2+xβ65+xβdx=
- \displaystyle \frac{1}{5} \ln \left| \frac{x +3}{x β 2} \right| + C51βlnβ£β£β£β£β£βxβ2x+3ββ£β£β£β£β£β+C
- \ln |x β 2| β \ln |x+3| + Clnβ£xβ2β£βlnβ£x+3β£+C
- \displaystyle \frac{7}{5} \ln \left| \frac{x β 2}{x+3} \right| + C57βlnβ£β£β£β£β£βx+3xβ2ββ£β£β£β£β£β+C
- \displaystyle \ln |x^2 β x + 6| + \frac{11}{\sqrt{23}} \arctan\frac{2x β 1}{\sqrt{23}} + Clnβ£x2βx+6β£+23β11βarctan23β2xβ1β+C
- \displaystyle \frac{2}{5} \ln |x β 2| β \frac{7}{5} \ln |x+3| + C52βlnβ£xβ2β£β57βlnβ£x+3β£+C
- \displaystyle \frac{7}{5} \ln |x β 2| β \frac{2}{5} \ln |x+3| + C57βlnβ£xβ2β£β52βlnβ£x+3β£+C
Q2. \displaystyle \int \frac{2x + 3}{6x^2 + 5x + 1} \, dx =β«6x2+5x+12x+3βdx=
- \displaystyle \frac{7}{3}\ln |3x + 1| β \frac{2}{3}\ln |2x + 1| + C37βlnβ£3x+1β£β32βlnβ£2x+1β£+C
- \displaystyle \frac{7}{3}\ln |2x + 1| β 4\ln |3x + 1| + C37βlnβ£2x+1β£β4lnβ£3x+1β£+C
- \displaystyle 7\ln |3x + 1| β \frac{2}{3}\ln |2x + 1| + C7lnβ£3x+1β£β32βlnβ£2x+1β£+C
- \displaystyle \frac{7}{3}\ln |3x + 1| β 2\ln |2x + 1| + C37βlnβ£3x+1β£β2lnβ£2x+1β£+C
- \displaystyle \frac{7}{3}\ln |2x + 1| β 2\ln |3x + 1| + C37βlnβ£2x+1β£β2lnβ£3x+1β£+C
- 7\ln |3x + 1| β 2\ln |2x + 1| + C7lnβ£3x+1β£β2lnβ£2x+1β£+C
Q3. \displaystyle \int \frac{x^2-x+5}{(x-2)(x-1)(x+3)} \, dx =β«(xβ2)(xβ1)(x+3)x2βx+5βdx=
- \displaystyle \frac{2}{5} \ln |x β 2| β \frac{3}{4} \ln |x-1| + C52βlnβ£xβ2β£β43βlnβ£xβ1β£+C
- \displaystyle -\frac{1}{4} \ln \left| \frac{x^2 + x β 6}{x-1} \right| + Cβ41βlnβ£β£β£β£β£βxβ1x2+xβ6ββ£β£β£β£β£β+C
- \displaystyle \frac{7}{5} \ln |x β 2| β \ln |x β 1| + \frac{1}{20} \ln |x + 3| + C57βlnβ£xβ2β£βlnβ£xβ1β£+201βlnβ£x+3β£+C
- \displaystyle \frac{7}{5} \ln |x β 2| β \frac{5}{4} \ln |x β 1| + \frac{17}{20} \ln |x + 3| + C57βlnβ£xβ2β£β45βlnβ£xβ1β£+2017βlnβ£x+3β£+C
- \displaystyle \frac{1}{4} \ln \left| \frac{x^2 + x β 6}{x-1} \right| + C41βlnβ£β£β£β£β£βxβ1x2+xβ6ββ£β£β£β£β£β+C
- \displaystyle \frac{37}{43} \ln |(x β 2)(x β 1)(x + 3)| + C4337βlnβ£(xβ2)(xβ1)(x+3)β£+C
Q4. \displaystyle \int \frac{2x-1}{x^3-x} \, dx =β«x3βx2xβ1βdx=
- \displaystyle \ln \left| \frac{x(x-1)}{x+1} \right| + Clnβ£β£β£β£β£βx+1x(xβ1)ββ£β£β£β£β£β+C
- \displaystyle \ln |x| β \frac{1}{2} \ln |x-2| β \frac{1}{2} \ln |x+1| + Clnβ£xβ£β21βlnβ£xβ2β£β21βlnβ£x+1β£+C
- \displaystyle \ln \left| \frac{x^2(x-1)}{(x+1)^3} \right| + Clnβ£β£β£β£β£β(x+1)3x2(xβ1)ββ£β£β£β£β£β+C
- \displaystyle \ln \left| \frac{x+1}{x(x-1)} \right| + Clnβ£β£β£β£β£βx(xβ1)x+1ββ£β£β£β£β£β+C
- \displaystyle \ln |x| + \frac{1}{2}\ln|x-1| β \frac{3}{2}\ln |x+1| + Clnβ£xβ£+21βlnβ£xβ1β£β23βlnβ£x+1β£+C
- \displaystyle \ln \left| \frac{(x+1)^3}{x^2(x-1)} \right| + Clnβ£β£β£β£β£βx2(xβ1)(x+1)3ββ£β£β£β£β£β+C
Q5. \displaystyle{\int} \frac{x^2-3}{x^2-4} \, dx =β«x2β4x2β3βdx=
Hint: start by performing long division of the numerator by the denominator.
- \displaystyle x+\frac{1}{4} \ln|x^2-4|+Cx+41βlnβ£x2β4β£+C
- \displaystyle x + \frac{1}{4}\ln\left|\frac{x-2}{x+2}\right| + Cx+41βlnβ£β£β£β£β£βx+2xβ2ββ£β£β£β£β£β+C
- \displaystyle \frac{3}{4} \ln\left|\frac{x+2}{x-2}\right|+C43βlnβ£β£β£β£β£βxβ2x+2ββ£β£β£β£β£β+C
- \displaystyle x-\frac{1}{x}-4x+Cxβx1ββ4x+C
- \displaystyle x-\frac{3}{4} \ln\left|\frac{x-2}{x+2}\right|+Cxβ43βlnβ£β£β£β£β£βx+2xβ2ββ£β£β£β£β£β+C
- \displaystyle x + \frac{1}{2}\ln|x-2| β \frac{1}{4}\ln|x+2| + Cx+21βlnβ£xβ2β£β41βlnβ£x+2β£+C
Q6. \displaystyle \int \frac{dx}{x^2 β 4x + 8} =β«x2β4x+8dxβ=
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}β«1+u2duβ.
- \displaystyle \frac{1}{4} \arctan \frac{x+2}{3} + C41βarctan3x+2β+C
- \displaystyle \frac{1}{2} \arctan \frac{x-2}{2} + C21βarctan2xβ2β+C
- \displaystyle \frac{1}{2} \arctan \left(1 + \frac{x}{2}\right) + C21βarctan(1+2xβ)+C
- \displaystyle \frac{1}{4} \arctan \left(1 + \frac{x}{2}\right) + C41βarctan(1+2xβ)+C
- \displaystyle \frac{1}{4} \arctan \frac{x-2}{2} + C41βarctan2xβ2β+C
- \displaystyle \frac{1}{2} \arctan \frac{x+2}{3} + C21βarctan3x+2β+C
Practice Quiz 04
Q1. \displaystyle \int \frac{x^3+10x^2+33x+36}{x^2+4x+3} \, dx =β«x2+4x+3x3+10x2+33x+36βdx=
- \ln |x + 1| β \ln |x+3| + Clnβ£x+1β£βlnβ£x+3β£+C
- \displaystyle \frac{8}{9} \ln |x + 1| + \frac{4}{9} \ln |x+3| + C98βlnβ£x+1β£+94βlnβ£x+3β£+C
- \displaystyle \frac{1}{2}x^2 + 6x + 6 \ln |x+1| + C21βx2+6x+6lnβ£x+1β£+C
- \displaystyle \frac{1}{2}x^2 + \ln |x^2 + 4x + 3| + C21βx2+lnβ£x2+4x+3β£+C
- x^2 + 6 \ln |x+1| + Cx2+6lnβ£x+1β£+C
- \displaystyle \frac{1}{2}x^2 + 6x + 3 \ln |x+1| + 3 \ln|x+3|+ C21βx2+6x+3lnβ£x+1β£+3lnβ£x+3β£+C
Q2. \displaystyle \int \frac{x+2}{(x-1)^2} \, dx =β«(xβ1)2x+2βdx=
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}(xβr)nP(x)β=xβrA1ββ+(xβr)2A2ββ+β―+(xβr)nAnββ
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)=A1β(xβr)nβ1+A2β(xβr)nβ2+β―+Anβ(β)
Notice that, in this case, the direct technique for finding the coefficients A_1, \ldots, A_nA1β,β¦,Anβ βsubstituting x=rx=r in (\astβ)β does not quite work: it only gives you the last coefficient, A_nAnβ. In order to find the other coefficients, A_1, \ldots, A_{n-1}A1β,β¦,Anβ1β, you can revert to equating the coefficients of each power of xx on both sides of the equation (\astβ).
- \displaystyle \ln |x-1| + \frac{3}{x-1} + Clnβ£xβ1β£+xβ13β+C
- \displaystyle \ln |x-1| β \frac{3}{x-1} + Clnβ£xβ1β£βxβ13β+C
- 3\ln|x+1| + C3lnβ£x+1β£+C
- \displaystyle \ln |x-1| + \frac{2}{x-1} + Clnβ£xβ1β£+xβ12β+C
- \ln|x+1| + 3\ln|x+1|^2 + Clnβ£x+1β£+3lnβ£x+1β£2+C
- \ln|x+1| β 3\ln|x+1|^2 + Clnβ£x+1β£β3lnβ£x+1β£2+C
Q3. Consider the integral
I(x) = \int \frac{dx}{x^4 β 6x^3 + 12x^2}I(x)=β«x4β6x3+12x2dxβ
Factoring the denominator will get you a repeated linear factor βwhich we have just seen how to deal withβ and an irreducible quadratic polynomial βwhich 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. Donβt worry! Start by plugging in x=0x=0 βthat 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)=36x1β[3βxarctan3βxβ3β+3]+C
- \displaystyle I(x) = -\frac{1}{36x} \left[ \sqrt{3} x \arctan\frac{x-3}{\sqrt{3}} +3\right] + CI(x)=β36x1β[3βxarctan3βxβ3β+3]+C
- \displaystyle I(x) = \frac{1}{24} \left[ \ln |x| β \frac{2}{x} β \frac{1}{2}\ln\left(x^2-6x+12\right) β \frac{1}{\sqrt{3}} \arctan\frac{x-3}{\sqrt{3}} \right] + CI(x)=241β[lnβ£xβ£βx2ββ21βln(x2β6x+12)β3β1βarctan3βxβ3β]+C
- \displaystyle I(x) = \frac{1}{24} \left[ \ln |x| β \frac{2}{x} β \frac{1}{2}\ln\left(x^2-6x+12\right) + \frac{1}{\sqrt{3}} \arctan\frac{x-3}{\sqrt{3}} \right] + CI(x)=241β[lnβ£xβ£βx2ββ21βln(x2β6x+12)+3β1βarctan3βxβ3β]+C
- \displaystyle I(x) = \frac{1}{48} \left[2 \ln |x| β \frac{1}{x} β \frac{1}{2}\ln\left(x^2-6x+12\right) β \frac{2}{3} \arctan{(x-3)} \right] + CI(x)=481β[2lnβ£xβ£βx1ββ21βln(x2β6x+12)β32βarctan(xβ3)]+C
- \displaystyle I(x) = \frac{1}{24} \left[ \ln |x| β \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)=241β[lnβ£xβ£βx2β+21βln(x2β6x+12)β3β1βarctan3βxβ3β]+C
Week 03:Calculus: Single Variable Part 3 β 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)_iβ«x=abβf(x)dx=P:Ξxβ0limβi=1βnβf(xiβ)(Ξx)iβ
Here PP is a partition of the interval [a, b][a,b] into nn intervals P_iPiβ, each of width (\Delta x)_i(Ξx)iβ. The point x_ixiβ is a sampling of P_iPiβ, that is, a point in the interval P_iPiβ. 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_iPiβ of common length (\Delta x)_i = (b-a)/n(Ξx)iβ=(bβa)/n. For the sampling, choose the right endpoint of each P_iPiβ; this gives you the formula
x_i = a + i \frac{b-a}{n}xiβ=a+inbβaβ
With these choices of partition and sampling, compute the Riemann sums for the integral
\int_{x=1}^2 \frac{dx}{x}β«x=12βxdxβ
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 0.693β«x=12βxdxβ=ln2β0.693
Compare this value to the ones you obtain from the Riemann sums.
- \displaystyle 1, \frac{3}{2} = 1.5, \frac{11}{6} \simeq 1.8331,23β=1.5,611ββ1.833
- \displaystyle \frac{1}{2} = 0.5, \frac{5}{6} \simeq 0.833, \frac{13}{12} \simeq 1.08321β=0.5,65ββ0.833,1213ββ1.083
- \displaystyle 1, \frac{1}{2} = 0.5, \frac{5}{6} \simeq 0.8331,21β=0.5,65ββ0.833
- \displaystyle \frac{1}{2} = 0.5, \frac{7}{12} \simeq 0.583, \frac{37}{60} \simeq 0.61721β=0.5,127ββ0.583,6037ββ0.617
- \displaystyle \frac{1}{2} = 0.5, \frac{5}{6} \simeq 0.833, \frac{5}{12} \simeq 0.41721β=0.5,65ββ0.833,125ββ0.417
- \displaystyle 1, \frac{5}{6} \simeq 0.833, \frac{47}{60} \simeq 0.7831,65ββ0.833,6047ββ0.783
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 \, dxβ«x=03βx2dx
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=1βnβi=2n(n+1)β,i=1βnβi2=6n(n+1)(2n+1)β,i=1βnβi3=4n2(n+1)2β
- \displaystyle 9 + \frac{9}{n} + \frac{9}{n^2}9+n9β+n29β
- \displaystyle 9 + \frac{9}{2n} + \frac{3}{2n^2}9+2n9β+2n23β
- \displaystyle 9 + \frac{2n}{9} + \frac{2n^2}{3}9+92nβ+32n2β
- \displaystyle 9 + \frac{27}{2n} + \frac{9}{2n^2}9+2n27β+2n29β
- \displaystyle 9 + \frac{2n}{27} + \frac{2n^2}{9}9+272nβ+92n2β
- 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=β«x=02βxdx=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) β IE(n)=RS(n)βI
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+β
- E(n)E(n) is in O(n^2)O(n2) as n \to +\inftynβ+β
- E(n)E(n) is in O(1/n)O(1/n) as n \to +\inftynβ+β
- E(n)E(n) is in O(1/n^2)O(1/n2) as n \to +\inftynβ+β
- E(n)E(n) is in O(n)O(n) as n \to +\inftynβ+β
- E(n)E(n) is in O(1)O(1) as n \to +\inftynβ+β
Q4. Evaluate \displaystyle \int_{x=-\pi/4}^{\pi/4} \left( x^2 + \ln|\cos x|\right) \sin \frac{x}{2} \, dxβ«x=βΟ/4Ο/4β(x2+lnβ£cosxβ£)sin2xβdx. 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 βcalled its even partβ and an odd function βits 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) β f(-x)}{2}feven(x)=2f(x)+f(βx)β,fodd(x)=2f(x)βf(βx)β
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) β f(x)}{2} = -f^{\mathrm{odd}}(x)feven(βx)=2f(βx)+f(x)β=feven(x),fodd(βx)=2f(βx)βf(x)β=βfodd(x)
and
f^{\mathrm{even}}(x) + f^{\mathrm{odd}}(x) = \frac{f(x) + f(-x)}{2} + \frac{f(x) β f(-x)}{2} = f(x)feven(x)+fodd(x)=2f(x)+f(βx)β+2f(x)βf(βx)β=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 =dxdββ«t=0arcsinxβlnβ£sint+costβ£dt=
- \displaystyle \ln \left| x + \sqrt{1-x^2} \right| \arcsin x lnβ£β£β£β£βx+1βx2ββ£β£β£β£βarcsinx
- \ln \left| x + \sqrt{1-x^2} \right|lnβ£β£β£βx+1βx2ββ£β£β£β
- \displaystyle \frac{1}{\sqrt{1-x^2}}\ln \left| x + \sqrt{1-x^2} \right|1βx2β1βlnβ£β£β£β£βx+1βx2ββ£β£β£β£β
- \ln \arcsin xlnarcsinx
- \arcsin\left( \ln \left| x + \sqrt{1-x^2} \right| \right)arcsin(lnβ£β£β£βx+1βx2ββ£β£β£β)
- 0
Q2. \displaystyle \frac{d}{dx} \int_{t=\sin x}^{\tan x} e^{-t^2} \, dt =dxdββ«t=sinxtanxβeβt2dt=
- \displaystyle \int_{2\sin x\, cos x}^{2 \tan x\, \sec^2 x} e^{-t^2} \, dtβ«2sinxcosx2tanxsec2xβeβt2dt
- e^{-4\tan^2 x \sec^4 x} β e^{-4\sin^2 x \cos^2 x}eβ4tan2xsec4xβeβ4sin2xcos2x
- 00
- e^{-\tan^2 x}\sec^2 x β e^{-\sin^2 x}\cos xeβtan2xsec2xβeβsin2xcosx
- 2e^{-\tan^2 x}\tan x \sec^2 x β 2e^{-\sin^2 x}\sin x \cos x2eβtan2xtanxsec2xβ2eβsin2xsinxcosx
- 2(\tan x \sec^2 x β \sin x \cos x)e^{-t^2}2(tanxsec2xβsinxcosx)eβt2
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)=β«0xβln(cosh(t))dt
Hint: yes, thereβs more than one way to do this problemβ¦ Try using the F.T.I.C. to compute the derivatives.
- \displaystyle \frac{x^2}{6}6x2β
- \displaystyle \frac{x^2}{2}2x2β
- \displaystyle -\frac{x^5}{60}β60x5β
- \displaystyle \frac{x^3}{2}2x3β
- \displaystyle \frac{x^3}{6}6x3β
- \displaystyle -\frac{x^3}{18}β18x3β
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=0β100βn3
- 2.5\times 10^72.5Γ107
- 2.5\times 10^82.5Γ108
- 6.6\times 10^76.6Γ107
- 1.0\times 10^71.0Γ107
- 3.3\times 10^73.3Γ107
- 1.0\times 10^81.0Γ108
Main Quiz 02
Q1. \displaystyle \int_{x=-1}^1 \frac{dx}{1+x^2} =β«x=β11β1+x2dxβ=
- \displaystyle \frac{\pi}{4}4Οβ
- \piΟ
- 2\pi2Ο
- \displaystyle \frac{\pi}{2}2Οβ
- 00
- \displaystyle \frac{\pi}{3}3Οβ
Q2. \displaystyle \int_{x=0}^{3} 5x \sqrt{x+1} \, dx =β«x=03β5xx+1βdx=
- \displaystyle \frac{116}{3}3116β
- 3030
- 12\sqrt{3}123β
- 00
- -12\sqrt{3}β123β
- \displaystyle -\frac{116}{3}β3116β
Q3. \displaystyle \int_{x=-\pi}^\pi \frac{d}{dx} ( x\cos x ) \, dx =β«x=βΟΟβdxdβ(xcosx)dx=
- x\cos x + Cxcosx+C
- 00
- 2\pi2Ο
- \cos x β x\sin x + Ccosxβxsinx+C
- x\cos xxcosx
- -2\piβ2Ο
Q4. \displaystyle \frac{d}{dx} \int_{x=-\pi}^\pi x\cos x \, dx =dxdββ«x=βΟΟβxcosxdx=
- \cos x β x\sin x + Ccosxβxsinx+C
- 2\pi2Ο
- x\cos x + Cxcosx+C
- x\cos xxcosx
- 00
- -2\piβ2Ο
Q5. \displaystyle \frac{d}{dx} \int_{t=0}^x \cos t \, dt =dxdββ«t=0xβcostdt=
- \sin xsinx
- \sin x β 1sinxβ1
- 1 β \sin x1βsinx
- 1 β \cos x1βcosx
- \cos x β 1cosxβ1
- \cos xcosx
Week 04:Calculus: Single Variable Part 3 β 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\thetaβ«ΞΈ=βΟ/4Ο/4βsecΞΈdΞΈ
- \displaystyle \int_{x=0}^1 \frac{dx}{\sqrt{x} β 10}β«x=01βxββ10dxβ
- \displaystyle \int_{x=0}^1 \frac{dx}{x-1}β«x=01βxβ1dxβ
- \displaystyle \int_{x=-1}^1 \ln(1+x^2) \, dxβ«x=β11βln(1+x2)dx
- \displaystyle \int_{t=0}^{+\infty} t^{x-1} e^{-t} \, dtβ«t=0+ββtxβ1eβtdt
- \displaystyle \int_{x=0}^1 \frac{dx}{x^2}β«x=01βx2dxβ
- \displaystyle \int_{x=1}^{2} \sqrt{2-x} \, dxβ«x=12β2βxβdx
- \displaystyle \int_{\theta=\pi/4}^{3\pi/4} \sec\theta \, d\thetaβ«ΞΈ=Ο/43Ο/4βsecΞΈdΞΈ
- \displaystyle \int_{x=-\infty}^{+\infty} e^{-x^2/2} \, dxβ«x=ββ+ββeβx2/2dx
- \displaystyle \int_{x=-1,000,000}^{1,000,000} e^{-x^2/2} \, dxβ«x=β1,000,0001,000,000βeβx2/2dx
Q2. Consider the integral
\int_{x=0}^1 \frac{e^{-x}}{x} \, dxβ«x=01βxeβxβdx
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)xeβxβ=1+O(x), so the integral converges.
- \displaystyle \frac{e^{-x}}{x} = 1 + O(x)xeβxβ=1+O(x), so the integral diverges.
- \displaystyle \frac{e^{-x}}{x} = e^{-x} + O(xe^{-x})xeβxβ=eβx+O(xeβx), so the integral converges.
- \displaystyle \frac{e^{-x}}{x} = \frac{1}{x} + O(1)xeβxβ=x1β+O(1), so the integral diverges.
- \displaystyle \frac{e^{-x}}{x} = \frac{1}{x} + O(1)xeβxβ=x1β+O(1), so the integral converges.
- \displaystyle \frac{e^{-x}}{x} = e^{-x} + O(xe^{-x})xeβxβ=eβx+O(xeβx), 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}} \, dxβ«x=01βxβcos2xβdx
- \displaystyle \frac{\cos^2 x}{\sqrt{x}} = 1 + O(x^4)xβcos2xβ=1+O(x4), so the integral diverges.
- \displaystyle \frac{\cos^2 x}{\sqrt{x}} = \sqrt{x} + O(x^{3/2})xβcos2xβ=xβ+O(x3/2), so the integral converges.
- \displaystyle \frac{\cos^2 x}{\sqrt{x}} = 1 + O(x^{4})xβcos2xβ=1+O(x4), so the integral converges.
- \displaystyle \frac{\cos^2 x}{\sqrt{x}} = \frac{1}{\sqrt{x}} + O(x^{3/2})xβcos2xβ=xβ1β+O(x3/2), so the integral converges.
- \displaystyle \frac{\cos^2 x}{\sqrt{x}} = \frac{1}{\sqrt{x}} + O(x^{3/2})xβcos2xβ=xβ1β+O(x3/2), so the integral diverges.
- \displaystyle \frac{\cos^2 x}{\sqrt{x}} = \sqrt{x} + O(x^{3/2})xβcos2xβ=xβ+O(x3/2), so the integral diverges.
Q4. Calculate
I = \int_{x=1}^2 \frac{dx}{\sqrt{x β 1}}I=β«x=12βxβ1βdxβ
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 β 1}}I=Tβ1+limββ«x=T2βxβ1βdxβ
- I = 1I=1
- \displaystyle I = \frac{\sqrt{2}}{2} β 1I=22βββ1
- I = -2I=β2
- The integral diverges.
- I = 2I=2
- I = \sqrt{2} β 1I=2ββ1
Q5. Calculate
I = \int_{x=0}^4 \frac{2 \, dx}{\sqrt{16 β x^2}}I=β«x=04β16βx2β2dxβ
- I = 0I=0
- \displaystyle I = -\frac{\pi}{2}I=β2Οβ
- I = -\piI=βΟ
- I = \piI=Ο
- \displaystyle I = \frac{\pi}{2}I=2Οβ
- The integral diverges
Q6. Determine the leading order term of the integrand as x \to +\inftyxβ+β to decide whether the following integral converges or diverges.
\int_{x=1}^{+\infty} \frac{\sqrt[3]{x + 3}}{x^3} \, dxβ«x=1+ββx33x+3ββdx
- \displaystyle \frac{\sqrt[3]{x + 3}}{x^3} = \frac{1}{x^3} + O\left(\frac{1}{x^4}\right)x33x+3ββ=x31β+O(x41β), 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+3ββ=x10/31β+O(x13/31β), 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+3ββ=x8/31β+O(x11/31β), so the integral diverges.
- \displaystyle \frac{\sqrt[3]{x + 3}}{x^3} = \frac{1}{x^3} + O\left(\frac{1}{x^4}\right)x33x+3ββ=x31β+O(x41β), 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+3ββ=x8/31β+O(x11/31β), 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+3ββ=x10/31β+O(x13/31β), so the integral converges.
Q7. Does the integral converge or diverge? \int_{x=1}^{+\infty} \frac{1-5^{-x}}{x} \, dxβ«x=1+ββx1β5βxβdx
- The leading order term in the integrand as x \to +\inftyxβ+β is 5^{-x}5βx, so the integral converges.
- The leading order term in the integrand as x \to +\inftyxβ+β is 11, so the integral diverges.
- The leading order term in the integrand as x \to +\inftyxβ+β is 5^{-x}5βx, so the integral diverges.
- The leading order term in the integrand as x \to +\inftyxβ+β is \displaystyle \frac{1}{x}x1β, so the integral converges.
- The leading order term in the integrand as x \to +\inftyxβ+β is 11, so the integral converges.
- The leading order term in the integrand as x \to +\inftyxβ+β is \displaystyle \frac{1}{x}x1β, so the integral diverges.
Q8. Calculate
I = \int_{x=0}^{+\infty} e^{-x} \sin x \, dxI=β«x=0+ββeβxsinxdx
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+βlimββ«x=0Tβeβxsinxdx
- I = 0I=0
- \displaystyle I = -\frac{1}{2}I=β21β
- I = -1I=β1
- \displaystyle I = \frac{1}{2}I=21β
- 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} } \, dxβ«x=1+ββx3β1β1βdx
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+β 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(xβ1)β1β+O(xβ1β).
- As x \to +\inftyxβ+β, the integrand is \displaystyle \frac{1}{x^{3/2}} + O\left(\frac{1}{x^{9/2}}\right)x3/21β+O(x9/21β).
- As x \to +\inftyxβ+β, the integrand is \displaystyle \frac{1}{x^{1/3}} + O(x^{2/3})x1/31β+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)(xβ1)3/21β+O((xβ1)1/21β).
Q2. Consider the following two integrals:
I_1 = \int_{x=2}^{+\infty} \frac{dx}{\sqrt{x^3 β 8}}, \qquad I_2 = \int_{x=2}^{+\infty} \frac{1}{\sqrt{(x-2)^3}} \, dxI1β=β«x=2+ββx3β8βdxβ,I2β=β«x=2+ββ(xβ2)3β1βdx
They look very much alike. In fact, both of their integrands blow up at x=2x=2, and the domain of integration βthe same one in both casesβ is unbounded. But which of the following statements is true?
- Both I_1I1β and I_2I2β diverge.
- Both I_1I1β and I_2I2β converge.
- I_2I2β converges, but I_1I1β diverges.
- I_1I1β converges, but I_2I2β 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 xxβ¦
Identify the leading order term as x \to +\inftyxβ+β of the integrand of
\int_{x=1}^{+\infty} \frac{1}{\sinh x} \, dxβ«x=1+ββsinhx1βdx
Use this information to decide whether it converges or diverges.
- \displaystyle \frac{1}{\sinh x} = \frac{e^{-x}}{x} + O(xe^{-x})sinhx1β=xeβxβ+O(xeβx), so the integral diverges.
- \displaystyle \frac{1}{\sinh x} = \frac{e^{-x}}{x} + O(xe^{-x})sinhx1β=xeβxβ+O(xeβx), so the integral converges.
- \displaystyle \frac{1}{\sinh x} = e^x + O(e^{3x})sinhx1β=ex+O(e3x), so the integral converges.
- \displaystyle \frac{1}{\sinh x} = e^x + O(e^{3x})sinhx1β=ex+O(e3x), so the integral diverges.
- \displaystyle \frac{1}{\sinh x} = 2e^{-x} + O(e^{-3x})sinhx1β=2eβx+O(eβ3x), so the integral converges.
- \displaystyle \frac{1}{\sinh x} = 2e^{-x} + O(e^{-3x})sinhx1β=2eβx+O(eβ3x), 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}Ipβ=β«x=1+ββlnpxdxβ Which of the following statements is true?
Two hints:
- Think about the growth of \ln^p xlnpx as x \to +\inftyxβ+β. Can you compare it to the growth of x^qxq for some value of qq ?
- Recall from Lecture 25 that if g(x) \geq f(x)g(x)β₯f(x) for every x \in [a,b]xβ[a,b], then\int_{x=a}^b g(x) \,dx \geq \int_{x=a}^b f(x) \,dxβ«x=abβg(x)dxβ₯β«x=abβf(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_pIpβ converges for any value of pp.
- I_pIpβ converges for p \gt 1p>1 and diverges for p \leq 1pβ€1.
- I_pIpβ converges for p \leq 1pβ€1 and diverges for p \gt 1p>1.
- I_pIpβ diverges for any value of pp.
- I_pIpβ 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}eβt2 is difficult. Letβs say I told you that
\int_{0}^\infty e^{-t^2}dt = \frac{\sqrt{\pi}}{2} .β«0ββeβt2dt=2Οββ.
Use this fact to compute \left(-\frac{1}{2}\right)!(β21β)! (that is, negative one-half, factorial).
Hint: use the \GammaΞ-function and a substitution.
- \displaystyle\frac{\pi}{2}2Οβ
- \pi^2Ο2
- \displaystyle\frac{\pi^2}{2}2Ο2β
- \piΟ
- \sqrt{\pi}Οβ
- \displaystyle\frac{\sqrt{\pi}}{2}2Οββ
Main Quiz 02
Q1. \displaystyle \int \sin ^2 x \cos ^2 x \, dx =β«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 \, dxβ«cosnxdx=ncosnβ1xsinxβ+nnβ1ββ«cosnβ2xdx
\int \sin^n x \, dx = -\frac{\sin^{n-1}x \cos x}{n} + \frac{n-1}{n} \int \sin^{n-2}x \, dxβ«sinnxdx=βnsinnβ1xcosxβ+nnβ1ββ«sinnβ2xdx
More than one answer may be correct.
- \displaystyle \frac{1}{8} x + \frac{1}{16} \sin 2x β \frac{1}{4} \sin x \cos^3 x + C81βx+161βsin2xβ41βsinxcos3x+C
- \displaystyle \frac{1}{8} x + \frac{1}{32} \cos 4x + C81βx+321βcos4x+C
- \displaystyle \frac{1}{8} x + \frac{1}{16} \cos 2x β \frac{1}{4} \sin x \cos^3 x + C81βx+161βcos2xβ41βsinxcos3x+C
- \displaystyle \frac{1}{8} x β \frac{1}{16} \cos 2x + \frac{1}{4} \sin^3 x \cos x + C81βxβ161βcos2x+41βsin3xcosx+C
- \displaystyle \frac{1}{8} x β \frac{1}{16} \sin 2x + \frac{1}{4} \sin^3 x \cos x + C81βxβ161βsin2x+41βsin3xcosx+C
- \displaystyle \frac{1}{8} x β \frac{1}{32} \sin 4x + C81βxβ321βsin4x+C
Q2. \displaystyle \int \sin ^3 \frac{x}{2} \cos ^3 \frac{x}{2} \, dx =β«sin32xβcos32xβdx=
(More than one answer may be correct.)
- \displaystyle -\frac{1}{3} \sin^6 \frac{x}{2} + \frac{1}{2} \sin^4 \frac{x}{2} + Cβ31βsin62xβ+21βsin42xβ+C
- \displaystyle \frac{1}{3} \cos^6 \frac{x}{2} -\frac{1}{2} \cos^4 \frac{x}{2} + C31βcos62xββ21βcos42xβ+C
- \displaystyle \frac{1}{24} \cos^3 x -\frac{1}{8} \cos x + C241βcos3xβ81βcosx+C
- \displaystyle \frac{1}{6} \sin^6 \frac{x}{2} β \frac{1}{4} \sin^4 \frac{x}{2} + C61βsin62xββ41βsin42xβ+C
- \displaystyle \frac{1}{96} \big( \cos x β 9\cos 3x \big) + C961β(cosxβ9cos3x)+C
- \displaystyle -2\sin^5 \frac{x}{2} + 2 \sin^3 \frac{x}{2} + Cβ2sin52xβ+2sin32xβ+C
Q3. \displaystyle \int \frac{x^3\, dx}{\sqrt{9-x^2}} =β«9βx2βx3dxβ=
Hint: use a trigonometric substitution to obtain a trigonometric integral.
- \displaystyle β \bigg( 6 + \frac{x^2}{3} \bigg) \sqrt{9-x^2} + Cβ(6+3x2β)9βx2β+C
- \displaystyle -3\sqrt{9-x^2} β \frac{1}{81} (9-x^2)^{3/2} + Cβ39βx2ββ811β(9βx2)3/2+C
- \displaystyle \frac{1}{6} (9-x^2)^{1/2} + C61β(9βx2)1/2+C
- 3 (9-x^2)^{3/2} + C3(9βx2)3/2+C
- \displaystyle \bigg( \frac{1}{3} + 9x^2 \bigg) \sqrt{9-x^2} + C(31β+9x2)9βx2β+C
- \displaystyle 9\sqrt{9-x^2} β \frac{1}{3} (9-x^2)^{3/2} + C99βx2ββ31β(9βx2)3/2+C
Q4. \displaystyle \int 5\tan ^5 x \sec ^3 x \, dx =β«5tan5xsec3xdx=
- \displaystyle\frac{5}{6} \tan^6 x + \frac{5}{8} \tan^8 x + C65βtan6x+85βtan8x+C
- \displaystyle\frac{5}{7} \sec^{14} x β 2\sec^{10} x + \frac{5}{3} \sec^6 x + C75βsec14xβ2sec10x+35βsec6x+C
- \displaystyle \sec^5 x β \frac{10}{3} \sec^3 x + 5\sec x + Csec5xβ310βsec3x+5secx+C
- \displaystyle\frac{4}{5} \sec^5 x \tan^5 x β \frac{4}{3} \sec^3 x \tan^3 x + C54βsec5xtan5xβ34βsec3xtan3x+C
- \displaystyle \frac{5}{7} \tan^7 x + \tan^5 x + C75βtan7x+tan5x+C
- \displaystyle \frac{5}{7} \sec^7 x β 2 \sec^5 x + \frac{5}{3} \sec^3 x + C75βsec7xβ2sec5x+35βsec3x+C
Q5. \displaystyle \int 7\tan ^4 x \sec ^4 x \, dx =β«7tan4xsec4xdx=
- \displaystyle \frac{1}{7} \tan^7 x + \frac{1}{5} \tan^5 x + C71βtan7x+51βtan5x+C
- \displaystyle \frac{7}{5} \tan^5 x + \frac{7}{3} \tan^3 x + C57βtan5x+37βtan3x+C
- \displaystyle \sec^{7} x + \frac{7}{5} \sec^{5} x + Csec7x+57βsec5x+C
- \displaystyle \tan^7 x + \frac{7}{5} \tan^5 x + Ctan7x+57βtan5x+C
- \displaystyle \frac{1}{2} \sec^{14} x + \frac{7}{10} \sec^{10} x + C21βsec14x+107βsec10x+C
- \displaystyle \tan^7 x β \frac{7}{5} \tan^5 x + Ctan7xβ57βtan5x+C
Q6. Assuming 0 \lt x \lt \pi0<x<Ο, compute
\displaystyle \int \cot x \sqrt{1 β \cos 2x} \, dx =β«cotx1βcos2xβdx=
Hint: remember the double-angle formula
\cos 2x = \cos^2 x β \sin^2 xcos2x=cos2xβsin2x
- \cos^2 x + Ccos2x+C
- \sqrt{2} \cos x + C2βcosx+C
- \displaystyle \frac{x}{2} + \frac{ \cos 2x}{2} + C2xβ+2cos2xβ+C
- \sqrt{2} \sin x + C2βsinx+C
- \cos x + Ccosx+C
- \sin x + Csinx+C
Main Quiz 03
Q1. \displaystyle \int \frac{(\arcsin x)^2}{\sqrt{1-x^2}}\, dx =β«1βx2β(arcsinx)2βdx=
- \displaystyle \frac{1}{3} \sin^3 x + C31βsin3x+C
- \displaystyle \frac{1}{3} \left( \arcsin\sqrt{1-x^2} \right)^3 + C31β(arcsin1βx2β)3+C
- \displaystyle \frac{1}{3} \sqrt{1-(\arcsin x)^3} + C31β1β(arcsinx)3β+C
- \displaystyle \frac{2}{3} \sqrt{1-\sin^3 x} + C32β1βsin3xβ+C
- \displaystyle \sqrt{\frac{1}{3}(\arcsin x)^3} + C31β(arcsinx)3β+C
- \displaystyle \frac{(\arcsin x)^3}{3\sqrt{1-x^2}} + C31βx2β(arcsinx)3β+C
- \displaystyle \frac{1}{3} (\arcsin x)^3 + C31β(arcsinx)3+C
- \displaystyle \frac{x^3}{3} + C3x3β+C
Q2. \displaystyle \int \frac{dx}{x^2-x-6} =β«x2βxβ6dxβ=
- \displaystyle \frac{1}{3} \ln\left| \frac{x-3}{x+2} \right| + C31βlnβ£β£β£β£β£βx+2xβ3ββ£β£β£β£β£β+C
- \displaystyle -\frac{1}{3} \ln\left| \frac{x-3}{x+2} \right| + Cβ31βlnβ£β£β£β£β£βx+2xβ3ββ£β£β£β£β£β+C
- \displaystyle -\frac{1}{3} \ln\left| \frac{x+3}{x-2} \right| + Cβ31βlnβ£β£β£β£β£βxβ2x+3ββ£β£β£β£β£β+C
- \displaystyle \frac{1}{5} \ln\left| \frac{x-3}{x+2} \right| + C51βlnβ£β£β£β£β£βx+2xβ3ββ£β£β£β£β£β+C
- \displaystyle -\frac{1}{5} \ln\left| \frac{x-3}{x+2} \right| + Cβ51βlnβ£β£β£β£β£βx+2xβ3ββ£β£β£β£β£β+C
- \displaystyle -\frac{1}{5} \ln\left| \frac{x+3}{x-2} \right| + Cβ51βlnβ£β£β£β£β£βxβ2x+3ββ£β£β£β£β£β+C
- \displaystyle \frac{1}{3} \ln\left| \frac{x+3}{x-2} \right| + C31βlnβ£β£β£β£β£βxβ2x+3ββ£β£β£β£β£β+C
- \displaystyle \frac{1}{5} \ln\left| \frac{x+3}{x-2} \right| + C51βlnβ£β£β£β£β£βxβ2x+3ββ£β£β£β£β£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}dxβ«x=01βx2cosxβdx
- \displaystyle \int_{x=0}^3 \frac{dx}{x-1}β«x=03βxβ1dxβ
- \displaystyle \int_{x=0}^1 \frac{\sin(\pi x)}{x^{3/2}} dxβ«x=01βx3/2sin(Οx)βdx
- \displaystyle \int_{x=1}^2 \frac{2}{\sqrt[3]{x-1}} dxβ«x=12β3xβ1β2βdx
- \displaystyle \int_{x=0}^{+\infty} \frac{2x + 3}{x^2 + 3x + 6}\, dxβ«x=0+ββx2+3x+62x+3βdx
- \displaystyle \int_{x=2}^{+\infty} \frac{e^{-x}}{x}dxβ«x=2+ββxeβxβdx
- \displaystyle \int_{x=2}^{+\infty} \frac{dx}{x^{1/2}(x-1)^{3/2}}β«x=2+ββx1/2(xβ1)3/2dxβ
- \displaystyle \int_{x=2}^{+\infty} \frac{x}{1+x^2}dxβ«x=2+ββ1+x2xβdx
Q4. \displaystyle \int_{x=0}^{\pi/2} x \sin 2x \,dx =β«x=0Ο/2βxsin2xdx=
- \displaystyle \frac{1}{4}41β
- \displaystyle -\frac{\pi}{4}β4Οβ
- \displaystyle -\frac{1}{4}β41β
- \displaystyle \frac{\pi}{2}2Οβ
- \displaystyle -\frac{1}{2}β21β
- \displaystyle -\frac{\pi}{2}β2Οβ
- \displaystyle \frac{\pi}{4}4Οβ
- \displaystyle \frac{1}{2}21β
Q5. Compute \displaystyle \int \frac{dx}{x^2-2x+10}β«x2β2x+10dxβ. Hint: partial fractions doesnβt seem to work on this problem, since you donβt have real roots for the denominatorβ¦is there another technique you could use?
- \displaystyle \frac{1}{3} \arctan\frac{x+1}{5} + C31βarctan5x+1β+C
- \displaystyle \frac{1}{2} \arctan\frac{x+1}{2} + C21βarctan2x+1β+C
- \displaystyle \frac{1}{3} \arctan\frac{x+1}{3} + C31βarctan3x+1β+C
- \displaystyle \frac{1}{3} \arctan\frac{x-1}{5} + C31βarctan5xβ1β+C
- \displaystyle \frac{1}{2} \arctan\frac{x-1}{4} + C21βarctan4xβ1β+C
- \displaystyle \frac{1}{2} \arctan\frac{x+1}{4} + C21βarctan4x+1β+C
- \displaystyle \frac{1}{2} \arctan\frac{x-1}{2} + C21βarctan2xβ1β+C
- \displaystyle \frac{1}{3} \arctan\frac{x-1}{3} + C31βarctan3xβ1β+C
Q6. Solve the following separable differential equation \displaystyle \frac{dx}{dt} = e^{t-x}dtdxβ=etβx.
- \displaystyle x = -\ln\left(C-e^t\right)x=βln(Cβet)
- \displaystyle x = \left( \frac{t^2}{2} + C \right) e^{-2t}x=(2t2β+C)eβ2t
- \displaystyle x = \ln\left(Ce^{-2t}\right)x=ln(Ceβ2t)
- \displaystyle x = \frac{t}{2} β \frac{1}{4} + Cx=2tββ41β+C
- \displaystyle x = \frac{t^2}{4} + \frac{1}{2} + Ce^{-2t}x=4t2β+21β+Ceβ2t
- \displaystyle x = -\frac{t}{2} β \frac{1}{4} + Ce^{2t}x=β2tββ41β+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 \,dxβ«cos34xdx .
- \displaystyle \frac{1}{16} \cos^4 4x + C161βcos44x+C
- \displaystyle \cos 4x + \frac{1}{4}\cos^4 4x + Ccos4x+41βcos44x+C
- \displaystyle \frac{1}{4} \cos^4 4x + C41βcos44x+C
- \displaystyle \frac{1}{4}\cos 4x + \frac{1}{12}\cos^4 4x + C41βcos4x+121βcos44x+C
- \displaystyle \sin 4x β \frac{1}{3}\sin^3 4x + Csin4xβ31βsin34x+C
- \displaystyle \frac{1}{4}\sin 4x β \frac{1}{12}\sin^3 4x + C41βsin4xβ121βsin34x+C
- 12\cos^2 4x + C12cos24x+C
- \displaystyle \frac{1}{4}\sin 4x + \frac{1}{12}\sin^3 4x + C41βsin4x+121βsin34x+C
Q8. \displaystyle \left.\frac{d}{dx}\right|_{x=\pi} \int_{t=0}^{x} \frac{\cos 3t}{\sqrt{1+t}} \,dt =dxdββ£β£β£β£β£βx=Οββ«t=0xβ1+tβcos3tβdt=
- \displaystyle \frac{\pi}{\sqrt{1+\pi}}1+ΟβΟβ
- 00
- -\piβΟ
- \displaystyle -\frac{1}{\sqrt{1+\pi}}β1+Οβ1β
- \displaystyle \int_0^{\pi} \frac{d}{dt}\!\left( \frac{\cos 3t}{\sqrt{1+t}} \right) dtβ«0Οβdtdβ(1+tβcos3tβ)dt
- \displaystyle -1β1
- \displaystyle \frac{\cos(3\ln \pi)}{\pi\sqrt{1+\ln \pi}}Ο1+lnΟβcos(3lnΟ)β
- \displaystyle \frac{1}{\sqrt{1+\pi}}1+Οβ1β
Q9. Which of the following is the integrating factor, II, used to solve the linear differential equation
e^{3t}\frac{dx}{dt}=2-e^txe3tdtdxβ=2βetx
- 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=e21βe2t, that is, I= {\rm exp}\left(\frac{1}{2}e^{2t}\right)I=exp(21βe2t)
- I = e^{-e^{-t}}I=eβeβt, that is, I= {\rm exp}\left( -e^{-t}\right)I=exp(βeβt)
- I = 2e^{-3t}I=2eβ3t, that is, I= 2\,{\rm exp }\left( -3t\right)I=2exp(β3t)
- I = e^{\frac{1}{3}e^{3t}}I=e31βe3t, that is, I= {\rm exp}\left( \frac{1}{3}e^{3t}\right)I=exp(31βe3t)
- I = e^{-\frac{1}{3}e^{-3t}}I=eβ31βeβ3t, that is, I= {\rm exp}\left( -\frac{1}{3}e^{-3t}\right)I=exp(β31βeβ3t)
- I = e^{-\frac{1}{2}e^{-2t}}I=eβ21βeβ2t, that is, I= {\rm exp}\left( -\frac{1}{2}e^{-2t}\right)I=exp(β21βeβ2t)
Q10. The size z(t)z(t) of a hailstone evolves according to the differential equation
\frac{dz}{dt} = A \sqrt{z} β B\sqrt{z^3}dtdzβ=AzββBz3β
where AA and BB are positive constants. Without solving the differential equation, determine the limiting size \displaystyle \lim_{t\to+\infty}z(t)tβ+βlimβz(t) in the case where z(0)=1z(0)=1. Hint: think in terms of equilibriaβ¦
- \displaystyle\frac{\sqrt{A}}{B}BAββ
- \displaystyle-\frac{A}{B}βBAβ
- \displaystyle\sqrt{\frac{A}{B}}BAββ
- \displaystyle\frac{A}{B}BAβ
- \displaystyle-\frac{B}{A}βABβ
- \displaystyle\frac{B}{A}ABβ
- \displaystyle\frac{A}{\sqrt{B}}BβAβ
- \displaystyle\sqrt{\frac{B}{A}}AB
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