# Quiz 4: Integration by Parts

Question

## Question 1

Recall that integration by parts is a technique to re-express the integral of a product of two functions $u$ and $\frac{dv}{dx}$ in a form which allows it to be more easily evaluated. The formula is $\int \phantom{\rule{1em}{0ex}}u\phantom{\rule{2.77695pt}{0ex}}\frac{dv}{dx}\phantom{\rule{2.77695pt}{0ex}}dx=uv-\int \phantom{\rule{1em}{0ex}}v\phantom{\rule{2.77695pt}{0ex}}\frac{du}{dx}\phantom{\rule{2.77695pt}{0ex}}dx$. When applying the method of integration by parts to find $\int \phantom{\rule{1em}{0ex}}x{e}^{x}\phantom{\rule{1em}{0ex}}dx$, the best choice of $u$ and $\frac{dv}{dx}$ is
 a) $u={e}^{x},\phantom{\rule{1em}{0ex}}\frac{dv}{dx}=x$ b) $u=x{e}^{x},\phantom{\rule{1em}{0ex}}\frac{dv}{dx}=1$ c) $u=1,\phantom{\rule{1em}{0ex}}\frac{dv}{dx}=x{e}^{x}$ d) $u=x,\phantom{\rule{1em}{0ex}}\frac{dv}{dx}={e}^{x}$ e) None of the above

Not correct. Choice (a) is false.
This expresses the original integral in terms of a new integral which is even harder! After applying integration by parts as suggested, we have $\int \phantom{\rule{1em}{0ex}}x{e}^{x}\phantom{\rule{1em}{0ex}}dx=\frac{1}{2}{x}^{2}{e}^{x}-\int \phantom{\rule{2.77695pt}{0ex}}\frac{1}{2}{x}^{2}{e}^{x}\phantom{\rule{2.77695pt}{0ex}}dx$.
Not correct. Choice (b) is false.
This expresses the original integral in terms of a new integral which is even harder! After applying integration by parts as suggested, we have $\int \phantom{\rule{1em}{0ex}}x{e}^{x}\phantom{\rule{1em}{0ex}}dx={x}^{2}{e}^{x}-\int \phantom{\rule{2.77695pt}{0ex}}\left({x}^{2}{e}^{x}+x{e}^{x}\right)\phantom{\rule{2.77695pt}{0ex}}dx$.
Not correct. Choice (c) is false.
The problem of finding $v$ if $dv=x{e}^{x}\phantom{\rule{2.77695pt}{0ex}}dx$ is exactly the integral we started with! So no progress in this case.
This gives genuine simplification and results in an easy integral. We obtain $\int \phantom{\rule{2.77695pt}{0ex}}x{e}^{x}\phantom{\rule{2.77695pt}{0ex}}dx=x{e}^{x}-\int \phantom{\rule{2.77695pt}{0ex}}{e}^{x}\phantom{\rule{2.77695pt}{0ex}}dx$.
Not correct. Choice (e) is false.

## Question 2

Find ${\int }_{0}^{1}\phantom{\rule{1em}{0ex}}x{e}^{x}\phantom{\rule{1em}{0ex}}dx$ using integration by parts, and enter your answer.

Choosing $u=x,\phantom{\rule{1em}{0ex}}\frac{dv}{dx}={e}^{x}$ gives ${\int }_{0}^{1}\phantom{\rule{1em}{0ex}}x{e}^{x}\phantom{\rule{1em}{0ex}}dx={\left[x{e}^{x}\right]}_{0}^{1}-{\int }_{0}^{1}\phantom{\rule{2.77695pt}{0ex}}{e}^{x}\phantom{\rule{2.77695pt}{0ex}}dx={\left[{e}^{x}\left(x-1\right)\right]}_{0}^{1}=1.$
Not correct. You may try again.
Try choosing $u=x,\phantom{\rule{1em}{0ex}}\frac{dv}{dx}={e}^{x}$.

## Question 3

The reduction formula for ${I}_{n}=\int \phantom{\rule{2.77695pt}{0ex}}x{\left(lnx\right)}^{n}\phantom{\rule{2.77695pt}{0ex}}dx$ is ${I}_{n}=\frac{1}{2}{x}^{2}{\left(lnx\right)}^{n}-\frac{n}{2}{I}_{n-1}$. Given that ${I}_{0}=\frac{1}{2}{x}^{2}+C$, find ${\int }_{1}^{e}\phantom{\rule{2.77695pt}{0ex}}x\left(lnx\right)\phantom{\rule{2.77695pt}{0ex}}dx$. Give your answer correct to three decimal places.

The integral ${I}_{1}$ is $\frac{1}{2}{x}^{2}lnx-\frac{1}{4}{x}^{2}+C$, using the reduction formula with $n=1$. Evaluating this between $1$ and $e$ gives 2.097 to three decimal places.
Not correct. You may try again.
Substitute $n=1$ into the reduction formula to obtain ${I}_{1}$.

## Question 4

Which option is an antiderivative of ${x}^{4}{e}^{x}$ ? Use the reduction formula ${I}_{n}=\int \phantom{\rule{2.77695pt}{0ex}}{x}^{n}{e}^{x}\phantom{\rule{2.77695pt}{0ex}}dx={x}^{n}{e}^{x}-n{I}_{n-1}$ to help answer this question.
 a) ${e}^{x}{x}^{4}+4{x}^{3}{e}^{x}-12{x}^{2}{e}^{x}+24x{e}^{x}-24{e}^{x}$ b) ${e}^{x}\left({x}^{4}-4{x}^{3}-12{x}^{2}-24x+24\right)$ c) ${e}^{x}{x}^{4}-4{x}^{3}{e}^{x}+12{x}^{2}{e}^{x}-24x{e}^{x}+24{e}^{x}+C{e}^{x}$ d) ${e}^{x}\left({x}^{4}-4{x}^{3}+12{x}^{2}-24x+24\right)+15$ e) None of the above

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Not correct. Choice (e) is false.

## Question 5

Which option equals ${\int }_{0}^{1}\phantom{\rule{2.77695pt}{0ex}}x{tan}^{-1}x\phantom{\rule{2.77695pt}{0ex}}dx$ ? (Hint: use integration by parts with $u={tan}^{-1}x$ and $\frac{dv}{dx}=x$.)
 a) $\frac{\pi }{4}-\frac{1}{2}$ b) $-\frac{1}{2}$ c) $\frac{\pi }{8}-\frac{1}{2}$ d) $\frac{1}{2}\left(\frac{\pi }{4}+1\right)$ e) $\frac{\pi }{4}$

The indefinite integral is $\frac{1}{2}{x}^{2}{tan}^{-1}x-\frac{1}{2}x+\frac{1}{2}{tan}^{-1}x+C$, which, when evaluated between 0 and 1, matches this option.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.
Not correct. Choice (e) is false.

## Question 6

Which option equals ${\int }_{0}^{\frac{1}{2}}\phantom{\rule{2.77695pt}{0ex}}{sin}^{-1}x\phantom{\rule{2.77695pt}{0ex}}dx$ ? (Hint: use integration by parts with $u={sin}^{-1}x$ and $\frac{dv}{dx}=1$.)
 a) $\frac{\pi }{12}-\frac{\sqrt{3}}{2}+1$ b) $\frac{\pi }{4}+\frac{\sqrt{3}}{2}$ c) $\frac{\pi }{12}+\frac{\sqrt{3}}{2}-1$ d) $\frac{\pi }{3}-1$ e) $\frac{\pi }{12}-\frac{\sqrt{3}}{2}-1$

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
The indefinite integral is $x{sin}^{-1}x+\sqrt{1-{x}^{2}}+C$, which, when evaluated between $0$ and $\frac{1}{2}$, matches this option.
Not correct. Choice (d) is false.
Not correct. Choice (e) is false.

## Question 7

The finite area bounded by the curve $y=lnx$, the line $y=1$ and the tangent line to $y=lnx$ at $x=1$ is given as an integral with respect to $x$ by ${\int }_{1}^{2}\phantom{\rule{2.77695pt}{0ex}}\left(x-1-lnx\right)\phantom{\rule{2.77695pt}{0ex}}dx+{\int }_{2}^{e}\phantom{\rule{2.77695pt}{0ex}}\left(1-lnx\right)\phantom{\rule{2.77695pt}{0ex}}dx$. Which option equals the same area given as an integral with respect to $y$ ? (You must draw a sketch to help you with this question.)
 a) ${\int }_{1}^{e}\phantom{\rule{2.77695pt}{0ex}}{e}^{y}-\left(y+1\right)\phantom{\rule{2.77695pt}{0ex}}dy$ b) ${\int }_{0}^{1}\phantom{\rule{2.77695pt}{0ex}}lny-y+1\phantom{\rule{2.77695pt}{0ex}}dy$ c) ${\int }_{0}^{e}\phantom{\rule{2.77695pt}{0ex}}{e}^{y}-y+1\phantom{\rule{2.77695pt}{0ex}}dy$ d) ${\int }_{0}^{1}\phantom{\rule{2.77695pt}{0ex}}{e}^{y}-\left(y+1\right)\phantom{\rule{2.77695pt}{0ex}}dy$ e) None of the above

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Not correct. Choice (e) is false.

## Question 8

Which option equals $\int \phantom{\rule{2.77695pt}{0ex}}x{sec}^{2}x\phantom{\rule{2.77695pt}{0ex}}dx$ ?
 a) $xsecxtanx-ln\left(cosx\right)+C$ b) $xtanx+ln|cosx|+C$ c) $x{tan}^{2}x-ln|cosx|+C$ d) $xtanx-ln\left(cosx\right)+C$ e) None of the above

Not correct. Choice (a) is false.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.
Not correct. Choice (e) is false.

## Question 9

In some problems you need to apply the integration by parts method twice in order to obtain the required answer. The integral $\int \phantom{\rule{2.77695pt}{0ex}}sin\left(lnx\right)\phantom{\rule{2.77695pt}{0ex}}dx$ is one such problem. Which of the following options gives the expression obtained after one application of integration by parts?
 a) $xsin\phantom{\rule{0.3em}{0ex}}\left(lnx\right)-\int \phantom{\rule{2.77695pt}{0ex}}cos\phantom{\rule{0.3em}{0ex}}\left(lnx\right)\phantom{\rule{2.77695pt}{0ex}}dx$ b) $-cosxlnx+\int \phantom{\rule{2.77695pt}{0ex}}\frac{cosx}{x}\phantom{\rule{2.77695pt}{0ex}}dx$ c) $-xsin\phantom{\rule{0.3em}{0ex}}\left(lnx\right)+\int \phantom{\rule{2.77695pt}{0ex}}xcos\phantom{\rule{0.3em}{0ex}}\left(lnx\right)\phantom{\rule{2.77695pt}{0ex}}dx$ d) $xsin\phantom{\rule{0.3em}{0ex}}\left(lnx\right)-\int \phantom{\rule{2.77695pt}{0ex}}\frac{sin\phantom{\rule{0.3em}{0ex}}\left(lnx\right)}{x}\phantom{\rule{2.77695pt}{0ex}}dx$ e) $cosxlnx+\int \phantom{\rule{2.77695pt}{0ex}}\frac{cosx}{x}\phantom{\rule{2.77695pt}{0ex}}dx$

Not correct. Choice (b) is false.
Try $u=sin\phantom{\rule{0.3em}{0ex}}\left(lnx\right)$ and $\frac{dv}{dx}=1$ in the integration by parts formula.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.
Not correct. Choice (e) is false.

## Question 10

Which option equals ${\int }_{1}^{e}\phantom{\rule{2.77695pt}{0ex}}sin\left(lnx\right)\phantom{\rule{2.77695pt}{0ex}}dx$  ?
 a) $\frac{1}{2}\left(esin1+ecos1\right)$ b) $sin1-cos1$ c) $\frac{1}{2}\left(esin1-ecos1+1\right)$ d) $\frac{1}{2}sin1-2ecos1+1$ e) $esin1-ecos1$

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.