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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{d v}{d x}

in a form which allows it to be more easily evaluated. The formula is

\int u \frac{d v}{d x} d x = u v - \int v \frac{d u}{d x} d x

. When applying the method of integration by parts to find

\int x e^{x} d x

, the best choice of

u

and

\frac{d v}{d x}

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 x e^{x} d x = \frac{1}{2} x^{2} e^{x} - \int \frac{1}{2} x^{2} e^{x} d x

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 x e^{x} d x = x^{2} e^{x} - \int (x^{2} e^{x} + x e^{x}) d x

Not correct. Choice (c) is false.

The problem of finding

v

d v = x e^{x} d x

is exactly the integral we started with! So no progress in this case.

Your answer is correct.

This gives genuine simplification and results in an easy integral. We obtain

\int x e^{x} d x = x e^{x} - \int e^{x} d x

Not correct. Choice (e) is false.

Question 2

Find

\int_{0}^{1} x e^{x} d x

using integration by parts, and enter your answer.

Your answer is correct

Choosing

u = x, \frac{d v}{d x} = e^{x}

gives

\int_{0}^{1} x e^{x} d x = {[x e^{x}]}_{0}^{1} - \int_{0}^{1} e^{x} d x = {[e^{x} (x - 1)]}_{0}^{1} = 1 .

Not correct. You may try again.

Try choosing

u = x, \frac{d v}{d x} = e^{x}

Question 3

The reduction formula for

I_{n} = \int x {(ln x)}^{n} d x

I_{n} = \frac{1}{2} x^{2} {(ln x)}^{n} - \frac{n}{2} I_{n - 1}

. Given that

I_{0} = \frac{1}{2} x^{2} + C

, find

\int_{1}^{e} x (ln x) d x

. Give your answer correct to three decimal places.

Your answer is correct

The integral

I_{1}

\frac{1}{2} x^{2} ln x - \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 x^{n} e^{x} d x = x^{n} e^{x} - n I_{n - 1}

to help answer this question.

Not correct. Choice (a) is false.

Not correct. Choice (b) is false.

Not correct. Choice (c) is false.

Your answer is correct.

Not correct. Choice (e) is false.

Question 5

Which option equals

\int_{0}^{1} x {tan}^{- 1} x d x

? (Hint: use integration by parts with

u = {tan}^{- 1} x

and

\frac{d v}{d x} = x

Your answer is correct.

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}} {sin}^{- 1} x d x

? (Hint: use integration by parts with

u = {sin}^{- 1} x

and

\frac{d v}{d x} = 1

Not correct. Choice (a) is false.

Not correct. Choice (b) is false.

Your answer is correct.

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 = ln x

, the line

y = 1

and the tangent line to

y = ln x

x = 1

is given as an integral with respect to

x

\int_{1}^{2} (x - 1 - ln x) d x + \int_{2}^{e} (1 - ln x) d x

. 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.)

Not correct. Choice (a) is false.

Not correct. Choice (b) is false.

Not correct. Choice (c) is false.

Your answer is correct.

Not correct. Choice (e) is false.

Question 8

Which option equals

\int x {sec}^{2} x d x

Not correct. Choice (a) is false.

Your answer is correct.

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 sin (ln x) d x

is one such problem. Which of the following options gives the expression obtained after one application of integration by parts?

Your answer is correct.

Not correct. Choice (b) is false.

Try

u = sin (ln x)

and

\frac{d v}{d x} = 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} sin (ln x) d x

Not correct. Choice (a) is false.

Not correct. Choice (b) is false.

Your answer is correct.

Not correct. Choice (d) is false.

Not correct. Choice (e) is false.

ABN: 15 211 513 464. CRICOS number: 00026A. Phone: +61 2 9351 2222.

Authorised by: Head, School of Mathematics and Statistics.

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Questions:

right first attempt
right
wrong

a)	$u = e^{x}, \frac{d v}{d x} = x$	b)	$u = x e^{x}, \frac{d v}{d x} = 1$
c)	$u = 1, \frac{d v}{d x} = x e^{x}$	d)	$u = x, \frac{d v}{d x} = e^{x}$
e)	None of the above

a)	$e^{x} x^{4} + 4 x^{3} e^{x} - 12 x^{2} e^{x} + 24 x e^{x} - 24 e^{x}$	b)	$e^{x} (x^{4} - 4 x^{3} - 12 x^{2} - 24 x + 24)$
c)	$e^{x} x^{4} - 4 x^{3} e^{x} + 12 x^{2} e^{x} - 24 x e^{x} + 24 e^{x} + C e^{x}$	d)	$e^{x} (x^{4} - 4 x^{3} + 12 x^{2} - 24 x + 24) + 15$
e)	None of the above

a)	$\frac{π}{4} - \frac{1}{2}$	b)	$- \frac{1}{2}$
c)	$\frac{π}{8} - \frac{1}{2}$	d)	$\frac{1}{2} (\frac{π}{4} + 1)$
e)	$\frac{π}{4}$

a)	$\frac{π}{12} - \frac{\sqrt{3}}{2} + 1$	b)	$\frac{π}{4} + \frac{\sqrt{3}}{2}$
c)	$\frac{π}{12} + \frac{\sqrt{3}}{2} - 1$	d)	$\frac{π}{3} - 1$
e)	$\frac{π}{12} - \frac{\sqrt{3}}{2} - 1$

a)	$\int_{1}^{e} e^{y} - (y + 1) d y$	b)	$\int_{0}^{1} ln y - y + 1 d y$
c)	$\int_{0}^{e} e^{y} - y + 1 d y$	d)	$\int_{0}^{1} e^{y} - (y + 1) d y$
e)	None of the above

a)	$x sec x tan x - ln (cos x) + C$	b)	$x tan x + ln \| cos x \| + C$
c)	$x {tan}^{2} x - ln \| cos x \| + C$	d)	$x tan x - ln (cos x) + C$
e)	None of the above

a)	$x sin (ln x) - \int cos (ln x) d x$	b)	$- cos x ln x + \int \frac{cos x}{x} d x$
c)	$- x sin (ln x) + \int x cos (ln x) d x$	d)	$x sin (ln x) - \int \frac{sin (ln x)}{x} d x$
e)	$cos x ln x + \int \frac{cos x}{x} d x$

a)	$\frac{1}{2} (e sin 1 + e cos 1)$	b)	$sin 1 - cos 1$
c)	$\frac{1}{2} (e sin 1 - e cos 1 + 1)$	d)	$\frac{1}{2} sin 1 - 2 e cos 1 + 1$
e)	$e sin 1 - e cos 1$

The University of Sydney - School of Mathematics and Statistics

Quiz 4: Integration by Parts

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

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