## MATH1111 Quizzes

Integration by Substitution Quiz
Web resources available Questions

This quiz tests the work covered in lecture on integration by substitution and corresponds to Section 7.1 of the textbook Calculus: Single and Multivariable (Hughes-Hallett, Gleason, McCallum et al.).

There are more web quizzes at Wiley, select Section 1.

The MATH1011 Quiz 11 should also be appropriate to try. Only questions 4, 5, 8, 9 and 10 involve integration by substitution. The other questions would be good revision. These and other sources use $u$ rather than $w$ as the letter for the substituted function.

Visual calculus again has good examples at http://archives.math.utk.edu/visual.calculus/4/substitutions.3/ and there are plenty of exercises at http://archives.math.utk.edu/visual.calculus/4/substitutions.1/index.html with hints. The visual calculus quiz has some very hard questions.

If you are looking for another explanation then try the pdf file at http://www.mathcentre.ac.uk/students.php/all_subjects/integration/by_substitution/resources/79.

Finally the Maths Learning Centre Booklet Introduction to Integration is a good source of explanations and exercises.

Which of the following substitutions are correct? (Zero or more options can be correct)
 a) $\int t{e}^{{t}^{2}+4}\phantom{\rule{0.3em}{0ex}}dt=\int {e}^{w}\phantom{\rule{0.3em}{0ex}}dw$ where $w={t}^{2}+4\phantom{\rule{0.3em}{0ex}}.$ b) $\int \frac{3{x}^{2}+4x+1}{{x}^{3}+2{x}^{2}+x-3}\phantom{\rule{0.3em}{0ex}}dx=\int \frac{1}{w}\phantom{\rule{0.3em}{0ex}}dw$ where $w={x}^{3}+2{x}^{2}+x-3\phantom{\rule{0.3em}{0ex}}.$ c) $\int \frac{cos\sqrt{x}}{2\sqrt{x}}\phantom{\rule{0.3em}{0ex}}dx=\int cosw\phantom{\rule{0.3em}{0ex}}dw$ where $w=\sqrt{x}\phantom{\rule{0.3em}{0ex}}.$ d) $\int \frac{1}{{x}^{2}+4}\phantom{\rule{0.3em}{0ex}}dx=\int \frac{1}{w}\phantom{\rule{0.3em}{0ex}}dw$ where $w={x}^{2}+4\phantom{\rule{0.3em}{0ex}}.$ e) $\int \left({z}^{2}+2z\right){\left({z}^{3}+3{z}^{2}-4\right)}^{4}\phantom{\rule{0.3em}{0ex}}dz=\int \frac{{w}^{4}}{3}\phantom{\rule{0.3em}{0ex}}dw$ where $w={z}^{3}+3{z}^{2}-4\phantom{\rule{0.3em}{0ex}}.$

There is at least one mistake.
For example, choice (a) should be False.
The constant is not correct
$\int x{e}^{{x}^{2}+4}\phantom{\rule{0.3em}{0ex}}dx=\int \frac{1}{2}{e}^{w}\phantom{\rule{0.3em}{0ex}}dw$ where $w={x}^{2}+4\phantom{\rule{0.3em}{0ex}}.$
There is at least one mistake.
For example, choice (b) should be True.
$w={x}^{3}+2{x}^{2}+x-3$ so $dw=\left(3{x}^{2}+4x+1\right)\phantom{\rule{0.3em}{0ex}}dx$ as required.
There is at least one mistake.
For example, choice (c) should be True.
$w=\sqrt{x}={x}^{\frac{1}{2}}$ so $dw=\frac{1}{2}{x}^{-\frac{1}{2}}\phantom{\rule{0.3em}{0ex}}dw=\frac{1}{2\sqrt{x}}\phantom{\rule{0.3em}{0ex}}dx$ as required.
There is at least one mistake.
For example, choice (d) should be False.
This integral cannot be solved using this method of substitution.
There is at least one mistake.
For example, choice (e) should be True.
$w={z}^{3}+3{z}^{2}-4$ and $dw=\left(3{z}^{2}+6z\right)\phantom{\rule{0.3em}{0ex}}dx$ so $\left({z}^{2}+2z\right)\phantom{\rule{0.3em}{0ex}}dx=\frac{1}{3}\phantom{\rule{0.3em}{0ex}}dw$ as required
Correct!
1. False The constant is not correct
$\int x{e}^{{x}^{2}+4}\phantom{\rule{0.3em}{0ex}}dx=\int \frac{1}{2}{e}^{w}\phantom{\rule{0.3em}{0ex}}dw$ where $w={x}^{2}+4\phantom{\rule{0.3em}{0ex}}.$
2. True $w={x}^{3}+2{x}^{2}+x-3$ so $dw=\left(3{x}^{2}+4x+1\right)\phantom{\rule{0.3em}{0ex}}dx$ as required.
3. True $w=\sqrt{x}={x}^{\frac{1}{2}}$ so $dw=\frac{1}{2}{x}^{-\frac{1}{2}}\phantom{\rule{0.3em}{0ex}}dw=\frac{1}{2\sqrt{x}}\phantom{\rule{0.3em}{0ex}}dx$ as required.
4. False This integral cannot be solved using this method of substitution.
5. True $w={z}^{3}+3{z}^{2}-4$ and $dw=\left(3{z}^{2}+6z\right)\phantom{\rule{0.3em}{0ex}}dx$ so $\left({z}^{2}+2z\right)\phantom{\rule{0.3em}{0ex}}dx=\frac{1}{3}\phantom{\rule{0.3em}{0ex}}dw$ as required
Find the indefinite integral $\int {x}^{2}{\left({x}^{3}+4\right)}^{2}\phantom{\rule{0.3em}{0ex}}dx\phantom{\rule{0.3em}{0ex}}.$ Exactly one option must be correct)
 a) $\int {x}^{2}{\left({x}^{3}+4\right)}^{2}\phantom{\rule{0.3em}{0ex}}dx={\left({x}^{3}+4\right)}^{3}+C$ b) $\int {x}^{2}{\left({x}^{3}+4\right)}^{2}\phantom{\rule{0.3em}{0ex}}dx=\frac{2{\left({x}^{3}+4\right)}^{3}}{3}+C$ c) $\int {x}^{2}{\left({x}^{3}+4\right)}^{2}\phantom{\rule{0.3em}{0ex}}dx=\frac{{\left({x}^{3}+4\right)}^{3}}{9}+C$ d) $\int {x}^{2}{\left({x}^{3}+4\right)}^{2}\phantom{\rule{0.3em}{0ex}}dx=\frac{{\left({x}^{3}+4\right)}^{2}}{6}+C$

Choice (a) is incorrect
Try again, you have not dealt with the constants correctly.
Choice (b) is incorrect
Try again, you have not dealt with the constants correctly.
Choice (c) is correct!
Let $w={x}^{3}+4$ then $dw=3{x}^{2}\phantom{\rule{0.3em}{0ex}}dx$ and ${x}^{2}\phantom{\rule{0.3em}{0ex}}dx=\frac{1}{3}\phantom{\rule{0.3em}{0ex}}dw\phantom{\rule{0.3em}{0ex}}.$

$\int {x}^{2}{\left({x}^{3}+4\right)}^{2}\phantom{\rule{0.3em}{0ex}}dx=\int \frac{1}{3}{w}^{2}\phantom{\rule{0.3em}{0ex}}dw=\frac{1}{9}{w}^{3}+C=\frac{{\left({x}^{3}+4\right)}^{3}}{9}+C\phantom{\rule{0.3em}{0ex}}.$
Choice (d) is incorrect
Try again, you have not integrated correctly.
Find the indefinite integral $\int \left({x}^{3}+2\right)cos\phantom{\rule{0.3em}{0ex}}\left({x}^{4}+8x\right)\phantom{\rule{0.3em}{0ex}}dx\phantom{\rule{0.3em}{0ex}}.$ Exactly one option must be correct)
 a) $\int \left({x}^{3}+2\right)cos\phantom{\rule{0.3em}{0ex}}\left({x}^{4}+8x\right)\phantom{\rule{0.3em}{0ex}}dx=\frac{-sin\phantom{\rule{0.3em}{0ex}}\left({x}^{4}+8x\right)}{4}+C$ b) $\int \left({x}^{3}+2\right)cos\phantom{\rule{0.3em}{0ex}}\left({x}^{4}+8x\right)\phantom{\rule{0.3em}{0ex}}dx=\frac{sin\phantom{\rule{0.3em}{0ex}}\left({x}^{4}+8x\right)}{4}+C$ c) $\int \left({x}^{3}+2\right)cos\phantom{\rule{0.3em}{0ex}}\left({x}^{4}+8x\right)\phantom{\rule{0.3em}{0ex}}dx=\frac{cos\phantom{\rule{0.3em}{0ex}}w}{4}+C$ d) This integral cannot be integrated using substitution.

Choice (a) is incorrect
Try again, you have not integrated $cosw$ correctly.
Choice (b) is correct!
Let $w={x}^{4}+8x$ then $dw=\left(4{x}^{3}+8\right)dx=4\left({x}^{3}+2\right)dx$ and $\left({x}^{3}+2\right)dx=\frac{1}{4}\phantom{\rule{0.3em}{0ex}}dw\phantom{\rule{0.3em}{0ex}}.$
$\begin{array}{llll}\hfill \int \left({x}^{3}+2\right)cos\left({x}^{4}+8x\right)\phantom{\rule{0.3em}{0ex}}dx& =\int \frac{1}{4}cosw\phantom{\rule{0.3em}{0ex}}dw=\frac{1}{4}sinw\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{1}{4}sin\phantom{\rule{0.3em}{0ex}}\left({x}^{4}+8x\right)+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
Choice (c) is incorrect
Try again, you have not finished the integral.
Choice (d) is incorrect
Try again, let $w={x}^{4}+8x\phantom{\rule{0.3em}{0ex}}.$
Which of the following integrals have been evaluated correctly? (Zero or more options can be correct)
 a) ${\int }_{\mathit{0}}^{2}{x}^{2}{\left({x}^{3}+1\right)}^{1∕2}={\int }_{1}^{9}\frac{{w}^{1∕2}}{3}\phantom{\rule{0.3em}{0ex}}dw={\frac{2}{9}{w}^{3∕2}|}_{1}^{9}=\frac{52}{9}$ b) ${\int }_{\mathit{0}}^{\pi }\theta sin\phantom{\rule{0.3em}{0ex}}\left({\theta }^{2}\right)\phantom{\rule{0.3em}{0ex}}d\theta ={\int }_{\mathit{0}}^{\pi }\frac{w}{2}\phantom{\rule{0.3em}{0ex}}dw={\frac{{w}^{2}}{4}|}_{\mathit{0}}^{\pi }=\frac{{\pi }^{2}}{4}$ c) ${\int }_{1}^{2}\frac{dx}{{\left(3-5x\right)}^{2}}=-\frac{1}{5}{\int }_{-7}^{-2}\frac{du}{{u}^{2}}={\frac{-1}{5u}|}_{-7}^{-2}=\frac{1}{14}$ d) ${\int }_{\mathit{0}}^{4}\sqrt{3x+4}\phantom{\rule{0.3em}{0ex}}dx=\frac{1}{3}{\int }_{4}^{16}\sqrt{w}\phantom{\rule{0.3em}{0ex}}dw={\frac{1}{3}\frac{2}{3}{w}^{3∕2}|}_{4}^{16}=\frac{112}{9}$ e) ${\int }_{\mathit{0}}^{1}{t}^{2}{e}^{-{t}^{3}}\phantom{\rule{0.3em}{0ex}}dt={\int }_{\mathit{0}}^{-1}\frac{-{e}^{w}}{3}\phantom{\rule{0.3em}{0ex}}dw={-\frac{{e}^{w}}{3}|}_{\mathit{0}}^{-1}=\frac{1}{3}-\frac{1}{3e}$

There is at least one mistake.
For example, choice (a) should be True.
There is at least one mistake.
For example, choice (b) should be False.
There are a variety of errors. The limits of integration have not been changed and the integration has not been performed correctly. $\begin{array}{llll}\hfill {\int }_{\mathit{0}}^{\pi }\theta sin\phantom{\rule{0.3em}{0ex}}\left({\theta }^{2}\right)\phantom{\rule{0.3em}{0ex}}d\theta & ={\int }_{\mathit{0}}^{{\pi }^{2}}\frac{sinw}{2}\phantom{\rule{0.3em}{0ex}}dw=\frac{-cosw}{2}|{}_{\mathit{0}}^{{\pi }^{2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{1}{2}\left(cos{\pi }^{2}-1\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
There is at least one mistake.
For example, choice (c) should be False.
The answer is correct but the steps are incorrect. The limits of integration must stay in the correct order.
${\int }_{1}^{2}\frac{dx}{{\left(3-5x\right)}^{2}}=-\frac{1}{5}{\int }_{-2}^{-7}\frac{du}{{u}^{2}}={\frac{1}{5u}|}_{-2}^{-7}=\frac{1}{14}$
There is at least one mistake.
For example, choice (d) should be True.
There is at least one mistake.
For example, choice (e) should be True.
Correct!
1. True
2. False There are a variety of errors. The limits of integration have not been changed and the integration has not been performed correctly. $\begin{array}{llll}\hfill {\int }_{\mathit{0}}^{\pi }\theta sin\phantom{\rule{0.3em}{0ex}}\left({\theta }^{2}\right)\phantom{\rule{0.3em}{0ex}}d\theta & ={\int }_{\mathit{0}}^{{\pi }^{2}}\frac{sinw}{2}\phantom{\rule{0.3em}{0ex}}dw=\frac{-cosw}{2}|{}_{\mathit{0}}^{{\pi }^{2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{1}{2}\left(cos{\pi }^{2}-1\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
3. False The answer is correct but the steps are incorrect. The limits of integration must stay in the correct order.
${\int }_{1}^{2}\frac{dx}{{\left(3-5x\right)}^{2}}=-\frac{1}{5}{\int }_{-2}^{-7}\frac{du}{{u}^{2}}={\frac{1}{5u}|}_{-2}^{-7}=\frac{1}{14}$
4. True
5. True