5E: Exercises
- Page ID
- 10317
This page is a draft and is under active development.
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Exercise \(\PageIndex{1}\)
\(\displaystyle\int_{-1}^{4} \frac{x}{\sqrt{x+4}} \, dx\)
- Answer
-
Let \(u= x+4\), then \(du=dx\) and \(x=u-4\).
Now \(\displaystyle\int_{-1}^{4} \frac{x}{\sqrt{x+4}} \, dx = \displaystyle \int_{-1}^{4} \frac{u-4}{\sqrt{u}} du\)
\(= \displaystyle \int_{-1}^{4} {(u-4)}u^{\frac{-1}{2}} du\)
\(=\displaystyle \int_{-1}^{4} u^{\frac{1}{2}} -4u^{\frac{-1}{2} }du\)
\(=\left(\displaystyle\frac{2}{3} u^{\frac{3}{2}} -4 \frac{2}{1}u^{\frac{1}{2}} \right) \left |_{x=-1}^{x=4} \right.\)
\(= \left(\displaystyle\frac{2}{3} (x+4)^{\frac{3}{2}} -8 (x+4)^{\frac{1}{2}} \right) \left |_{x=-1}^{x=4} \right.\)
\(=\frac{2}{3} ( 8^{\frac{3}{2} }- 3 ^{\frac{3}{2}} )-8( 8^{\frac{1}{2}}- 3^{\frac{1}{2}} )\)
\(=6\sqrt{3}-\frac{16}{3}\sqrt{2}.\)
Exercise \(\PageIndex{2}\)
\(\displaystyle\int_{0}^{1} x^3\sqrt{x^2+3}\, dx\)
- Answer
-
Let \(u= x^2+3\), then \(du=2xdx\) and \(x^2=u-3\). Now,
\(\displaystyle\int_{0}^{1} x^3\sqrt{x^2+3} \, dx = \displaystyle \frac{1}{2}\int_{0}^{1} x^2\sqrt{x^2+3}\, \,2xdx\)
\(= \displaystyle \frac{1}{2}\int_{0}^{1} (u-3) \sqrt{u}\, du)\)
\(= \displaystyle \frac{1}{2}\int_{0}^{1} (u^{\frac{3}{2}}-3 u^{\frac{-1}{2} }) \, du)\)
\(= \displaystyle \frac{1}{2} ( \frac{2}{5} u^{\frac{5}{2}}-3 \frac{2}{1} u^{\frac{1}{2} }) \left |_{x=0}^{x=1} \right.)\)
\(= \displaystyle \frac{1}{2} ( \frac{2}{5} (x^2+3)^{\frac{5}{2}}-3 \frac{2}{1} (x^2+3)^{\frac{1}{2} }) \left |_{x=0}^{x=1} \right.)\)
\(= \displaystyle \frac{1}{2} \left( \frac{2}{5} ( (1^2+3)^{\frac{5}{2}} -(0^2+3)^{\frac{5}{2}} \right)-6\left((1^2+3)^{\frac{1}{2} }-(0^2+3)^{\frac{1}{2} } \right) \)
\(= \displaystyle \frac{6\sqrt{3}}{5} -\frac{8}{5} \).
Exercise \(\PageIndex{3}\)
\(\displaystyle\int_{1}^{\sqrt{2}} \frac{x}{x^4+3}\, dx\)
- Answer
-
Let \(u= x^2\), then \(du=2xdx\). Now,
\(\displaystyle\int_{1}^{\sqrt{2}} \frac{x}{x^4+3}\, dx=\displaystyle \frac{1}{2}\int_{1}^{\sqrt{2}} \frac{1}{u^2+3}\, du\)
\( =\displaystyle \frac{1}{2 \sqrt{3}} \tan^{-1} \left( \frac{u}{\sqrt{3}} \right) \left|_{x=1}^{x=\sqrt{2}} \right.\)
\( =\displaystyle \frac{1}{2 \sqrt{3}} \tan^{-1} \left( \frac{x^2}{\sqrt{3}} \right) \left|_{x=1}^{x=\sqrt{2}} \right.\)
\( =\displaystyle \frac{1}{2 \sqrt{3}} \left(\tan^{-1} \left( \frac{(\sqrt{2})^2}{\sqrt{3}} \right)-\tan^{-1} \left( \frac{1^2}{\sqrt{3}}\right) \right) \)
\( =\displaystyle \frac{1}{2 \sqrt{3}} \left(\tan^{-1} \left( \frac{2}{\sqrt{3}} \right)-\tan^{-1} \left( \frac{1}{\sqrt{3}}\right) \right) \)
\( =\displaystyle \frac{1}{2 \sqrt{3}} \left(\tan^{-1} \left( \frac{2}{\sqrt{3}} \right)-\frac{\pi}{6} \right) \)
\( =\displaystyle \frac{\pi-6\tan^{-1} \left( \frac{2}{\sqrt{3}} \right)}{12\sqrt{3}}\).
Exercise \(\PageIndex{4}\)
\(\displaystyle\int_{1}^{\sqrt{2}} xe^{-x^{2}}\, dx\)
- Answer
-
Let \(u= -x^2\), then \(du=-2xdx\). Now,
\(\displaystyle\int_{1}^{\sqrt{2}} xe^{-x^{2}}\, dx= \displaystyle \frac{-1}{2}\int_{1}^{\sqrt{2} } e^u du\)
\( = \displaystyle \frac{-1}{2}e^u \left|_{x=1}^{x=\sqrt{2}} \right.\)
\( = \displaystyle \frac{-1}{2} e^{-x^{2}} \left|_{x=1}^{x=\sqrt{2}} \right.\)
\( = \displaystyle \frac{-1}{2} \left( e^{-(\sqrt{2})^{2}} - e^{-1}\right) \)
\( = \displaystyle \frac{-1}{2} \left( e^{-2} - e^{-1}\right) \)
\( = \displaystyle \frac{e-1}{2e^2}\).
Exercise \(\PageIndex{5}\)
\(\displaystyle\int \frac{\sin(3x)}{e^{1+\cos(3x)} }\, dx\).
- Answer
-
Let \(u=e^{-(1+\cos(3x))}\), then \(du= 3e^{-(1+\cos(3x))} \sin(3x) dx\). Now,
\(\displaystyle\int \frac{\sin(3x)}{e^{1+\cos(3x)} }\, dx= \frac{1}{3} \int du= \frac{u}{3}+C = \frac{e^{-(1+\cos(3x))}}{6} +C.\)