4.7E: Exercises
- Page ID
- 13774
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 \(\PageIndex{1}\)
In the following exercises, use a change of variables to evaluate the definite integral.
1. \(\displaystyle∫^1_0x\sqrt{1−x^2}dx\)
2. \(\displaystyle∫^1_0\frac{x}{\sqrt{1+x^2}}dx\)
- Answer
-
\(\displaystyleu=1+x^2,du=2xdx,\frac{1}{2}∫^2_1u^{−1/2}du=\sqrt{2}−1\)
3. \(\displaystyle∫^2_0\frac{t}{\sqrt{5+t^2}}dt\)
4. \(\displaystyle∫^1_0\frac{t}{\sqrt{1+t^3}}dt\)
- Answer
-
\(\displaystyleu=1+t^3,du=3t^2,\frac{1}{3}∫^2_1u^{−1/2}du=\frac{2}{3}(\sqrt{2}−1)\)
5. \(\displaystyle∫^{π/4}_0sec^2θtanθdθ\)
6. \(\displaystyle∫^{π/4}_0\frac{sinθ}{cos^4θ}dθ\)
- Answer
-
\(\displaystyleu=cosθ,du=−sinθdθ,∫^1_{1/\sqrt{2}}u^{−4}du=\frac{1}{3}(2\sqrt{2}−1)\)
Exercise \(\PageIndex{2}\)
In the following exercises, evaluate the definite integral.
1. \(\displaystyle ∫^2_1\frac{1+2x+x^2}{3x+3x^2+x^3}dx\)
- Answer
-
\(\displaystyle \frac{1}{3}ln(\frac{26}{7})\)
2. \(\displaystyle ∫^{π/4}_0tanxdx\)
3. \(\displaystyle ∫^{π/3}_0\frac{sinx−cosx}{sinx+cosx}dx\)
- Answer
-
\(\displaystyle ln(\sqrt{3}−1)\)
4. \(\displaystyle ∫^{π/2}_{π/6}cscxdx\)
5. \(\displaystyle ∫^{π/3}_{π/4}cotxdx\)
- Answer
-
\(\displaystyle \frac{1}{2}ln\frac{3}{2}\)
Exercise \(\PageIndex{3}\)
In the following exercises, use a change of variables to show that each definite integral is equal to zero.
- \(\displaystyle∫^π_0cos^2(2θ)sin(2θ)dθ\)
- \(\displaystyle∫^\sqrt{π}_0tcos(t^2)sin(t^2)dt\)
- Answer
-
\(\displaystyle u=sin(t^2);\) the integral becomes \(\displaystyle\frac{1}{2}∫^0_0udu.\)
3. \(\displaystyle∫^1_0(1−2t)dt\)
4. \(\displaystyle∫^1_0\frac{1−2t}{(1+(t−\frac{1}{2})^2)}dt\)
- Answer
-
\(\displaystyleu=(1+(t−\frac{1}{2})^2);\) the integral becomes \(\displaystyle−∫^{5/4}_{5/4}\frac{1}{u}du\).
5. \(\displaystyle∫^π_0sin((t−\frac{π}{2})^3)cos(t−\frac{π}{2})dt\)
6. \(\displaystyle∫^2_0(1−t)cos(πt)dt\)
- Answer
-
\(\displaystyle∫^{−1}_1ucos(π(1−u))du=∫^{−1}_1u[cosπcosu−sinπsinu]du=−∫^{−1}_1ucosudu=∫^{1−}_1ucosudu=0\)
since the integrand is odd.
7. \(\displaystyle∫^{3π/4}_{π/4}sin^2tcostdt\)