7.4: Homework- u-substitution
- Page ID
- 88687
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- As a review of the chain rule from derivatives, find \(\frac{d}{dx} e^{-x^2 + 1}\).
\(2x e^{x^2 + 1}\)ans
- Read through section 6A again and then read through section 6B.
- Watch the Khan Academy video u-substitution.
- Compute the following indefinite integral using the method of \(u\)-substitution.
\(\int (5x^4 + 2x) e^{x^5 + x^2}dx\)
\(e^{x^5 + x^2} + C\)ans - Watch another example of \(u\)-substitution: u-substitution 2.
- Compute the following indefinite integral using the method of \(u\)-substitution.
\(\int \frac{2x}{x^2 - 5}dx\)
\(\ln(x^2 - 5) + C\)ans - Reread the part about the chain rule shortcut for \(u\)-substitution in chapter 6 of the online notes, and reread Example 6B.2. Then try the following problems.
- \(\int e^{-3x}dx\).
\(\frac{1}{-3} e^{-3x} + C\)ans
- \(\int \left( \frac{1}{2}x - 1 \right)^4dx\)
\(.4 \left( \frac{1}{2} x - 1 \right)^5 + C\)ans
.
- \(\int_0^2 (5x - 3)^3dx\).
\(116\)ans
- \(\int \frac{1}{7x - 2}dx\)
\(\frac{1}{7} \ln(7x - 2) + C\)ans
- \(\int \frac{1}{\sqrt{0.5x - 1}}dx\)
\(4 \sqrt{0.5x - 1} + C\)ans
- \(\int e^{-3x}dx\).
- Try some more \(u\)-substitution integrals.
- \(\int (8x^3) (x^4 + 1)^2dx\)
- We see that \(u = x^4 + 1\) in this case.
- We see \(\frac{du}{dx} = 4x^3\), so \(dx = \frac{1}{4x^3} du\).
- We have
\[\begin{align*} & = \int (8x^3) u^2 \frac{1}{4x^3} du \\ & = \int 8 u^2 \frac{1}{4}du \\ & = 2 \int u^2du \end{align*}\]
- Integrating we have
\(2 \int u^2du = 2 \left( \frac{1}{3} u^3 \right ) + C = \frac{2}{3} u^3 + C\)
- Substitution of \(u = x^4 + 1\) yields
\(\frac{2}{3} (x^4 + 1)^3 + C\)
ans - \(\int (x^4 + 1)^2 (8x^3)dx\)
This is exactly the same as the previous problem, just written a different way. No need to redo work.ans
- \(\int (3x^2 - 1) e^{x^3 - x}dx\)
- We see that \(u = x^3 - x\) in this case.
- We see \(\frac{du}{dx} = 3x^2 - 1\), so \(dx = \frac{1}{3x^2 - 1} du\).
- We have
\[\begin{align*} & = \int (3x^2 - 1) e^u \frac{1}{3x^2 - 1} du \\ & = \int e^udu \end{align*}\]
- Integrating we have
\(\int e^udu = e^u + C\)
- Substitution of \(u = x^3 - x\) yields
\(e^{x^3 - x} + C\)
ans - \(\int (x^2 - \frac{1}{3}) e^{x^3 - x}dx\)
\(\frac{1}{3} e^{x^3 - x} + C\)ans
- \(\int (e^x+ x)^5 (e^x+1)dx\)
\(\frac{1}{6} (e^x + x)^6 + C\)ans
- \(\int \frac{2x}{x^2 - 1}dx\)
\(\ln(x^2 -1) + C\)ans
- \(\int e^x \sqrt{e^x + 1}dx\)
ans
- \(\int \frac{\sin(x)}{\cos(x)}dx\)
\(-\ln(\cos(x)) + C\)ans
- \(\int \frac{\sqrt{ \ln(x) }}{x}dx\)
ans
- \(\int \frac{x}{(x^2 - 5)^3}dx\)
- We see that \(u = x^2 - 5\) in this case.
- We see \(\frac{du}{dx} = 2x\), so \(dx = \frac{1}{2x} du\).
- We have
\[\begin{align*} & = \int \frac{x}{u^3}\frac{1}{2x}du \\ & = \int \frac{1}{2u^3}du \\ & = \int \frac{1}{2}u^{-3}du \end{align*}\]
- Integrating we have
\(\int \frac{1}{2}u^{-3}du = - \frac{1}{4} u^{-2} + C\)
- Substitution of \(u = x^2 - 5\) yields
\(- \frac{1}{4} (x^2 - 5)^{-2} + C\)
ans - \(\int \frac{\cos(x)}{\sqrt{\sin(x) + 1}}dx\)
\(2\sqrt{\sin(x) + 1} + C\)ans
- \(\int \frac{1}{x (2 \ln(x) + 1)^4}dx\).
\(-\frac{1}{6(2 \ln(x) + 1)^3} + C\)ans
- \(\int (e^{5x} + 1)^9 (e^{5x} )dx\).
\(\frac{1}{2} (e^{5x} + 1)^{10} + C\)ans
- \(\int (8x^3) (x^4 + 1)^2dx\)