Skip to main content
Mathematics LibreTexts

4.8E: AntiDerivative & Indefinite Integral Exercises

  • Page ID
    13802
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)

    4.8: Antiderivatives

    For the following exercises, find the antiderivative \(F(x)\) of each function \(f(x).\)

    470) \(f(x)=\frac{1}{x^2}+x\)

    471) \(f(x)=e^x−3x^2+sinx\)

    Answer:
    \(F(x)=e^x−x^3−cos(x)+C\)

    472) \(f(x)=e^x+3x−x^2\)

    473) \(f(x)=x−1+4sin(2x)\)

    Answer:
    \(F(x)=\frac{x^2}{2}−x−2cos(2x)+C\)

    474) \(f(x)=5x^4+4x^5\)

    475) \(f(x)=x+12x^2\)

    Answer:
    \(F(x)=\frac{1}{2}x^2+4x^3+C\)

    476) \(f(x)=\frac{1}{\sqrt{x}}\)

    477) \(f(x)=(\sqrt{x})^3\)

    Answer:
    \(F(x)=\frac{2}{5}(\sqrt{x})^5+C\)

    478) \(f(x)=x^{1/3}+(2x)^{1/3}\)

    479) \(f(x)=\frac{x^{1/3}}{x^{2/3}}\)

    Answer:
    \((F(x)=\frac{3}{2}x^{2/3}+C\)

    480) \(f(x)=2sin(x)+sin(2x)\)

    481) \(f(x)=sec^2(x)+1\)

    Answer:
    (F(x)=x+tan(x)+C\)

    482) \(f(x)=sinxcosx\)

    483) \(f(x)=sin^2(x)cos(x)\)

    Answer:
    \(F(x)=\frac{1}{3}sin^3(x)+C\)

    484) \(f(x)=0\)

    485) \(f(x)=\frac{1}{2}csc^2(x)+\frac{1}{x^2}\)

    Answer:
    \(F(x)=−\frac{1}{2}cot(x)−\frac{1}{x}+C\)

    486) \(f(x)=cscxcotx+3x\)

    487) \(f(x)=4cscxcotx−secxtanx\)

    Answer:
    \(F(x)=−secx−4cscx+C\)

    488) \(f(x)=8secx(secx−4tanx)\)

    489) \(f(x)=\frac{1}{2}e^{−4x}+sinx\)

    Answer:
    \(F(x)=−\frac{1}{8}e^{−4x}−cosx+C\)

    For the following exercises, evaluate the integral.

    490) \(∫(−1)dx\)

    491) \(∫sinxdx\)

    Answer:
    \(−cosx+C\)

    492) \(∫(4x+\sqrt{x})dx\)

    493) \(∫\frac{3x^2+2}{x^2}dx\)

    Answer:
    \(3x−\frac{2}{x}+C\)

    494) \(∫(secxtanx+4x)dx\)

    495) \(∫(4\sqrt{x}+\sqrt[4]{x})dx\)

    Answer:
    \(\frac{8}{3}x^{3/2}+\frac{4}{5}x^{5/4}+C\)

    496) \(∫(x^{−1/3}−x^{2/3})dx\)

    497) \(∫\frac{14x^3+2x+1}{x^3}dx\)

    Answer:
    \(14x−\frac{2}{x}−\frac{1}{2x^2}+C\)

    498) \(∫(e^x+e^{−x})dx\)

    For the following exercises, solve the initial value problem.

    499) \(f′(x)=x^{−3},f(1)=1\)

    Answer:
    \(f(x)=−\frac{1}{2x^2}+\frac{3}{2}\)

    500) \(f′(x)=\sqrt{x}+x^2,f(0)=2\)

    501) \(f′(x)=cosx+sec^2(x),f(\frac{π}{4})=2+\frac{\sqrt{2}}{2}\)

    Answer:
    \(f(x)=sinx+tanx+1\)

    502) \(f′(x)=x^3−8x^2+16x+1,f(0)=0\)

    503) \(f′(x)=\frac{2}{x^2}−\frac{x^2}{2},f(1)=0\)

    Answer:
    \(f(x)=−\frac{1}{6}x^3−\frac{2}{x}+\frac{13}{6}\)

    J4.8.1) not here yet

    Answer:
    \(8\)

    J4.8.2) not here yet

    J4.8.3) not here yet

    Answer:
    \(8\)

    For the following exercises, find two possible functions \(f\) given the second- or third-order derivatives

    504) \(f''(x)=x^2+2\)

    505) \(f''(x)=e^{−x}\)

    Solution: Answers may vary; one possible answer is \(f(x)=e^{−x}\)

    506) \(f''(x)=1+x\)

    507) \(f'''(x)=cosx\)

    Solution: Answers may vary; one possible answer is \(f(x)=−sinx\)

    508) \(f'''(x)=8e^{−2x}−sinx\)

    509) A car is being driven at a rate of \(40\) mph when the brakes are applied. The car decelerates at a constant rate of \(10\) \(ft/sec^2\). How long before the car stops?

    Solution: \(5.867\) sec

    510) In the preceding problem, calculate how far the car travels in the time it takes to stop.

    511) You are merging onto the freeway, accelerating at a constant rate of \(12\) ft/sec2. How long does it take you to reach merging speed at \(60\) mph?

    Solution: \(7.333\) sec

    512) Based on the previous problem, how far does the car travel to reach merging speed?

    513) A car company wants to ensure its newest model can stop in \(8\) sec when traveling at \(75\) mph. If we assume constant deceleration, find the value of deceleration that accomplishes this.

    Solution: \(13.75 ft/sec^2\)

    514) A car company wants to ensure its newest model can stop in less than \(450\) ft when traveling at \(60\) mph. If we assume constant deceleration, find the value of deceleration that accomplishes this.

    For the following exercises, find the antiderivative of the function, assuming \(F(0)=0.\)

    515) [T] \(f(x)=x^2+2\)

    Solution: \(F(x)=\frac{1}{3}x^3+2x\)

    516) [T] \(f(x)=4x−\sqrt{x}\)

    517) [T] \(f(x)=sinx+2x\)

    Solution: \(F(x)=x^2−cosx+1\)

    518) \([T] f(x)=e^x\)

    519) \([T] f(x)=\frac{1}{(x+1)^2}\)

    Solution: \(F(x)=−\frac{1}{(x+1)}+1\)

    520) [T] \(f(x)=e^{−2x}+3x^2\)

    For the following exercises, determine whether the statement is true or false. Either prove it is true or find a counterexample if it is false.

    521) If \(f(x)\) is the antiderivative of \(v(x)\), then \(2f(x)\) is the antiderivative of \(2v(x).\)

    Solution: True

    522) If \(f(x)\) is the antiderivative of \(v(x)\), then \(f(2x)\) is the antiderivative of \(v(2x).\)

    523) If \(f(x)\) is the antiderivative of \(v(x),\) then \(f(x)+1\) is the antiderivative of \(v(x)+1.\)

    Solution: False

    524) If \(f(x)\) is the antiderivative of \(v(x)\), then \((f(x))^2\) is the antiderivative of \((v(x))^2.\)


    4.8E: AntiDerivative & Indefinite Integral Exercises is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by LibreTexts.

    • Was this article helpful?