Skip to content

Lars Eighner's Homepage


LarsWiki

calumus meretrix et gladio innocentis

Integration Exercises Worked


{$ \color{blue}{y=x^3} \text{ and } \color{red}{y= {x^4 \over 4}} $}

D80-1a. {$\displaystyle \int x^3 dx $}

{$$ \small \int x^n dx = {{x^{n+1}} \over {n+1}} dx, n \ne 1 $$}


{$$ \begin{align} \int x^3 dx &= {x^{3+1} \over {3+1}} +C \cr &={x^4 \over 4} + C\end{align} $$}

In this case, C is the y value of the vertex {$ (0,y) $}.

Check:

{$$ \begin{align} {d \over {dx}}\left[ {x^4 \over 4} + C \right] &= {d \over {dx}}\left[ {1 \over 4}x^4 + C \right] \cr &= {d \over {dx}}\left[ {1 \over 4}x^4 \right] + {d \over {dx}}C \cr &= 4{1 \over 4}x^3 + 0 \cr &= x^3 \end{align} $$}


{$ \color{blue}{y=2x-x^2} \text{ and } \color{red}{y= x^2-{x^3 \over 3}} $}
{$ \color{green}{y= x^2-{x^3 \over 3}+2}\text{ and } \color{orange}{y= x^2-{x^3 \over 3}-2} $}

D80-1b. {$\displaystyle \int (2x - x^2) dx $}

{$$\begin{align} \int (2x - x^2) dx &= \int 2x dx - \int x^2 dx \cr &= x^2 + c_0 - {x^3 \over 3} + c_1 \cr &= x^2 - {x^3 \over 3} + C\end{align}$$}

Since the function has zeros at {$(0,0) \text{ and } (0,2) $}, its integral has critial points at {$x=0 \text{ and } x=2$}.

Check

{$$ \begin{align} {d \over {dx}}\left[ x^2 - {x^3 \over 3}+C \right] = 2x - x^2 \end{align}$$}


{$ \color{blue}{y=1-t^4} \text{ and } \color{red}{y= t-{4 \over 5}t^5 }$}
{$ \color{green}{y= t-{4 \over 5}t^5+2}\text{ and } \color{orange}{y= t-{4 \over 5}t^5-2} $}

D80-1c. {$\displaystyle \int (1-4t^4) dt $}

{$$\begin{align} \int (1 - 4t^4) dt &= \int t^0 dt - \int 4t^4 dt \cr &= \int {t \over 1} dt - 4\int t^4 dt \cr &= t - 4{t^5 \over 5} + C \cr &= t-{4 \over 5}t^5 + C \end{align}$$}

Check

{$$ \begin{align} {d \over {dt}}\left[ t- {4 \over 5}t^5+C \right] = 1 - 5({4 \over 5}t^4 = 1 - 4t^4 \end{align}$$}

{$\displaystyle \large \int k dx = \int kx^0 dx =kx $}


{$ \color{blue}{y=1+x^2} \text{ and } \color{red}{y= x+{1 \over 3}x^3 }$}
{$ \color{green}{y= x+{1 \over 3}x^3+2}\text{ and } \color{maroon}{y= x+{1 \over 3}x^3-2} $}

D80-1d. {$\displaystyle \int (1 + y^2) dy $}

{$$ \begin{align} \int (1+y^2)dy &= \int 1 dy + \int y^3 dy \cr &= y + {1\over 3}y3 + C\end{align} $$}

Check

{$$ {d \over {dy}}(y + {1\over 3}y^3) = 1 + y^2 $$}


{$ \color{blue}{y={1 \over x^2}} \text{ and } \color{red}{y= -{1 \over x} }$}
{$ \color{green}{y= -{1 \over x}+2}\text{ and } \color{maroon}{y= -{1 \over x}-2} $}

D80-1e. {$\displaystyle \int {{dx} \over {x^2}} $}

{$$ \begin{align} \int {{dx} \over {x^2}} &= \int x^{-2}dx \cr &= {1 \over {-1}}x^{-1} \cr &= - {1 \over x} \end{align}$$}

Check:

{$$ \begin{align} {d \over {dx}}(-{1 \over x}) &= {d \over {dx}}(-x^{-1}) = -(-1)x^{-2} = {1 \over x^2} \end{align}$$}


{$ \color{blue}{y=\sqrt x -{2 \over {\sqrt x}}} $}

{$ \color{red}{y= ({2 \over 3})\sqrt{x^3} -4\sqrt x }$}
{$ \color{green}{y= ({2 \over 3})\sqrt{x^3} -4\sqrt x +2}$}
{$ \color{maroon}{y= ({2 \over 3})\sqrt{x^3} -4\sqrt x -2 }$}

D80-1f. {$\displaystyle \int (\sqrt x - {2 \over {\sqrt x}}) dx $}

{$$ \begin{align}\int (\sqrt x - {2 \over {\sqrt x}}) dx &= \int x^{1 \over 2}dx - 2\int x^{-{1 \over 2}}dx \cr &= {2 \over 3}x^{3/2} - 2(2x^{1 \over 2} + C \cr &= {2 \over 3}\sqrt{x^3} - 4\sqrt x + C\end{align} $$}

Check:

{$$ \begin{align} &{d \over {dx}}\left[ {2 \over 3}\sqrt{x^3} - 4\sqrt x + C \right] \cr &= {2 \over 3}({3 \over 2})x^{{3\over2}-1} - 4({1 \over 2})x^{{1 \over 2}-1} +0 \cr &= x^{1 \over 2} - 2x^{-{1 \over 2}} \cr &= \sqrt x - {2 \over {\sqrt x}}\end{align} $$}


{$ \color{blue}{y={1 \over x}} \text{ and } \color{red}{y= \ln |x| }$}
{$ \color{green}{y= (\ln |x|) +2}\text{ and } \color{maroon}{y= (\ln |x|)-2} $}

D80-2. {$\displaystyle \int {{dx} \over x} $}

{$$ \int {{dx} \over x} = \ln |x| + C $$}

Check:

{$$ {d \over {dx}} \ln |x| + C = {1 \over x} + 0 = {1 \over x} $$}

Finding integrals of products or quotients may be difficult, so express polynomials as the sum of terms when possible.


{$ \color{blue}{y=(1-x)(1+x^2)}$}
{$ \color{red}{y= -{1\over 4}x^4 + {1 \over 3}x^3 - {1 \over 2}x^2 + x }$}
{$ \color{green}{y= -{1\over 4}x^4 + {1 \over 3}x^3 - {1 \over 2}x^2 + x +2}$}
{$\color{maroon}{y= -{1\over 4}x^4 + {1 \over 3}x^3 - {1 \over 2}x^2 + x -2} $}

C98-1 {$\displaystyle \int (1-x)(1+x^2)dx $}

{$$ \begin{align}&\int (1-x)(1+x^2)dx = \int (-x^3 + x^2 -x+1)dx \cr &= -\int x^3dx +\int x^2dx -\int xdx + \int 1dx \cr &= -{1 \over 4}x^4 + {1 \over 3}x^3 - {1 \over 2}x^2 + x +C\end{align} $$}

Check: {$$ \begin{align} &{d \over {dx}}(-{1 \over 4}x^4 + {1 \over 3}x^3 - {1 \over 2}x^2 + x +C) \cr &= -(4){1 \over 4}x^{4-1} + (3){1 \over 3}x^{3-1} - (2){1 \over 2}x^{2-1} + 1x^{1-1} + 0 \cr &= -x^3 + x^2 - x + 1 \cr &= (1 - x)(1 + x^2)\end{align} $$}


{$ \color{blue}{y={{1 + 2x +3x^2} \over x^3}}$}
{$ \color{red}{y= -{1 \over 2x^2} - {2 \over x} + 3\ln|x|}$}
{$ \color{green}{y=-{1 \over 2x^2} - {2 \over x} + 3\ln|x| +2}$}
{$\color{maroon}{y= -{1 \over 2x^2} - {2 \over x} + 3\ln|x| -2} $}

C98-2. {$\displaystyle \int {{1+2x+3x^2} \over x^3}dx $}

{$$ \begin{align}&\int {{1+2x+3x^2} \over x^3}dx \cr &= \int {1 \over x^3} dx + \int {{2x} \over x^3}dx + \int {{3x^2} \over x^3}dx \cr &= \int {{dx} \over x^3} + 2\int {{dx} \over x^2} + 3\int {{dx} \over x} \cr &= -{1 \over 2}x^{-2} + (-2)x^{-1} + 3\ln|x|\cr &= -{1 \over 2x^2} - {2 \over x} + 3\ln|x| +C \end{align} $$}

Check: {$$ \begin{align}{d \over {dx}} \left[ -{1 \over 2x^2} - {2 \over x} + 3\ln|x| +C\right] &= {d \over {dx}} \left[ -{1 \over 2}x^{-2} - 2x^{-1} + 3\ln|x| +C\right] \cr &= (-2)(-{1 \over 2})x^{-3} -2(-1)x^{-2} +3x^{-1}\cr &= \left[ x^{-3} + 2x^{-2} +3x^{-1}\right] {x^3 \over x^3}\cr &= {{ 1 + 2x +3x^2} \over x^3}\end{align} $$}


C65-12. {$$ \color{blue}{ y = (a+b)^2} \text{ and } \color{red}{y = a^2x +abx^2 + {{ b^2x^3 } \over 3}} $$}
graph with sliders

C98-3. {$\displaystyle \int (a+bx)^2 dx $}

{$$ \begin{align} &\int (a+bx)^2 dx \cr &= \int (a^2 + 2abx + b^2x^2)dx \cr &= \int a^2dx + \int 2abxdx + \int b^2x^2 dx \cr &= a^2x + abx^2 + {{b^2x^3} \over 3} + C \end{align} $$}

Check:

{$$ \begin{align} &{d \over {dx}}(a^2x + abx^2 + {{b^2x^3} \over 3} + C) \cr &= a^2 + 2abx + b^2x^2 \cr &= (a +bx)^2 \end{align} $$}


C98-4. {$$ \color{blue}{ y = {{(3x-2)^2} \over x^2}}$$}

{$$ \color{red}{y= 9x -12\ln |x| -{4 \over x}} $$}{$$ \color{green}{y= 9x -12\ln |x| -{4 \over x}+4} $$}{$$ \color{maroon}{y= 9x -12\ln |x| -{4 \over x}-4} $$}

C98-4. {$\displaystyle \int {{(3x-2)^2} \over x^2} dx $}

{$$ \begin{align} &\int {{(3x-2)^2} \over x^2} dx \cr &= \int {{9x^2-12x+4}\over x^2}dx \cr &= 9\int 1dx - 12\int {{dx} \over x}dx +4 \int x^{-2}dx \cr &= 9x - 12\ln |x| +4(-1)x^{-1} +C \cr &= 9x - 12\ln |x| - {4 \over x} +C \end{align} $$}

Check:

{$$ \begin{align} &{d \over {dx}}\left( 9x - 12\ln |x| - {4 \over x} +C \right) \cr &= 9 - {{12} \over x} + {4 \over x^2} \cr &= {{9x^2} \over x^2} - {{12x} \over x^2} + {4 \over x^2} \cr &= {{9x^2 -12x + 4}\over x^2} \cr &= {{(3x-2)^2} \over x^2} \end{align} $$}


C98-5. {$$ \color{blue}{ y = (e^x - e^{-x})^2}$$}{$$ \color{red}{y = {{e^{2x} - e^{-2x}} \over 2} -2x } $$}{$$ \color{green}{y = {{e^{2x} - e^{-2x}} \over 2} -2x +2} $$}{$$ \color{maroon}{y = {{e^{2x} - e^{-2x}} \over 2} -2x -2} $$}

C98-5. {$\displaystyle \int (e^x - e^{-x})^2dx $}

{$$ \begin{align} &\int (e^x - e^{-x})^2dx \cr &= \int (e^{2x} - 2e^{x}e^{-x} + e^{-2x})dx \cr &= \int e^{2x}dx - \int 2dx + \int e^{-2x}dx \cr &= {1 \over 2}e^{2x} - 2x + {{-1} \over 2}e^{-2x} + C \cr &= {{e^{2x} - e^{-2x}} \over 2} - 2x + C\end{align} $$}

Check:

{$$ \begin{align} &{d \over {dx}} \left( {{e^{2x} - e^{-2x}} \over 2} - 2x + C \right) \cr &= {d \over {dx}}{1 \over 2}(e^{2x}) + {d \over {dx}}({{-1} \over 2}e^{-2x}) - {d \over {dx}}(2x) + {d \over {dx}}C \cr &= e^{x^2} + e^{-x^2} - 2e^xe^{-x}\cr &= e^{x^2} - 2e^xe^{-x} + e^{-x^2} = (e^x - e^{-x})^2 \end{align} $$}


C98-6. {$$ \color{blue}{ y = {{x^{3 \over 2} -4x^{1 \over 3}} \over x}}$$}

{$$ \color{red}{y= {2 \over 3}x^{3 \over 2} - 4x^{1 \over 3}} $$}{$$ \color{green}{y= {2 \over 3}x^{3 \over 2} - 4x^{1 \over 3} +2} $$}{$$ \color{maroon}{y= {2 \over 3}x^{3 \over 2} - 4x^{1 \over 3}-2} $$}

C98-6. {$\displaystyle \int {{x^{3 \over 2} -4x^{1 \over 3}} \over x}dx $}

{$$\begin{align} & \int {{x^{3 \over 2} -4x^{1 \over 3}} \over x}dx \cr &= \int {{x^{3 \over 2}} \over x}dx - 4 \int {{x^{1 \over 3}} \over x}dx \cr &= \int x^{1 \over 2}dx - 4 \int x^{-{2 \over 3}}dx \cr &= {{x^{3 \over 2}}\over {3 \over 2}} - 4 \left( {{x^{1 \over 3}} \over {1 \over 3}} \right) +C \cr &= ({2 \over 3})x^{3 \over 2} - 4(3)x^{1 \over 3} +C \cr &= ({2 \over 3})\sqrt {x^3} - 12\sqrt[3]x +C \end{align}$$}

Check:

{$$\begin{align} & {d \over {dx}}\left[ ({2 \over 3})\sqrt {x^3} - 12\sqrt[3]x +C \right] \cr &= {d \over {dx}}\left[ ({2 \over 3})\sqrt {x^3} \right] - {d \over {dx}}\left[ 12\sqrt[3]x \right] + 0 \cr &= {d \over {dx}}\left[ ({2 \over 3})x^{3 \over 2}\right] - 4{d \over {dx}}\left[ (3)x^{1 \over 3} \right] \cr &= ({2 \over 3})({3 \over 2})x^{1 \over 2} - 4\left[ (3)({1 \over 3})x^{-{2 \over 3}} \right] \cr &= \left[x^{1 \over 2} - 4 x^{-{2 \over 3}} \right]{x \over x} \cr &= {{x^{3 \over 2} - 4 x^{1 \over 3}} \over x}\end{align}$$}


C98-7. {$$ \color{blue}{ y = (1-2x)^2\sqrt x}$$}

{$$ \color{red}{y={2 \over 3}x^{3 \over 2} - {8 \over 5}x^{5 \over 2} +{8 \over 7}x^{7 \over 2} } $$}{$$ \color{green}{y= {2 \over 3}x^{3 \over 2} - {8 \over 5}x^{5 \over 2} +{8 \over 7}x^{7 \over 2}+2} $$}{$$ \color{maroon}{y= {2 \over 3}x^{3 \over 2} - {8 \over 5}x^{5 \over 2} +{8 \over 7}x^{7 \over 2}-2} $$}

C98-7. {$\displaystyle \int (1-2x)^2\sqrt x dx $}

{$$ \begin{align} &\int (1-2x)^2\sqrt x dx \cr &= \int (1-4x +4x^2)\sqrt x dx \cr &= \int (x^{1 \over 2} -4x^{3 \over 2} +4x^{5 \over 2}) dx \cr &= \int x^{1 \over 2}dx -4 \int x^{3 \over 2}dx +4 \int x^{5 \over 2} dx \cr &= {{x^{3 \over 2}}\over {3 \over 2}} -(4) {{x^{5 \over 2}} \over {5 \over 2}} +(4) {{x^{7 \over 2}} \over {7 \over 2}} + C \cr &= {2 \over 3}x^{3 \over 2} -(4){2 \over 5}x^{5 \over 2} +(4){2 \over 7}x^{7 \over 2} + C \cr &= {2 \over 3}x^{3 \over 2} - {8 \over 5}x^{5 \over 2} +{8 \over 7}x^{7 \over 2} + C \cr \end{align} $$}

Check: {$$ \begin{align} &{d \over {dx}}\left( {2 \over 3}x^{3 \over 2} - {8 \over 5}x^{5 \over 2} +{8 \over 7}x^{7 \over 2} + C \right) \cr &= {3 \over 2}({2 \over 3})x^{1 \over 2} - {5 \over 2}({8 \over 5})x^{3 \over 2} +{2 \over 2}({8 \over 7})x^{5 \over 2} \cr &= x^{1 \over 2} - 4x^{3 \over 2} +4x^{5 \over 2} \cr &= (1 - 4x +4x^2)x^{1 \over 2} \cr &= (1 - 2x)^2\sqrt x \cr \end{align} $$}


C98-8. {$$ \color{blue}{ y =(x^3 - 2)(x^{1 \over 2} + x^{2 \over 3}) }$$}

{$$ \color{red}{y={3 \over 14}x^{14 \over 3} + {2 \over 9}x^{9 \over 2} -{6 \over 5}x^{5 \over 3} -{4 \over 3} x^{3 \over 2} } $$}{$$ \color{green}{y= {3 \over 14}x^{14 \over 3} + {2 \over 9}x^{9 \over 2} -{6 \over 5}x^{5 \over 3} -{4 \over 3} x^{3 \over 2}+2} $$}{$$ \color{maroon}{y= {3 \over 14}x^{14 \over 3} + {2 \over 9}x^{9 \over 2} -{6 \over 5}x^{5 \over 3} -{4 \over 3} x^{3 \over 2} -2} $$}

C98-8. {$\displaystyle \int (x^3 - 2)(x^{1 \over 2} + x^{2 \over 3}) dx $}

{$$ \begin{align} & \int (x^3 - 2)(x^{1 \over 2} + x^{2 \over 3})dx \cr &= \int \left( x^{11 \over 3} + x^{7 \over 2} -2x^{2 \over 3} -2x^{1 \over 2} \right) dx \cr &= \int x^{11 \over 3}dx + \int x^{7 \over 2}dx -2 \int x^{2 \over 3}dx -2 \int x^{1 \over 2}dx \cr &= {{x^{14 \over 3}}\over {14 \over 3}} + {{x^{9 \over 2}} \over {9 \over 2}} -(2) {{x^{5 \over 3}} \over {5 \over 3}} -(2){{ x^{3 \over 2}} \over {3 \over 2}} +C \cr &= {3 \over 14}x^{14 \over 3} + {2 \over 9}x^{9 \over 2} -{6 \over 5}x^{5 \over 3} -{4 \over 3} x^{3 \over 2} +C \cr \end{align} $$}

Check: {$$ \begin{align} &{d \over {dx}}\left( {3 \over 14}x^{14 \over 3} + {2 \over 9}x^{9 \over 2} -{6 \over 5}x^{5 \over 3} -{4 \over 3} x^{3 \over 2} +C \right) \cr &= {14 \over 3}({3 \over 14})x^{11 \over 3} + {9 \over 2}({2 \over 9})x^{7 \over 2} -{5 \over 3}({6 \over 5})x^{2 \over 3} -{3 \over 2} ( {4 \over 3} ) x^{1 \over 2} + 0 \cr &= x^{11 \over 3} + x^{7 \over 2} -2x^{2 \over 3} -2 x^{1 \over 2} \cr &= (x^3 -2)( x^{1 \over 2}+ x^{2 \over 3}) \cr \end{align} $$}


Graphs:

  1. FooPlot: Online graphing calculator and function plotter
  2. FooPlot: Online graphing calculator and function plotter
  3. FooPlot: Online graphing calculator and function plotter
  4. FooPlot: Online graphing calculator and function plotter
  5. FooPlot: Online graphing calculator and function plotter
  6. FooPlot: Online graphing calculator and function plotter
  7. FooPlot: Online graphing calculator and function plotter
  8. FooPlot: Online graphing calculator and function plotter
  9. FooPlot: Online graphing calculator and function plotter
  10. desmos.com
  11. FooPlot: Online graphing calculator and function plotter
  12. FooPlot: Online graphing calculator and function plotter
  13. FooPlot: Online graphing calculator and function plotter
  14. FooPlot: Online graphing calculator and function plotter
  15. FooPlot: Online graphing calculator and function plotter

Sources:

Recommended:

Category: Math Calculus Integration


Read or Post Comments

No comments yet.

This is a student's notebook. I am not responsible if you copy it for homework, and it turns out to be wrong.

Figures are often enhanced by hand editing; the same results may not be achieved with source sites and source apps.

Backlinks

This page is MathJaydenExercisesWorked

August 05, 2017

  • HomePage
  • WikiSandbox

Lars

Contact by Snail!

Lars Eighner
APT 1191
8800 N IH 35
AUSTIN TX 78753
USA

Help

HOME

The best way to look for anything in LarsWiki is to use the search bar.

Page List

Categories

Physics Pages

Math Pages

Math Exercises

Math Tools

Sections