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### Integration Exercises Worked

##### Contents

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

#### 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:

Sources:

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Category: Math Calculus Integration

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

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### August 05, 2017

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