Table Practice

Integration by Parts Problem

Solution

Find the indefinite integral:

 

\(\int{{x{{e}^{{-x}}}\,dx}}\)

 

 

Since \(dv={{e}^{{-x}}}\,dx\) is the most complicated part, and \(u=x\) has a very simple derivative, pick \(u=x\). Also, if we use LIATE, an algebraic function comes before an exponential one for picking u, so again it makes sense to pick \(u=x\).

\(\begin{align}{l}u=x\\du=1dx=dx\end{array}\)         \(\begin{array}{l}dv={{e}^{{-x}}}\,dx\\v=\int{{dv=\int{{{{e}^{{-x}}}\,dx}}}}=-{{e}^{{-x}}}\end{align}\)   

                   

\(\displaystyle \begin{align}{c}\int{{u\,dv}}=uv-\int{{v\,du}}\\\int{{x{{e}^{{-x}}}\,dx=x\cdot -{{e}^{{-x}}}-}}\,\int{{-{{e}^{{-x}}}}}dx=-{{e}^{{-x}}}x-\left( {{{e}^{{-x}}}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=-x{{e}^{{-x}}}-{{e}^{{-x}}}+C=-{{e}^{{-x}}}\left( {x+1} \right)+C\end{align}\)$      

Find the indefinite integral:

 

\(\int{{x\sqrt{{x-4}}\,}}dx\)

 

 

Since the second part is more complicated, let’s try to set that to dv, and set the x to u:

\(\begin{align}{l}u=x\\du=1dx=dx\end{array}\)        \(\displaystyle \begin{array}{l}dv=\sqrt{{x-4}}\,dx={{\left( {x-4} \right)}^{{\frac{1}{2}}}}dx\\v=\int{{dv=\int{{{{{\left( {x-4} \right)}}^{{\frac{1}{2}}}}dx}}}}=\frac{2}{3}{{\left( {x-4} \right)}^{{\frac{3}{2}}}}\end{align}\)

 

 

\(\displaystyle \begin{align}{c}\,\int{{u\,dv}}=uv-\int{{v\,du}}\\\int{{x\sqrt{{x-4}}\,dx=}}\,x\cdot \frac{2}{3}{{\left( {x-4} \right)}^{{\frac{3}{2}}}}-\int{{\frac{2}{3}\,{{{\left( {x-4} \right)}}^{{\frac{3}{2}}}}dx=\frac{{2x}}{3}{{{\left( {x-4} \right)}}^{{\frac{3}{2}}}}-\frac{2}{3}\cdot \frac{2}{5}\,{{{\left( {x-4} \right)}}^{{\frac{5}{2}}}}+C}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{{2x}}{3}{{\left( {x-4} \right)}^{{\frac{3}{2}}}}-\frac{4}{{15}}\,{{\left( {x-4} \right)}^{{\frac{5}{2}}}}+C=\frac{2}{{15}}{{\left( {x-4} \right)}^{{\frac{3}{2}}}}\left[ {5x-2\left( {x-4} \right)} \right]+C\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{2}{{15}}{{\left( {x-4} \right)}^{{\frac{3}{2}}}}\left( {3x+8} \right)+C\end{align}\)

 

 

 

Find the indefinite integral: 

 

\(\int{{\frac{{x{{e}^{x}}}}{{{{{\left( {x+1} \right)}}^{2}}}}}}\,dx\)

 

Let’s try to make the \(\frac{1}{{{{{\left( {x+1} \right)}}^{2}}}}dx\) the dv part, since we can easily take the integral of this. (Remember if the choices for u and dv don’t work out the first time you try it, you can try to rearrange differently and try again.)

 

\(\begin{array}{l}u=x{{e}^{x}}\\du=\left( {x\cdot {{e}^{x}}+{{e}^{x}}\cdot 1} \right)dx\\={{e}^{x}}\left( {x+1} \right)dx\end{array}\)           \(\displaystyle \begin{array}{l}dv=\int{{\frac{1}{{{{{\left( {x+1} \right)}}^{2}}}}}}\,dx={{\left( {x+1} \right)}^{{-2}}}dx\\v=\int{{dv=\int{{{{{\left( {x+1} \right)}}^{{-2}}}}}dx}}=-{{\left( {x+1} \right)}^{{-1}}}\end{array}\)

 

\(\begin{array}{c}\int{{u\,dv}}=uv-\int{{v\,du}}\\\int{{\frac{{x{{e}^{x}}}}{{{{{\left( {x+1} \right)}}^{2}}}}}}\,dx\,=\left[ {x{{e}^{x}}\cdot -{{{\left( {x+1} \right)}}^{{-1}}}} \right]-\int{{\left[ {-{{{\left( {x+1} \right)}}^{{-1}}}\cdot {{e}^{x}}\left( {x+1} \right)} \right]dx}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{{-x{{e}^{x}}}}{{x+1}}-\int{{\left[ {-\frac{{{{e}^{x}}\left( {x+1} \right)}}{{x+1}}} \right]dx}}=\frac{{-x{{e}^{x}}}}{{x+1}}+\int{{{{e}^{x}}dx=}}\frac{{-x{{e}^{x}}}}{{x+1}}+{{e}^{x}}+C\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{{-x{{e}^{x}}}}{{x+1}}+\frac{{{{e}^{x}}\left( {x+1} \right)}}{{x+1}}+C=\frac{{-x{{e}^{x}}+x{{e}^{x}}+{{e}^{x}}}}{{x+1}}+\frac{{{{e}^{x}}\left( {x+1} \right)}}{{x+1}}=\frac{{{{e}^{x}}}}{{x+1}}+C\end{array}\)

Integration by Parts

 

For functions u and v that are functions of x with continuous derivatives, then:

                            \(\int{{u\,dv=uv\,\,-}}\,\int{{v\,}}du\)

 

\[\begin{array}{l}\int {u\,dv = uv – } \,\int {v\,} du\\\int {{x^2}\sin x\,dx = {x^2} \cdot – \cos x – } \,\int { – \cos x \cdot 2x\,} dx\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = – {x^2}\cos x + \int {2x\cos x\,} dx\,\,\,\,\,\,\end{array}\]

 

 

Integration Area Problem

Integration Area Problem

Sketch the region bounded by the graphs and find the area, with respect to y:

\(f\left( y \right)=y\left( {4-y} \right),\,\,\,\,g\left( y \right)=-y\)

 

Solution:  Find points of intersection: \(y\left( {4-y} \right)=-y;\,\,\,\,4y-{{y}^{2}}+y=0;\,\,\,y\left( {5-y} \right)=0;\,\,\,y=0,\,5\), so points of intersection are \((-5,\,5)\) and \((0,\,0)\).

 

Integration By Parts Twice

Now let’s graph.  If you’re not sure how to graph, you can always make t-charts.
Remember we go down to up for the interval, and right to left for the subtraction of functions:

 

 

Now let’s integrate:

\(\begin{array}{l}\int\limits_{0}^{5}{{\left[ {\left( {4y-{{y}^{2}}} \right)-\left( {-y} \right)} \right]dy}}=\int\limits_{0}^{5}{{\left( {5y-{{y}^{2}}} \right)dy}}\\\,\,\,=\left[ {\frac{5}{2}{{y}^{2}}-\frac{1}{3}{{y}^{3}}} \right]_{0}^{5}=\left( {\frac{5}{2}{{{\left( 5 \right)}}^{2}}-\frac{1}{3}{{{\left( 5 \right)}}^{3}}} \right)-0=\frac{{125}}{6}\end{array}\)

Sketch the region bounded by the graphs and find the area, with respect to y:

 

\(f\left( y \right)={{y}^{2}}+2,\,\,\,g\left( y \right)=0,\,\,\,y=-1,\,\,\,y=2\)

 

Solution:  Find points of intersection.  We’ll actually end up using y = –1 and y = 2 for the limits of integration, as we can see from graphing below. 

 

Now let’s graph.  If you’re not sure how to graph, you can always make t-charts.

 

Remember we go down to up for the interval, and right to left for the subtraction of functions:

 

Now let’s integrate:

\(\begin{array}{l}\int\limits_{0}^{5}{{\left[ {\left( {4y-{{y}^{2}}} \right)-\left( {-y} \right)} \right]dy}}=\int\limits_{0}^{5}{{\left( {5y-{{y}^{2}}} \right)dy}}\\\,\,\,=\left[ {\frac{5}{2}{{y}^{2}}-\frac{1}{3}{{y}^{3}}} \right]_{0}^{5}=\left( {\frac{5}{2}{{{\left( 5 \right)}}^{2}}-\frac{1}{3}{{{\left( 5 \right)}}^{3}}} \right)-0=\frac{{125}}{6}\end{array}\)

 

 

 

 

 

 

 

   

. \(y\left( {4-y} \right)=-y;\,\,\,\,4y-{{y}^{2}}+y=0;\,\,\,y\left( {5-y} \right)=0;\,\,\,y=0,\,5\), so points of intersection are \((-5,\,5)\) and \((0,\,0)\).

\(f\left( y \right)=y\left( {4-y} \right),\,\,\,\,g\left( y \right)=-y\)

Now let’s graph.  If you’re not sure how to graph, you can always make t-charts.
Remember we go down to up for the interval, and right to left for the subtraction of functions:

 

 

Now let’s integrate:

 

Sketch the region bounded by the graphs and find the area, with respect to y:

 

 

 

Solution:  Find points of intersection.  We’ll actually end up using y = –1 and y = 2 for the limits of integration, as we can see from graphing below. 

 

Now let’s graph.  If you’re not sure how to graph, you can always make t-charts.

 

Remember we go down to up for the interval, and right to left for the subtraction of functions:

 

Now let’s integrate:

 

\begin{align}
\int {u\,dv}& = uv – \int {v\,du} \\
\int {\frac{{x{e^x}}}{{{{\left( {x + 1} \right)}^2}}}} \,dx\, &= \left[ {x{e^x} \cdot – {{\left( {x + 1} \right)}^{ – 1}}} \right] – \int {\left[ { – {{\left( {x + 1} \right)}^{ – 1}} \cdot {e^x}\left( {x + 1} \right)} \right]dx} \\
\,\,\,\,& = \frac{{ – x{e^x}}}{{x + 1}} – \int {\left[ { – \frac{{{e^x}\left( {x + 1} \right)}}{{x + 1}}} \right]dx} = \frac{{ – x{e^x}}}{{x + 1}} + \int {{e^x}dx = } \frac{{ – x{e^x}}}{{x + 1}} + {e^x} + C \\
\,\,\,\, &= \frac{{ – x{e^x}}}{{x + 1}} + \frac{{{e^x}\left( {x + 1} \right)}}{{x + 1}} + C = \frac{{ – x{e^x} + x{e^x} + {e^x}}}{{x + 1}} \\
\end{align}

\[\begin{gathered}This is {\color{red} highlighted, and this is not.} \int {u\,dv = uv – } \,\\{v\,} du \hfill \\ \int {{x^2}\sin x\,dx = {x^2} \cdot – \cos x – } \,\int { – \cos x \cdot 2x\,} dx \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = – {x^2}\cos x + \int {2x\cos x\,} dx\,\,\,\,\,\, \hfill \\ \end{gathered} \]

 

 

\[\begin{align}{c}\int {{x^4}} {e^{ – 2x}}dx &= {x^4} \cdot – \frac{1}{2}{e^{ – 2x}} – 4{x^3} \cdot \frac{1}{4}{e^{ – 2x}} + 12{x^2} \cdot – \frac{1}{8}{e^{ – 2x}} – 24x \cdot \frac{1}{{16}}{e^{ – 2x}} + 24 \cdot – \frac{1}{{32}}{e^{ – 2x}} + C\\ &= – \frac{1}{2}{x^4}{e^{ – 2x}} – {x^3}{e^{ – 2x}} – \frac{3}{2}{x^2}{e^{ – 2x}} – \frac{3}{2}x{e^{ – 2x}} + – \frac{3}{4}{e^{ – 2x}} + C\\ = – \frac{1}{4}{e^{ – 2x}}\left( {2{x^4} + 4{x^3} + 6{x^2} + 6x + 3} \right) + C\end{align}\]

Integration Problem

Solution

 

 

 

 

Find the indefinite integral:

\[\int {{x^2}\sin x\,} dx\]

 

 

 

Since trig functions usually follow algebraic functions in determining u, let’s pick  and :

 

\(\begin{align}u &= {x^2}\\du &= 2x\,dx\end{align}\)         \(\begin{align}dv &= \sin x\,dx\\v &= \int {dv = \int {\sin x\,dx} } = – \cos x\end{align}\)

\[\begin{align}\int {u\,dv} &= \int {uv} \,\,\, – \,\int {v\,} du\\\int {{x^2}\sin x\,dx} &= {x^2} \cdot – \cos x\,\, – \,\int { – \cos x \cdot 2x\,} dx\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= {\color{red}{- {x^2}\cos x} } +{\color{blue}{\,\int {2x\cos xdx}}}\,\,\end{align}\]

 

Integration by parts second time:

\(\begin{align}u &= {\color{blue}{2x}}\\du &= 2\,dx\end{align}\)     \(\begin{align}dv &= {\color{blue}{\cos x\,dx}}\\v &= \int {dv = \int {\cos x\,dx} } = \sin x\end{align}\)

 

\[\begin{align}
\int {u\,dv} &= \,uv\,\, – \int {vdu\,} \\
{\color{blue}{ \int {2x\cos xdx}}} &= 2x \cdot \sin x\,\, – \int {\sin x \cdot 2dx} = 2x\sin x – 2( – \cos x) + C\, \\
&= {\color{blue}{\,2x\sin x + 2\cos x + C}} \\
\end{align} \]

 

Now put it all together:

\[\int {{x^2}\sin x\,} dx = {\color{red}{- {x^2}\cos x} } + {\color{blue}{\,2x\sin x + 2\cos x + C}}\, = \left( {2 – {x^2}} \right)\cos x + 2x\sin x + C\]