Calculus II - Integration by Parts

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Section 7.1 : Integration by Parts

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2. Evaluate \( \displaystyle \int_{6}^{0}{{\left( {2 + 5x} \right){{\bf{e}}^{\frac{1}{3}x}}\,dx}}\) .

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The first step here is to pick \(u\) and \(dv\). We want to choose \(u\) and \(dv\) so that when we compute \(du\) and \(v\) and plugging everything into the Integration by Parts formula the new integral we get is one that we can do.

With that in mind it looks like the following choices for \(u\) and \(dv\) should work for us.

\[u = 2 + 5x\hspace{0.5in}dv = {{\bf{e}}^{\frac{1}{3}x}}\,dx\] Show Step 2

Next, we need to compute \(du\) (by differentiating \(u\)) and \(v\) (by integrating \(dv\)).

\[\begin{align*}u & = 2 + 5x & \hspace{0.5in} & \to & \hspace{0.25in}du & = 5dx\\ dv & = {{\bf{e}}^{\frac{1}{3}x}}\,dx & \hspace{0.5in} & \to & \hspace{0.25in}v & = 3{{\bf{e}}^{\frac{1}{3}x}}\end{align*}\] Show Step 3

We can deal with the limits as we do the integral or we can just do the indefinite integral and then take care of the limits in the last step. We will be using the later way of dealing with the limits for this problem.

So, plugging \(u\), \(du\), \(v\) and \(dv\) into the Integration by Parts formula gives,

\[ \int{{\left( {2 + 5x} \right){{\bf{e}}^{\frac{1}{3}x}}}} = \left( {2 + 5x} \right)\left( {3{{\bf{e}}^{\frac{1}{3}x}}} \right) - \int{{5\left( {3{{\bf{e}}^{\frac{1}{3}x}}} \right)\,dx}} = 3{{\bf{e}}^{\frac{1}{3}x}}\left( {2 + 5x} \right) - 15\int{{{{\bf{e}}^{\frac{1}{3}x}}\,dx}}\] Show Step 4

Okay, the new integral we get is easily doable so let’s evaluate it to get,

\[ \int{{\left( {2 + 5x} \right){{\bf{e}}^{\frac{1}{3}x}}}} = 3{{\bf{e}}^{\frac{1}{3}x}}\left( {2 + 5x} \right) - 45{{\bf{e}}^{\frac{1}{3}x}} + c = 15x{{\bf{e}}^{\frac{1}{3}x}} - 39{{\bf{e}}^{\frac{1}{3}x}} + c\] Show Step 5

The final step is then to take care of the limits.

\[ \int_{6}^{0}{{\left( {2 + 5x} \right){{\bf{e}}^{\frac{1}{3}x}}\,dx}} = \left. {\left( {15x{{\bf{e}}^{\frac{1}{3}x}} - 39{{\bf{e}}^{\frac{1}{3}x}}} \right)} \right|_6^0 = \require{bbox} \bbox[2pt,border:1px solid black]{{ - 39 - 51{{\bf{e}}^2} = - 415.8419}}\]

Do not get excited about the fact that the lower limit is larger than the upper limit. This can happen on occasion and in no way affects how the integral is evaluated.