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Section 5-7 : Computing Definite Integrals

16. Evaluate the following integral, if possible. If it is not possible clearly explain why it is not possible to evaluate the integral.

\(\displaystyle \int_{{ - 6}}^{1}{{g\left( z \right)\,dz}}\) where \(g\left( z \right) = \left\{ {\begin{array}{*{20}{c}}{2 - z}&{z > - 2}\\{4{{\bf{e}}^z}}&{z \le - 2}\end{array}} \right.\)

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Hint : Recall that integrals we can always “break up” an integral as follows, \[\int_{a}^{b}{{f\left( x \right)\,dx}} = \int_{a}^{c}{{f\left( x \right)\,dx}} + \int_{c}^{b}{{f\left( x \right)\,dx}}\]

See if you can find a good choice for “\(c\)” that will make this integral doable.

Start Solution

This integral can’t be done as a single integral give the obvious change of the function at \(z = - 2\) which is in the interval over which we are integrating. However, recall that we can always break up an integral at any point and \(z = - 2\) seems to be a good point to do this.

Breaking up the integral at \(z = - 2\) gives,

\[\int_{{ - 6}}^{1}{{g\left( z \right)\,dz}} = \int_{{ - 6}}^{{ - 2}}{{g\left( z \right)\,dz}} + \int_{{ - 2}}^{1}{{g\left( z \right)\,dz}}\]

So, in the first integral we have \( - 6 \le z \le - 2\) and so we can use \(g\left( z \right) = 4{{\bf{e}}^z}\) in the first integral. Likewise, in the second integral we have \( - 2 \le z \le 1\) and so we can use \(g\left( z \right) = 2 - z\) in the second integral.

Making these function substitutions gives,

\[\int_{{ - 6}}^{1}{{g\left( z \right)\,dz}} = \int_{{ - 6}}^{{ - 2}}{{4{{\bf{e}}^z}\,dz}} + \int_{{ - 2}}^{1}{{2 - z\,dz}}\] Show Step 2

All we need to do at this point is evaluate each integral. Here is that work.

\[\begin{align*}\int_{{ - 6}}^{1}{{g\left( z \right)\,dz}} & = \int_{{ - 6}}^{{ - 2}}{{4{{\bf{e}}^z}\,dz}} + \int_{{ - 2}}^{1}{{2 - z\,dz}} = \left. {\left( {4{{\bf{e}}^z}} \right)} \right|_{ - 6}^{ - 2} + \left. {\left( {2z - \frac{1}{2}{z^2}} \right)} \right|_{ - 2}^1\\ & = \left[ {4{{\bf{e}}^{ - 2}} - 4{{\bf{e}}^{ - 6}}} \right] + \left[ {\frac{3}{2} - \left( { - 6} \right)} \right] = \require{bbox} \bbox[2pt,border:1px solid black]{{4{{\bf{e}}^{ - 2}} - 4{{\bf{e}}^{ - 6}} + \frac{{15}}{2}}}\end{align*}\]