I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.
Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
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 \(g\left( z \right) = \left\{ {\begin{array}{*{20}{c}}{2 - z}&{z > - 2}\\{4{{\bf{e}}^z}}&{z \le - 2}\end{array}} \right.\)Show All Steps Hide All Steps
See if you can find a good choice for “\(c\)” that will make this integral doable.
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 2All we need to do at this point is evaluate each integral. Here is that work.
\[\begin{align*}\int_{{ - 6}}^{1}{{g\left( z \right)\,dz}} & = \int_{{ - 6}}^{{ - 2}}{{4{{\bf{e}}^z}\,dz}} + \int_{{ - 2}}^{1}{{2 - z\,dz}} = \left. {\left( {4{{\bf{e}}^z}} \right)} \right|_{ - 6}^{ - 2} + \left. {\left( {2z - \frac{1}{2}{z^2}} \right)} \right|_{ - 2}^1\\ & = \left[ {4{{\bf{e}}^{ - 2}} - 4{{\bf{e}}^{ - 6}}} \right] + \left[ {\frac{3}{2} - \left( { - 6} \right)} \right] = \require{bbox} \bbox[2pt,border:1px solid black]{{4{{\bf{e}}^{ - 2}} - 4{{\bf{e}}^{ - 6}} + \frac{{15}}{2}}}\end{align*}\]