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.
∫1−6g(z)dz where g(z)={2−zz>−24ezz≤−2Show 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,
∫1−6g(z)dz=∫−2−6g(z)dz+∫1−2g(z)dzSo, in the first integral we have −6≤z≤−2 and so we can use g(z)=4ez in the first integral. Likewise, in the second integral we have −2≤z≤1 and so we can use g(z)=2−z in the second integral.
Making these function substitutions gives,
∫1−6g(z)dz=∫−2−64ezdz+∫1−22−zdz 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*}