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.
∫1−6g(z)dz=∫−2−64ezdz+∫1−22−zdz=(4ez)|−2−6+(2z−12z2)|1−2=[4e−2−4e−6]+[32−(−6)]=4e−2−4e−6+152