Answer
a. $-\frac{1}{2}$
b. $1$
c. $\frac{1}{4}$
Work Step by Step
a. Separate the integration into regions $[-1,0],[0,1]$; we have
$\int_{-1}^1 g(x)dx=\int_{-1}^0 g(x)dx+\int_{0}^1 g(x)dx=\int_{-1}^0 (-x-1)dx+\int_{0}^1 (x-1)dx=(-\frac{x^2}{2}-x)|_{-1}^0+(\frac{x^2}{2}-x)|_{0}^1=(\frac{(-1)^2}{2}-1)+(\frac{1^2}{2}-1)=-1$.
Thus the average can be obtained as
$\bar g_{[-1,1]}=av(g)=\frac{-1}{1-(-1)}=-\frac{1}{2}$
b. No separation is needed; we have
$\int_{1}^3 g(x)dx=\int_{1}^3 (x-1)dx=(\frac{x^2}{2}-x)|_{1}^3=(\frac{(3)^2}{2}-3)-(\frac{1^2}{2}-1)=2$.
Thus the average can be obtained as
$\bar g_{[1,3]}=av(g)=\frac{2}{3-1}=1$
c. Combine the above results; we have
$\int_{-1}^3 g(x)dx=\int_{-1}^1 g(x)dx+\int_{1}^3 g(x)dx=2-1=1$
and the average is
$\bar g_{[-1,3]}=av(g)=\frac{1}{3-(-1)}=\frac{1}{4}$
