Answer
a. $-\frac{1}{2}$
b. $-\frac{1}{2}$
c. $-\frac{1}{2}$
Work Step by Step
a. Using the figure to guide the calculations, we have
$\int_{-1}^0 h(x)dx=\int_{-1}^0 (x)dx=(\frac{x^2}{2})|_{-1}^0=0-(\frac{(-1)^2}{2})=-\frac{1}{2}$.
Thus the average can be obtained as
$\bar h_{[-1,0]}=av(h)=\frac{-\frac{1}{2}}{0-(-1)}=-\frac{1}{2}$
b. $\int_{0}^1 h(x)dx=\int_{0}^1 (-x)dx=(-\frac{x^2}{2})|_{0}^1=-(\frac{1^2}{2})-0=-\frac{1}{2}$.
Thus the average can be obtained as
$\bar h_{[-1,0]}=av(h)=\frac{-\frac{1}{2}}{1-0}=-\frac{1}{2}$
c. Combining the above results, we have
$\int_{-1}^1 h(x)dx=\int_{-1}^0 h(x)dx+\int_{0}^1 h(x)dx=-1$
and the average is
$\bar h_{[-1,1]}=av(h)=\frac{-1}{1-(-1)}=-\frac{1}{2}$
