Answer
a. 8
b. $\displaystyle \frac{4}{3}$
c. 20,$\displaystyle \quad \frac{10}{3}$
Work Step by Step
$a.$
$\displaystyle \int_{1}^{7}f(x)dx=$ the area beneath the graph,
which is the sum of areas of
* a rectangle 6$\times$1
* a triangle b=1, h=2
* a triangle b=2, h=1
(see image below)
$\displaystyle \int_{1}^{7}f(x)dx=6+\frac{1}{2}(1)(2)+\frac{1}{2}(2)(1)=8$
$b.$
Average value $=\displaystyle \frac{1}{b-a}\int_{a}^{b}f(x)dx=\frac{1}{7-1}(8)$
$=\displaystyle \frac{8}{6}=\frac{4}{3}$
$c.$
We would add another rectangle ($6\times 2$) to the existing area,
so
$\displaystyle \int_{1}^{7}f(x)dx$=$8+12=20$, and
Average value = $\displaystyle \frac{20}{6}=\frac{10}{3}$
