Answer
$\int_{0}^{5}f(x)~dx = 17$
Work Step by Step
We can evaluate the integral:
$\int_{0}^{5}f(x)~dx$
$=\int_{0}^{3}f(x)~dx+\int_{3}^{5}f(x)~dx$
$=\int_{0}^{3}3~dx+\int_{3}^{5}x~dx$
$=3(3-0)+(\frac{5^2-3^2}{2})$
$= 9+8$
$= 17$