Answer
$17$
Work Step by Step
Since $3$ is between $0$ and $5$ then by the Integral of the Function over Adjacent Intervals it follows that:
$$\begin{align*}
\int_{0}^{5}f(x)dx&=\int_{0}^{3}f(x)dx+\int_{3}^{5}f(x)dx\\
&=\int_{0}^{3}3dx+\int_{3}^{5}xdx\\
&=[3x]_{0}^{3}+[\frac{x^{2}}{2}]_{3}^{5}\\
&=(3\cdot 3-3\cdot 0)+\left[\frac{5^{2}}{2}-\frac{3^{2}}{2}\right]\\
&=9+8\\
&=17.
\end{align*}$$
