Answer
$$\frac{2}{15}$$
Work Step by Step
Since
$$\int_{1}^{b} f(x) d x =1-b^{-1} $$
Then
\begin{aligned}
\int_{1}^{5} f(x) d x&=\int_{1}^{3} f(x) d x+\int_{3}^{5} f(x) d x\\
\end{aligned}
Solve for the last integral:
\begin{aligned}
& \int_{3}^{5} f(x) d x=\int_{1}^{5} f(x) d x-\int_{1}^{3} f(x) d x\\
&=\left(1-5^{-1}\right)-\left(1-3^{-1}\right)\\
&=\left(1-\frac{1}{5}\right)-\left(1-\frac{1}{3}\right)\\
&=\frac{4}{5}-\frac{2}{3}\\
&=\frac{2}{15}
\end{aligned}
