Answer
$\int_{-1}^{5}f(x)~dx$
Work Step by Step
We can write the expression as a single integral:
$\int_{-2}^{2}f(x)~dx+\int_{2}^{5}f(x)~dx-\int_{-2}^{-1}f(x)~dx$
$= \int_{-2}^{2}f(x)~dx-\int_{-2}^{-1}f(x)~dx+\int_{2}^{5}f(x)~dx$
$= \int_{-1}^{2}f(x)~dx+\int_{2}^{5}f(x)~dx$
$=\int_{-1}^{5}f(x)~dx$
