## Calculus: Early Transcendentals 8th Edition

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