What is this technique called?:

Suppose f is continuous and $\displaystyle \displaystyle \int_{-2}^{2}f(x)dx = 4$ and

$\displaystyle \displaystyle \int_{-2}^{5}g(x)dx = 2$

Determine $\displaystyle \displaystyle \int_{5}^{2}f(y)dy $

I am just doing this out of memory, tell me if its right

$\displaystyle \displaystyle \int_{5}^{2}f(y)dy = -(\int_{-2}^{5}g(x)dx) - (-(\int_{-2}^{2}f(x)dx)) $

$\displaystyle \displaystyle \int_{5}^{2}f(y)dy = - 2 - (-4) = 2 $