If f is continuous on [-1,1] and f(-x) + f(x) = x^2 -2 then the integral from -1 to 1 of f(x)dx = ?
$\displaystyle I = \int_{-1}^{1} f(x) \, dx = \int_{-1}^{0} f(x) \, dx + \int_{0}^{1} f(x) \, dx$.
Let $\displaystyle x \rightarrow -x$ in $\displaystyle \int_{-1}^{0} f(x) \, dx$ and you get $\displaystyle \int_{-1}^{0} f(x) \, dx = \int_{1}^{0} f(-x) \, (-dx) = \int_{0}^{1} f(-x) \, dx $.
Therefore $\displaystyle I = \int_{0}^{1} f(-x) \, dx + \int_{0}^{1} f(x) \, dx = \int_{0}^{1} f(-x) + f(x) \, dx = ......$