Calculus: Early Transcendentals (2nd Edition)

$\int_{-a}^a f(x) dx = 2\int_0^a f(x) dx$ due to symmetry. Due to symmetry, the integral from $-a$ to $0$ is equal to the integral from $0$ to $a$ because an even function only consists of positive values. Thus, we can simply evaluate the integral from $0$ to $a$ and multiply by 2.
Because the function $f$ is even: $\int_{-a}^a f(x) dx = \int_{-a}^0 f(x) dx + \int_0^a f(x) dx = n + n = 2n$, where $n$ is the value obtained from evaluating the integral.