How to understand $\int_{-a}^{a} f(|x|) dx$

Question

Find $\int_{-6}^{6} f(|x|) dx$.

My (possibly incorrect?) solution:

\begin{align*} \int_{-6}^{6} f(|x|) dx &= \int_{-6}^{0} f(|x|) dx + \int_{0}^{6} f(|x|) dx \\ &= \int_{6}^{0} f(x) dx + \int_{0}^{6} f(x) dx \\ &= -\int_{0}^{6} f(x) dx + \int_{0}^{6} f(x) dx \\ &=0 \end{align*}

In which case, the function itself is irrelevant. So is it always the case that $\int_{-a}^a f(|x|)=0$?

Answer 1

Your second line is wrong. The first integral should instead be: $$\int_{-6}^{0} f(|x|) dx = \int_{-6}^{0} f(-x) dx$$ Now perform a $u$-substitution with $u=-x$: $$\int_{-6}^{0} f(-x) dx = -\int_{6}^{0} f(u) du = \int_{0}^{6} f(u) du$$ So the original integral is: \begin{align} \int_{-6}^6 f(|x|) dx &= 2 \int_0^6 f(x) dx \\ &= 2 \left(\int_0^1 f(x) dx + \int_1^4 f(x) dx + \int_4^5 f(x) dx + \int_5^6 f(x) dx \right) \\ &= 2 (2 + 12 + 2 - 2) = 28 \end{align}

Answer 2

$f(|x|)$ is even function, so for integral in symmetric limits holds $$\int\limits_{-a}^{a}f(|x|)\,dx=2\int\limits_{0}^{a}f(x)\,dx$$