# Show $\int_{-a}^{a} [f(x)-f(-x)] dx=0$ for any continuous function f(x)

Show that for any continuous function $f(x)$ on $[-a, a]$, where $a>0$,

$$\int_{-a}^{a} [f(x)-f(-x)] dx=0$$

I've tried to do something like:

Let the primitive function of $f(x)$ be $F(x)$.

$\int_{-a}^{a} [f(x)-f(-x)] dx$

$=[F(x)]_{-a}^{a}-[F(-x)]_{-a}^{a}$

$=F(-a)-F(a)-F(a)+F(-a)$

$=2F(-a)-2F(a)$

but I don't know how to continue.

• define the function $g(x)=f(x)-f(-x)$ now $g(-x)=-g(x)$ which shows that $g(x)$ is antisymmetric function. Integrating such a function over a symmetric interval naturally yields zero ("areas below and above the $x$-axis cancel"). – tired Nov 22 '16 at 14:29
• @tired Ah I understand! Thanks! :) – He Yifei 何一非 Nov 22 '16 at 14:34
• Continuity isn't required for this, just integrability – MPW Nov 22 '16 at 14:40

Hint:

$$I=\int_a^bf(x) \ dx=\int_a^bf(a+b-x)\ dx$$

$$I+I=\int_a^b\{f(x) +f(a+b-x)\}\ dx$$

Hint

If $g$ is an odd function then

$$\int_{-b}^b g(t)dt=0$$

If $g$ is continuous, if $\phi$ and $\psi$ are differentiable, and if $$G(x) = \int_{\psi(x)}^{\phi(x)} g(t)\, dt,$$ the fundamental theorem of calculus and the chain rule give $$G'(x) = g\bigl(\phi(x)\bigr)\, \phi'(x) - g\bigl(\psi(x)\bigr)\, \psi'(x). \tag{1}$$
Define $$F(x) = \int_{-x}^{x} \bigl[f(t) - f(-t)\bigr]\, dt.$$ The integrand $g(x) = f(x) - f(-x)$ is continuous (since $f$ is continuous), so (1) becomes $$F'(x) = \bigl[f(x) - f(-x)\bigr] - \bigl[f(-x) - f(x)\bigr](-1) = 0$$ for all $x$ in $[-a, a]$. By the identity theorem, $F(x) = F(0) = 0$ for all $x$ in $[-a, a]$.
we have: $$\int_{-a}^a [f(x)-f(-x)]dx=\int_{-a}^af(x)dx-\int_{-a}^af(-x)dx$$ and, substituting $-x=t \rightarrow dx=-dt$ in the second integral: $$\int_{-a}^a [f(x)-f(-x)]dx=\int_{-a}^af(x)dx+\int_{a}^{-a}f(t)dt$$ that is (since the variable in the integral is ''free''): $$\int_{-a}^a [f(x)-f(-x)]dx=\int_{-a}^af(x)dx+\int_{a}^{-a}f(x)dx =0$$
Hint: Substitute $x=-t$, \begin{align} \underbrace{\int_{-a}^a\left[f(x)-f(-x)\right]\,\mathrm{d}x}_{I} &=\int_a^{-a}\left[f(-t)-f(t)\right]\,\mathrm{d}(-t)\\ &=\underbrace{\int_{-a}^a\left[f(-t)-f(t)\right]\,\mathrm{d}t}_{-I} \end{align}