# Symmetrical Areas

## Key Questions

• Well, we need to be careful when you say "under the graph" since $f \left(x\right) = {x}^{3}$ goes below the x-axis when $x < 0$, but you meant the region between the graph and the x-axis, then the area of the region is 1/2.

Since there are two regions: one from $x = - 1$ to $x = 0$ and the other from $x = 0$ to $x = 1$, the area $A$ can be found by
$A = {\int}_{- 1}^{0} \left(0 - {x}^{3}\right) \mathrm{dx} + {\int}_{0}^{1} \left({x}^{3} - 0\right) \mathrm{dx}$
by Power Rule,
$= {\left[- {x}^{4} / 4\right]}_{- 1}^{0} + {\left[{x}^{4} / 4\right]}_{0}^{1} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$

• If $f$ is an even function (symmetric about the y-axis), then

${\int}_{- a}^{a} f \left(x\right) \mathrm{dx} = 2 {\int}_{0}^{a} f \left(x\right) \mathrm{dx}$.

If $f$ is an odd function (symmetric about the origin), then

${\int}_{- a}^{a} f \left(x\right) \mathrm{dx} = 0$.

Symmetries can be used to simplify computation of definite integrals. Let us look at the following examples.

Example 1 (Even Function)

int_{-1}^1(3x^2+1) dx =2int_0^1(3x^2+1) dx=2[x^3+x]_0^1=2(2-0)=4

Example 2 (Odd Function)

${\int}_{- \frac{\pi}{3}}^{\frac{\pi}{3}} \frac{\sin \theta}{\sqrt{{\cos}^{2} \theta + 1}} d \theta = 0$

I hope that this was helpful.