Area — Question 6

Question 6 - Solution

First find the points of intersection: $$x = x^3 \quad\Rightarrow\quad x(x^2-1)=0.$$

Thus, $$x=-1,\;0,\;1.$$

On the interval $(0,1)$, we have $$x \ge x^3,$$ and by symmetry, the same configuration occurs on $(-1,0)$.

Therefore, the total enclosed area is twice the area on $[0,1]$: $$A = 2\int_{0}^{1} (x - x^3)\,dx.$$

Evaluate the integral: $$\int_{0}^{1} (x - x^3)\,dx = \left[\frac{x^2}{2}-\frac{x^4}{4}\right]_0^1 = \frac12-\frac14 = \frac14.$$

Thus, $$A = 2\cdot\frac14=\frac12.$$

$$\boxed{\frac12}$$