Area — Question 5

Question 5 - Solution

First find the points of intersection by solving $$|x| = x^2.$$

For $x \ge 0$, this becomes $$x = x^2 \quad\Rightarrow\quad x=0 \text{ or } x=1.$$

For $x<0$, we have $$-x = x^2 \quad\Rightarrow\quad x=0 \text{ or } x=-1.$$

Thus, the curves intersect at $$x=-1,\;0,\;1.$$

On $[0,1]$, we have $|x| = x \ge x^2$, and by symmetry the same holds on $[-1,0]$. Therefore, the total area is twice the area on $[0,1]$: $$A = 2\int_{0}^{1} (x - x^2)\,dx.$$

Compute the integral: $$\int_{0}^{1} (x - x^2)\,dx = \left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1 = \frac12-\frac13 = \frac16.$$

Thus, $$A = 2\cdot\frac16 = \frac13.$$

$$\boxed{\frac13}$$