Area Between Curves — Question 10

Question 10 – Solution

1. Find all intersections.

$$x^3=x\quad\Longrightarrow\quad x(x-1)(x+1)=0 \quad\Longrightarrow\quad x=-1,\;0,\;1.$$

2. Split the area where the upper curve changes.

On $(-1,0)$, $x^3>x$; on $(0,1)$, $x>x^3$. For example, at $x=-1/2$, $x^3=-1/8>-1/2$; at $x=1/2$, $x=1/2>1/8=x^3$. Both enclosed regions contribute positive area:

3. Set up upper minus lower on each interval.

$$A=\int_{-1}^0(x^3-x)\,dx+\int_0^1(x-x^3)\,dx.$$

4. Integrate and evaluate each region.

The power rule gives $\int x^3\,dx=x^4/4$ and $\int x\,dx=x^2/2$. $$\begin{align*} A_1&=\left[\frac{x^4}{4}-\frac{x^2}{2}\right]_{-1}^0 =0-\left(\frac14-\frac12\right)=\frac14,\\[6pt] A_2&=\left[\frac{x^2}{2}-\frac{x^4}{4}\right]_0^1 =\frac12-\frac14=\frac14. \end{align*}$$

5. Add both areas.

$$A=A_1+A_2=\frac14+\frac14=\boxed{\frac12}.$$