Area Between Curves — Question 5

Question 5 – Solution

1. Find the intersections.

The square root requires $x\ge0$. Both sides are nonnegative, so squaring gives $$\sqrt x=x^2\quad\Longrightarrow\quad x=x^4 \quad\Longrightarrow\quad x(x^3-1)=0.$$ The factor $x^3-1=(x-1)(x^2+x+1)$ has only one real zero, $x=1$. The real solutions $x=0$ and $x=1$ both satisfy the original equation. The intersection points are $(0,0)$ and $(1,1)$.

2. Determine which curve is above the other.

For $0<x<1$, $x^{1/2}>x^2$. For example, at $x=1/4$, $\sqrt{x}=1/2$ while $x^2=1/16$.

3. Set up upper minus lower.

$$A=\int_0^1\left(\sqrt x-x^2\right)\,dx.$$

4. Integrate each power.

Use $\int x^n\,dx=x^{n+1}/(n+1)$ for $n\ne-1$: $$\int x^{1/2}\,dx=\frac{x^{3/2}}{3/2}=\frac23x^{3/2},\qquad \int x^2\,dx=\frac{x^3}{3}.$$

5. Apply the bounds and subtract.

$$\begin{align*} A&=\left[\frac23x^{3/2}-\frac{x^3}{3}\right]_0^1\\ &=\left(\frac23-\frac13\right)-0 =\boxed{\frac13}. \end{align*}$$