Area Between Curves — Question 8

Question 8 – Solution

1. Find the intersections.

$$2-x^2=x\quad\Longrightarrow\quad x^2+x-2=0 \quad\Longrightarrow\quad(x+2)(x-1)=0.$$ The zero-product property gives $x+2=0$ or $x-1=0$. Thus $x=-2$ and $x=1$, giving points $(-2,-2)$ and $(1,1)$.

2. Determine which curve is above the other.

On $[-2,1]$, $(2-x^2)-x=(1-x)(x+2)\ge0$. The parabola is above the line, even where both curves are below the $x$-axis.

3. Set up upper minus lower.

$$A=\int_{-2}^1(2-x-x^2)\,dx.$$

4. Find an antiderivative.

$$\int 2\,dx=2x,\qquad \int x\,dx=\frac{x^2}{2},\qquad \int x^2\,dx=\frac{x^3}{3}.$$

5. Apply the bounds, taking care with negative signs.

$$\begin{align*} A&=\left[2x-\frac{x^2}{2}-\frac{x^3}{3}\right]_{-2}^1\\ &=\left(2-\frac12-\frac13\right) -\left(-4-2+\frac83\right)\\ &=\frac76-\left(-\frac{10}{3}\right) =\boxed{\frac92}. \end{align*}$$