Review question

# Can we find the area between a parabola and a line? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9357

## Solution

The area of the region bounded by the curves $y = x^2$ and $y = x + 2$ equals

1. $\frac{7}{3}$,

2. $\frac{7}{2}$,

3. $\frac{9}{2}$,

4. $\frac{11}{2}$.

The curves intersect when $x^2 = x + 2$, or $(x - 2)(x + 1) = 0$, so the points of intersection are at $x = 2$ and $x = -1$.

From our sketch we can see that $x + 2 \geq x^2$ in the region $-1 \leq x \leq 2$, so
$\text{Area enclosed} = \left( \text{Area below } \; y = x + 2 \right) - \left(\text{Area below } \; y = x^2 \right) .$

The area below $y=x^2$ is calculated by integration, and the area below $y=x+2$ can be found using the formula for the area of a trapezium. We have $\text{Area below } \; y = x + 2 \; = 3 \times \frac{1+4}{2}=\frac{15}{2}, \text{while}$ $\text{Area below } \; y = x^2 \; = \int\limits_{-1}^2 x^2 \; dx=\left[\frac{1}{3} x^3\right] _{-1}^2=\frac{8}{3}-\frac{-1}{3}=3.$

So the area we seek is $\frac{15}{2}-3=\frac{9}{2}$

You could also calculate the area under the straight line $y=x+2$ using the integral $\int\limits_{-1}^2 x+2 \; dx$.