Find the area of the region bounded by the parabola y = x^2 and the line y = x + 2.

Step-by-step solution

1. 1

   Locate the intersection points. Setting $x^2=x+2$ gives $x^2-x-2=0$, which factors as $(x-2)(x+1)=0$, so the curves cross at $x=-1$ and $x=2$. These become the limits of integration.

2. 2

   Decide which curve is on top. Test the midpoint $x=0.5$: the line gives $2.5$ while the parabola gives $0.25$. The line is above the parabola throughout $(-1,2)$, so the integrand is line minus parabola.

3. 3

   Set up the integral. Area $=\displaystyle\int_{-1}^{2}\left[(x+2)-x^2\right]dx$. Taking top minus bottom guarantees a positive integrand, so no absolute values or case splits are needed.

4. 4

   Find the antiderivative. $\int\left(x+2-x^2\right)dx=\frac{x^2}{2}+2x-\frac{x^3}{3}$.

5. 5

   Evaluate at the limits. At $x=2$: $2+4-\frac{8}{3}=\frac{10}{3}$. At $x=-1$: $\frac{1}{2}-2+\frac{1}{3}=-\frac{7}{6}$. The difference is $\frac{10}{3}+\frac{7}{6}=\frac{27}{6}=\frac{9}{2}$.

6. 6

   Confirm numerically. Simpson's rule applied to $(x+2)-x^2$ over $[-1,2]$ returns $4.5000$, matching $\frac{9}{2}$ square units.