# How do you sketch the region enclosed by y=x+1, y=9-x^2, x=-1, x=2 and find the area?

Mar 13, 2018

$\textcolor{b l u e}{\frac{39}{2}}$ Units squared.

#### Explanation:

If we look at the two functions in the given interval we see that:

$y = 9 - {x}^{2}$ attains greater values than $y = x + 1$. therefore $y = 9 - {x}^{2}$ is above $y = x + 1$ in the given interval.

The area between these two functions in the given interval will be:

${\int}_{- 1}^{2} \left(9 - {x}^{2}\right) - \left(x + 1\right) \mathrm{dx}$

${\int}_{- 1}^{2} \left(- {x}^{2} - x + 8\right) \mathrm{dx} = {\left[- \frac{1}{3} {x}^{3} - \frac{1}{2} {x}^{2} + 8 x\right]}_{- 1}^{2}$

$\text{Area} = {\left[- \frac{1}{3} {x}^{3} - \frac{1}{2} {x}^{2} + 8 x\right]}^{2} - {\left[- \frac{1}{3} {x}^{3} - \frac{1}{2} {x}^{2} + 8 x\right]}_{- 1}$

Plugging in upper and lowers bounds:

$= {\left[- \frac{1}{3} {\left(2\right)}^{3} - \frac{1}{2} {\left(2\right)}^{2} + 8 \left(2\right)\right]}^{2} - {\left[- \frac{1}{3} {\left(- 1\right)}^{3} - \frac{1}{2} {\left(- 1\right)}^{2} + 8 \left(- 1\right)\right]}_{- 1}$

$\text{Area} = {\left[\frac{34}{3}\right]}^{2} - {\left[- \frac{49}{6}\right]}_{- 1} = \textcolor{b l u e}{\frac{39}{2}}$ Units squared.

GRAPH: