# How do you find the area between f(x)=x^2+2x+1 and g(x)=2x+5?

Nov 12, 2016

The area is $\frac{32}{3}$.

#### Explanation:

Start by finding the intersection points.

$\left\{\begin{matrix}y = {x}^{2} + 2 x + 1 \\ y = 2 x + 5\end{matrix}\right.$

Substitute the second equation into the first.

$2 x + 5 = {x}^{2} + 2 x + 1$

$0 = {x}^{2} + 2 x - 2 x + 1 - 5$

$0 = {x}^{2} - 4$

$0 = \left(x + 2\right) \left(x - 2\right)$

$x = - 2 \mathmr{and} 2$

$\therefore y = 2 x + 5$

$y = 2 \left(- 2\right) + 5 \mathmr{and} y = 2 \left(2\right) + 5$

$y = 1 \mathmr{and} y = 9$

The solution set is hence $\left\{- 2 , 1\right\}$ and $\left\{2 , 9\right\}$. The next step is doing a basic sketch of the graph. This is not what I'm going to do for this answer, but this is often a necessity when you don't have access to a graphing calculator.

So, our goal is to subtract the area of the lower graph, $y = {x}^{2} + 2 x + 1$, from the area of the upper graph, $y = 2 x + 5$.

We do this by integrating.

${\int}_{- 2}^{2} \left(2 x + 5\right) \mathrm{dx}$

$\implies \left({x}^{2} + 5 x\right) {|}_{- 2}^{2}$

$\implies {2}^{2} + 5 \left(2\right) - \left({\left(- 2\right)}^{2} + 5 \left(- 2\right)\right)$

$\implies 4 + 10 - 4 + 10$

$\implies 20$

Now for the lower graph.

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

$\implies \left(\frac{1}{3} {x}^{3} + {x}^{2} + x\right) {|}_{- 2}^{2}$

$\implies \frac{1}{3} {\left(2\right)}^{3} + {2}^{2} + 2 - \left(\frac{1}{3} {\left(- 2\right)}^{3} + {\left(- 2\right)}^{2} - 2\right)$

$\implies \frac{8}{3} + 4 + 2 + \frac{8}{3} - 4 + 2$

$\implies \frac{28}{3}$

Subtract:

$A = 20 - \frac{28}{3} = \frac{32}{3}$

Hence, the area between $f \left(x\right) = {x}^{2} + 2 x + 1$ and $g \left(x\right) = 2 x + 5$ is $\frac{32}{3} {\text{ units}}^{2}$.

Hopefully this helps!