# How do you find the area between f(x)=x^2+2x+1,g(x)=3x+3?

Nov 16, 2016

4.5 square units

#### Explanation:

The points of intersection of the two functions are at:
$\left(- 1 , 0\right)$ and $\left(2 , 9\right)$.

So, we have to find the area under the curve of $g \left(x\right)$, then subtract the area under the curve of $f \left(x\right)$ because $g \left(x\right)$ is above $f \left(x\right)$ in the interval $\left[- 1 , 2\right]$.
${\int}_{-} {1}^{2} g \left(x\right) \mathrm{dx} - {\int}_{-} {1}^{2} f \left(x\right) \mathrm{dx}$
${\int}_{-} {1}^{2} \left(3 x + 3\right) \mathrm{dx} - {\int}_{-} {1}^{2} \left({x}^{2} + 2 x + 1\right) \mathrm{dx}$

Simplify the are under the curve of the linear equation using geometry, and the area under the curve of the parabola using a fundamental theorem of calculus.

Let $F \left(x\right)$ be the antiderivative of $f \left(x\right)$
$\frac{1}{2} \left(3\right) \left(9\right) - \left[F \left(2\right) - F \left(- 1\right)\right]$
$\frac{27}{2} - \left[F \left(2\right) - F \left(- 1\right)\right]$

To simplify this, we need to know $F \left(x\right)$. Integrate $f \left(x\right)$:
$F \left(x\right) = \int {x}^{2} + 2 x + 1$
$F \left(x\right) = \frac{1}{3} {x}^{3} + {x}^{2} + x + c$
Solve $F \left(2\right)$ and $F \left(- 1\right)$:
$F \left(2\right) = \frac{1}{3} {\left(2\right)}^{3} + {2}^{2} + 2 + c = \frac{26}{3} + c$
$F \left(- 1\right) = \frac{1}{3} {\left(- 1\right)}^{3} + {\left(- 1\right)}^{2} - 1 + c = - \frac{1}{3} + c$

Now continue to simplify the expression above:
$\frac{27}{2} - \left[F \left(2\right) - F \left(- 1\right)\right]$
$13.5 - \left[\left(\frac{26}{3} \cancel{+ c}\right) - \left(- \frac{1}{3} \cancel{+ c}\right)\right]$
$= 13.5 - 9$
$= 4.5$ square units

Nov 16, 2016

Start by finding the intersection points of the two curves.

$\left\{\begin{matrix}y = {x}^{2} + 2 x + 1 \\ y = 3 x + 3\end{matrix}\right.$

$3 x + 3 = {x}^{2} + 2 x + 1$

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

$0 = \left(x - 2\right) \left(x + 1\right)$

$x = 2 \mathmr{and} - 1$

$y = 3 x + 3$

$y = 9 \mathmr{and} 0$

Hence, the points of intersection are $\left\{2 , 9\right\}$ and $\left\{- 1 , 0\right\}$

We now do the graph of each function (on the same grid). We find that $g \left(x\right)$ is above $f \left(x\right)$ in the area that they share, so we find the area under $g \left(x\right)$, and will subtract the area of $f \left(x\right)$ from that.

${\int}_{- 1}^{2} \left(3 x + 3\right) \mathrm{dx}$

$\implies \frac{3}{2} {x}^{2} + 3 x {|}_{- 1}^{2}$

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

$\implies \frac{3}{2} \left(4\right) + 6 - \frac{3}{2} + 3$

$\implies 6 + 6 - \frac{3}{2} + 3$

$\implies \frac{27}{2}$

Now for $f \left(x\right)$:

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

$\implies \frac{1}{3} {x}^{3} + {x}^{2} + x {|}_{- 1}^{2}$

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

$\implies \frac{8}{3} + 4 + 2 + \frac{1}{3} - 1 + 1$

$\implies 9$

Now subtract the two areas.

$\implies \frac{27}{2} - 9 = \frac{9}{2} = 4.5$

Hence, the area between the graphs of $f \left(x\right) = {x}^{2} + 2 x + 1$ and $g \left(x\right) = 3 x + 3$ is $4.5 {\text{ units}}^{2}$.

Hopefully this helps!