# How do you find the area between f(x)=-x^2+4x+2, g(x)=x+2?

Nov 27, 2016

We start by finding the intersection points of the two functions.

$\left\{\begin{matrix}y = - {x}^{2} + 4 x + 2 \\ y = x + 2\end{matrix}\right.$

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

${x}^{2} - 3 x = 0$

$x \left(x - 3\right) = 0$

$x = 0 \mathmr{and} 3$

$y = 0 + 2 \mathmr{and} y = 3 + 2$

$y = 2 \mathmr{and} y = 5$

Hence, the intersection points are $\left(0 , 2\right)$ and $\left(3 , 5\right)$.

We now do a rudimentary sketch of the two functions.

We always proceed in the following way: AREA BETWEEN CURVES = AREA OF CURVE ABOVE - AREA OF CURVE BELOW. We find this area using integration.

We will subtract the area under $y = x + 2$ from $y = - {x}^{2} + 4 x + 2$.

$\implies {\int}_{0}^{3} \left(- {x}^{2} + 4 x + 2 - \left(x + 2\right)\right) \mathrm{dx}$

$\implies {\int}_{0}^{3} \left(- {x}^{2} + 3 x\right)$

$\implies - \frac{1}{3} {x}^{3} + \frac{3}{2} {x}^{2} {|}_{0}^{3}$

$\implies - \frac{1}{3} {\left(3\right)}^{3} + \frac{3}{2} {\left(3\right)}^{2} - \left(- \frac{1}{3} {\left(0\right)}^{3} + \frac{3}{2} {\left(0\right)}^{2}\right)$

$\implies - \frac{1}{3} \left(27\right) + \frac{3}{2} \left(9\right)$

$\implies - 9 + \frac{27}{2}$

$\implies \frac{9}{2}$

Hence, the area between the curves is $\frac{9}{2} {\text{ u}}^{2}$.

Hopefully this helps!