# Find the area bounded by the curve y=2x^2-6x and y=-x^2+9?

Mar 15, 2018

$\text{Area} = 32$

#### Explanation:

First, it is a good idea to graph the given curves:

We need to evaluate the area bounded by these curves, i.e find the area in between them.

That's the area between their points of intersection, so we need to find that: their points of intersection:

$R i g h t a r r o w 2 {x}^{2} - 6 x = - {x}^{2} + 9$

$R i g h t a r r o w 3 {x}^{2} - 6 x - 9 = 0$

$R i g h t a r r o w {x}^{2} - 2 x - 3 = 0$

$R i g h t a r r o w {x}^{2} + x - 3 x - 3 = 0$

$R i g h t a r r o w x \left(x + 1\right) - 3 \left(x + 1\right) = 0$

$R i g h t a r r o w \left(x + 1\right) \left(x - 3\right) = 0$

$\therefore x = - 1 , 3$

We don't really need to find the points of intersection; the $x$-intercepts will suffice.

This is because the curves are already specified in terms of $y$.

Now we can start evaluating the definite integrals of these curves in the interval $\left[- 1 , 3\right]$:

$R i g h t a r r o w {\int}_{- 1}^{3}$ $\left(\left(- {x}^{2} + 9\right) - \left(2 {x}^{2} - 6 x\right)\right)$ $\mathrm{dx}$

$= {\int}_{- 1}^{3}$ $\left(- 3 {x}^{2} + 6 x + 9\right)$ $\mathrm{dx}$

$= | - {x}^{3} + 3 {x}^{2} + 9 x {|}_{- 1}^{3}$

$= \left(- {\left(3\right)}^{3} + 3 {\left(3\right)}^{2} + 9 \left(3\right)\right) - \left(- {\left(- 1\right)}^{3} + 3 {\left(- 1\right)}^{2} + 9 \left(- 1\right)\right)$

$= \left(- 27 + 27 + 27\right) - \left(1 + 3 - 9\right)$

$= \left(27\right) - \left(- 5\right)$

$= 32$

Therefore, the area bounded by the curves is $32$.