# How do you find the area cut off by the x axis, above the x-axis, and y=(3+x)(4-x)?

Jan 13, 2017

$\frac{343}{6}$ square units.

#### Explanation:

Start by expanding.

$y = - {x}^{2} + 4 x - 3 x + 12$

$y = - {x}^{2} + x + 12$

This is a parabola that opens downward. So, the area we need to find is between $a$ and $b$, where $a$ and $b$ are the zeroes of your function.

We started in factored form, so we can immediately say that our zeroes are $x = - 3$ and $x = 4$.

We need to evaluate the following integral:

${\int}_{-} {3}^{4} - {x}^{2} + x + 12 \mathrm{dx}$

$= {\int}_{-} {3}^{4} - \frac{1}{3} {x}^{3} + \frac{1}{2} {x}^{2} + 12 x$

Use the second fundamental theorem of calculus to evaluate.

$= - \frac{1}{3} {\left(4\right)}^{3} + \frac{1}{2} {\left(4\right)}^{2} + 12 \left(4\right) - \left(- \frac{1}{3} {\left(- 3\right)}^{3} + \frac{1}{2} {\left(- 3\right)}^{2} + 12 \left(- 3\right)\right)$

$= - \frac{64}{3} + 8 + 48 - 9 - \frac{9}{2} + 36$

$= 11 - \frac{9}{2} - \frac{64}{3}$

$= 83 - \frac{9}{2} - \frac{64}{3}$

$= \frac{343}{6}$ square units

Hopefully this helps!