# How do you find the area between the curves x+3y=21 and x+7=y^2?

Mar 29, 2015 Solve each function for $x$

$x = 21 - 3 y$

$x = {y}^{2} - 7$

Now find out where the two curves intersect.

${y}^{2} - 7 = 21 - 3 y$

${y}^{2} + 3 y - 28 = 0$

$\left(y + 7\right) \left(y - 4\right) = 0$

$y = - 7$ and $y = 4$

The integral for the area is

${\int}_{-} {7}^{4} 21 - 3 y - \left({y}^{2} - 7\right) \mathrm{dy}$

${\int}_{-} {7}^{4} 21 - 3 y - {y}^{2} + 7 \mathrm{dy}$

${\int}_{-} {7}^{4} 28 - 3 y - {y}^{2} \mathrm{dy}$

Integrating we get

$28 y - \frac{3}{2} {y}^{2} - \frac{1}{3} {y}^{3}$

Now evaluate

$\left[28 \left(4\right) - \frac{3}{2} {\left(4\right)}^{2} - \frac{1}{3} {\left(4\right)}^{3}\right]$
$- \left[28 \left(- 7\right) - \frac{3}{2} {\left(- 7\right)}^{2} - \frac{2}{3} {\left(- 7\right)}^{3}\right]$

$\left[112 - 24 - \frac{64}{3}\right] - \left[- 196 - \frac{147}{2} + \frac{343}{3}\right]$

$\left[\frac{200}{3}\right] - \left[- \frac{931}{6}\right] = \frac{200}{3} + \frac{931}{6} = \frac{1331}{6} \cong 221.83$