How do you find the area bounded by y=x+4 and y=x^2+2?

Oct 31, 2016

Please see the explanation. $A = 4.5 u n i t {s}^{2}$

Explanation:

Given:
$y = x + 4$ [1]
$y = {x}^{2} + 2$ [2]

We need to find the points of intersection.

Subtract equation [1] from equation [2]:

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

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

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

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

The line is greater than the parabola in the region $- 1 < = x \le 2$, therefore, the integral for the area, A, is:

$A = {\int}_{-} {1}^{2} \left(\left(x + 4\right) - \left({x}^{2} + 2\right)\right) \mathrm{dx}$
$A = {\int}_{-} {1}^{2} \left(- {x}^{2} + x + 2\right) \mathrm{dx}$
$A = - {x}^{3} / 3 + {x}^{2} / 2 + 2 x {|}_{-} {1}^{2}$
$A = 4.5 u n i t {s}^{2}$