# What is the area of the region in the first quadrant that is enclosed by the graphs of y=x^3+8 and y=x+8?

Aug 4, 2017

The area is $\frac{1}{4}$.

#### Explanation:

Th two graphs intersect where
${x}^{3} + 8 = x + 8$ which is at $x = 0$ and $x = 1$.

If we test at $x = \frac{1}{2}$ we see that $x + 8 > {x}^{3} + 8$ in the interval $\left[0 , 1\right]$

So,

The area is

${\int}_{0}^{1} \left(\left(x + 8\right) - \left({x}^{3} + 8\right)\right) \mathrm{dx} = {\int}_{0}^{1} \left(x - {x}^{3}\right) \mathrm{dx} = \frac{1}{4}$

Although I didn't do it, you can draw the graph if you like.

$y = x + 8$ is a line with slope $1$, through $\left(0 , 8\right)$

$y = {x}^{3} + 8$ is the graph of the cubic translated up $8$.