# What is the net area between f(x) = sqrt(x+3)-x^3  and the x-axis over x in [2, 4 ]?

Dec 27, 2017

$55.107$ square units ( 3 .d.p.)

#### Explanation:

Since we need the area between the x axis and the curve, in the interval $\left[2 , 4\right]$, our integral will be:

${\int}_{2}^{4} \left(\sqrt{x + 3} - {x}^{3}\right) \mathrm{dx} = \frac{2}{3} {\left(x + 3\right)}^{\frac{3}{2}} - \frac{1}{4} {x}^{4}$

$A r e a = {\left[\frac{2}{3} {\left(x + 3\right)}^{\frac{3}{2}} - \frac{1}{4} {x}^{4}\right]}^{4} - {\left[\frac{2}{3} {\left(x + 3\right)}^{\frac{3}{2}} - \frac{1}{4} {x}^{4}\right]}_{2}$

Plugging in upper and lowers bounds:

$A r e a = {\left[\frac{2}{3} {\left(4 + 3\right)}^{\frac{3}{2}} - \frac{1}{4} {\left(4\right)}^{4}\right]}^{4} - {\left[\frac{2}{3} {\left(2 + 3\right)}^{\frac{3}{2}} - \frac{1}{4} {\left(2\right)}^{4}\right]}_{2}$

$A r e a = \left[\frac{14}{3} \sqrt{7} - 64\right] - \left[\frac{10}{3} \sqrt{5} - 4\right] \approx - 55.107$

Area $= 55.107$ square units.

GRAPH: