# What is the net area between f(x) = sqrt(x+3)-x^3+6x  and the x-axis over x in [2, 4 ]?

Aug 5, 2016

$\frac{2}{3} \left(7 \sqrt{7} - 5 \sqrt{5} - 36\right) \approx - 19.107$

#### Explanation:

Note that the area between a curve defined by $f \left(x\right)$ on the interval $\left[a , b\right]$ can be expressed as:

${\int}_{a}^{b} f \left(x\right) \mathrm{dx}$

Thus, the area here is equivalent to:

${\int}_{2}^{4} \left(\sqrt{x + 3} - {x}^{3} + 6 x\right) \mathrm{dx}$

The integral can be split up:

$= {\int}_{2}^{4} \sqrt{x + 3} \mathrm{dx} - {\int}_{2}^{4} {x}^{3} \mathrm{dx} + 6 {\int}_{2}^{4} x \mathrm{dx}$

The second and third terms are more easily integrated: use the rule that ${\int}_{a}^{b} {x}^{n} \mathrm{dx} = {\left[{x}^{n + 1} / \left(n + 1\right)\right]}_{a}^{b}$.

$= {\int}_{2}^{4} \sqrt{x + 3} \mathrm{dx} - {\left[{x}^{4} / 4\right]}_{2}^{4} + 6 {\left[{x}^{2} / 2\right]}_{2}^{4}$

For the first integral, we can use substitution. Let $u = x + 3$. Thus, $\mathrm{du} = \mathrm{dx}$. Don't forget to plug the current bounds into $x + 3$ given the substitution:

$= {\int}_{5}^{7} \sqrt{u} \mathrm{du} - {\left[{x}^{4} / 4\right]}_{2}^{4} + 6 {\left[{x}^{2} / 2\right]}_{2}^{4}$

Integrate again using the aforementioned rule, recalling that $\sqrt{u} = {u}^{\frac{1}{2}}$:

$= {\left[{u}^{\frac{3}{2}} / \left(\frac{3}{2}\right)\right]}_{5}^{7} - {\left[{x}^{4} / 4\right]}_{2}^{4} + 6 {\left[{x}^{2} / 2\right]}_{2}^{4}$

Bring out multiplicative constants:

$= \frac{2}{3} {\left[{u}^{\frac{3}{2}}\right]}_{5}^{7} - \frac{1}{4} {\left[{x}^{4}\right]}_{2}^{4} + 3 {\left[{x}^{2}\right]}_{2}^{4}$

Evaluate:

$= \frac{2}{3} \left[{7}^{\frac{3}{2}} - {5}^{\frac{3}{2}}\right] - \frac{1}{4} \left[{4}^{4} - {2}^{4}\right] + 3 \left[{4}^{2} - {2}^{2}\right]$

$= \frac{2}{3} \left[7 \sqrt{7} - 5 \sqrt{5}\right] - \frac{1}{4} \left[240\right] + 3 \left[12\right]$

$= \frac{2}{3} \left[7 \sqrt{7} - 5 \sqrt{5}\right] - 24$

$= \frac{2}{3} \left(7 \sqrt{7} - 5 \sqrt{5} - 36\right)$

$\approx - 19.107$