# What is the net area between f(x) = x^2+1/x  and the x-axis over x in [2, 4 ]?

Apr 11, 2016

$A = \frac{56}{3} + \ln 2$ or $19.36$ units.

#### Explanation:

Over the interval $x \in \left[2 , 4\right]$, ${x}^{2} + \frac{1}{x}$ is always positive, so the area we will be computing is between this graph and the positive $x$-axis.

To actually find the area, integrate ${x}^{2} + \frac{1}{x}$ from $2$ to $4$:
$A = {\int}_{2}^{4} {x}^{2} + \frac{1}{x} \mathrm{dx}$

Using the sum rule,
$A = {\int}_{2}^{4} {x}^{2} \mathrm{dx} + {\int}_{2}^{4} \frac{1}{x} \mathrm{dx}$

And evaluating the integrals,
$A = {\left[{x}^{3} / 3\right]}_{2}^{4} + {\left[\ln x\right]}_{2}^{4}$

Doing some substitutions and algebra:
$A = \left({\left(4\right)}^{3} / 3 - {\left(2\right)}^{3} / 3\right) + \left(\ln 4 - \ln 2\right)$
$A = \left(\frac{64}{3} - \frac{8}{3}\right) + \ln \left(\frac{4}{2}\right)$
$A = \frac{56}{3} + \ln 2$

If we wanted a precise answer, we could leave it like this. An approximation to 2 decimal places is $A = 19.36$ units.