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

Feb 3, 2016

64.1589

Explanation:

Integrate $f \left(x\right)$ over the region setting 2 as the lower limit and 4 as the upper limit.

$A r e a = {\int}_{2}^{4} f \left(x\right) \mathrm{dx}$
$= {\int}_{2}^{4} {x}^{3} + \frac{6}{x} \mathrm{dx} = {\left[{x}^{4} / 4 + 6 \ln A b s \left(x\right)\right]}_{2}^{4}$
$= \left\{{\left(4\right)}^{4} / 4 + 6 \ln \left(4\right)\right\} - \left\{{\left(2\right)}^{4} / 4 + 6 \ln \left(2\right)\right\}$
$= 64 + 12 \ln \left(2\right) - 4 - 6 \ln \left(2\right)$

$= 60 + 6 \ln \left(2\right) \approx 64.1589$