# What is the net area between f(x) = (x-2)^3  and the x-axis over x in [1, 5 ]?

Jul 8, 2018

$20.5$

#### Explanation:

The integrand $f \left(x\right) = {\left(x - 2\right)}^{3}$ crosses 0 at $x = 2$. The area enclosed between $x = 1$ and $x = 2$ is given by

${A}_{1} = | {\int}_{1}^{2} {\left(x - 2\right)}^{3} \mathrm{dx} |$
qquadquad = |(x-2)^4/4]_1^2|
$q \quad q \quad = | - \frac{1}{4} | = \frac{1}{4}$

while, the area enclosed between $x = 2$ and $x = 5$ is given by

${A}_{2} = | {\int}_{2}^{5} {\left(x - 2\right)}^{3} \mathrm{dx} |$
qquadquad = |(x-2)^4/4]_2^5|
$q \quad q \quad = | \frac{81}{4} | = \frac{81}{4}$

Thus, the net area is $\frac{81}{4} + \frac{1}{4} = \frac{82}{4} = 20.5$