snajat3l

2022-02-11

Write the integral as a sum of integrals without absolute values and evaluate:

Amubbemibren3k

Compute the definite integral:
${\int }_{-6}^{3}{|x|}^{3}dx$
When $0, x is positive.
This means ${|x|}^{3}={x}^{3}={x}^{3}$.
When $-6, x is negative.
This means ${|x|}^{3}={\left(-x\right)}^{3}=-{x}^{3}$:
$={\int }_{-6}^{0}-{x}^{3}dx+{\int }_{0}^{3}{x}^{3}dx$
Factor out constants:
$=-{\int }_{-6}^{0}{x}^{3}dx+{\int }_{0}^{3}{x}^{3}dx$
Apply the fundamental theorem of calculus.
The antiderivative of ${x}^{3}$ is $\frac{{x}^{4}}{4}$:
$=\left(-\frac{{x}^{4}}{4}\right){\mid }_{-6}^{0}+{\int }_{0}^{3}{x}^{3}dx$
Evaluate the antiderivative at the limits and subtract.
$\left(-\frac{{x}^{4}}{4}\right){\mid }_{-6}^{0}=\left(-\frac{{0}^{4}}{4}\right)-\left(-\frac{1}{4}{\left(-6\right)}^{4}\right)=324:$
$=324+{\int }_{0}^{3}{x}^{3}dx$
Apply the fundamental theorem of calculus.
The antiderivative of ${x}^{3}$ is $\frac{{x}^{4}}{4}:$
$=324+\frac{{x}^{4}}{4}{\mid }_{0}^{3}$
Evaluate the antiderivative at the limits and subtract.
$\frac{{x}^{4}}{4}{\mid }_{0}^{3}=\frac{{3}^{4}}{4}-\frac{{0}^{4}}{4}=\frac{81}{4}:$
$=\frac{1377}{4}$