# Determine whether the integral is divergent or convergent. If it is convergent, evaluate it. If not, state your answer as "divergent

Determine whether the integral is divergent or convergent. If it is convergent, evaluate it. If not, state your answer as "divergent".
${\int }_{3}^{\mathrm{\infty }}\left(\frac{8}{{\left(x+4\right)}^{3}}/2\right)$

You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Luvottoq

Compute the definite integral:
${\int }_{3}^{\mathrm{\infty }}\frac{4}{{\left(x+4\right)}^{3}}dx$
Factor out constants:
$=4{\int }_{3}^{\mathrm{\infty }}\frac{1}{{\left(x+4\right)}^{3}}dx$
For the integrand $\frac{1}{{\left(x+4\right)}^{3}}$, substitute u = x + 4 and du = dx.
This gives a new lower bound $u=4+3=7$ and upper bound $u=\infty$:
$=4{\int }_{7}^{\mathrm{\infty }}\frac{1}{{u}^{3}}du$
Apply the fundamental theorem of calculus.
The antiderivative of $\frac{1}{{u}^{3}}is-\frac{1}{2{u}^{2}}$:
$=\underset{b\to \mathrm{\infty }}{lim}\left(-\frac{2}{{u}^{2}}\right){|}_{7}^{b}$
Evaluate the antiderivative at the limits and subtract.
$\left(-\frac{2}{{u}^{2}}\right){|}_{7}^{\mathrm{\infty }}=\left(-\frac{2}{{\mathrm{\infty }}^{2}}\right)-\left(-\frac{2}{{7}^{2}}\right)=\frac{2}{49}$:
$=\frac{2}{49}$