# Determine whether the integral is divergent or convergent. If it is convergent, evaluate it. If not, give the answer -1.

Determine whether the integral is divergent or convergent. If it is convergent, evaluate it. If not, give the answer -1.
${\int }_{4}^{\mathrm{\infty }}x{e}^{-2x}$

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Sally Cresswell

Compute the definite integral:
${\int }_{4}^{\mathrm{\infty }}{e}^{-2x}xdx$
For the integrand ${e}^{-2x}x$, integrate by parts, $\int fdg=fg-\int gdf$, where
$f=x,dg={e}^{-2x}dx,df=dx,g=-\frac{1}{2}{e}^{-2x}$:
$=\left(-\frac{1}{2}{e}^{-2x}x\right){|}_{4}^{\mathrm{\infty }}+\frac{1}{2}{\int }_{4}^{\mathrm{\infty }}{e}^{-2x}dx$
INTERMEDIATE STEPS:
Possible derivation:
$\frac{d}{dx}\left(x\right)$
Use the power rule, $\frac{d}{dx}\left({x}^{n}\right)=n{x}^{n-1}$, where n = 1.
$\frac{d}{dx}\left(x\right)=\frac{d}{dx}\left({x}^{1}\right)={x}^{0}$:
Evaluate the antiderivative at the limits and subtract.
$\left(-\frac{1}{2}{e}^{-2x}x\right){|}_{4}^{\mathrm{\infty }}=\left(\underset{b\to \mathrm{\infty }}{lim}-\frac{1}{2}{e}^{-2b}b\right)-\left(-\frac{1}{2}{e}^{-24}4\right)=\left(\underset{b\to \mathrm{\infty }}{lim}-\frac{1}{2}{e}^{-2b}b\right)-\left(-\frac{2}{{e}^{8}}\right)$:
$=\left(\underset{b\to \mathrm{\infty }}{lim}-\frac{1}{2}{e}^{-2b}b\right)+\frac{2}{{e}^{8}}+\frac{1}{2}{\int }_{4}^{\mathrm{\infty }}{e}^{-2x}dx$
$\underset{b\to \mathrm{\infty }}{lim}-\frac{1}{2}{e}^{-2b}b=0$:
$=\frac{2}{{e}^{8}}+\frac{1}{2}{\int }_{4}^{\mathrm{\infty }}{e}^{-2x}dx$
INTERMEDIATE STEPS:
Find the following limit:
$\underset{x\to \mathrm{\infty }}{lim}-\frac{1}{2}{e}^{-2x}x$
Hint: Factor a constant multiple out of the limit.
$\underset{x\to \mathrm{\infty }}{lim}-\frac{1}{2}x{e}^{-2x}=-\frac{1}{2}\underset{x\to \mathrm{\infty }}{lim}x{e}^{-2x}$:
$-\frac{\underset{x\to \mathrm{\infty }}{lim}{e}^{-2x}x}{2}$
Hint: | Linear functions grow asymptotically slower than exponential functions.
Since the polynomial x grows asymptotically slower than ${e}^{2x}$ as x approaches ∞,