Chapter 7.3, Problem 23E

To determine

To determine the integral ${\displaystyle \int x^3{e}^{-2x}dx}$

Answer to Problem 23E

The value of the integral ${\displaystyle \int x^3{e}^{-2x}dx}$ is $e^{-2x}\left[-\frac{x^3}{2}-\frac{3x^2}{4}-\frac{6x}{8}-\frac{6}{16}\right]$

Explanation of Solution

Given information: ${\displaystyle \int x^3{e}^{-2x}dx}$

Formula used: ${\displaystyle \int uvdx=u{\displaystyle \int vdx-{\displaystyle \int \left(u^{\prime }{\displaystyle \int vdx}\right)}}}dx$

Calculation :

Differentiation of $u(x)$	Signs	Integration of   $v(x)$
$x^3$	(+)	$e^{-2x}$
$3x^2$	(-)	$\frac{-e^{-2x}}{2}$
$6x$	(+)	$\frac{e^{-2x}}{4}$
$6$	(-)	$\frac{-e^{-2x}}{8}$
0	(+)	$\frac{e^{-2x}}{16}$

  ${\displaystyle \int x^3{e}^{-2x}dx}=(x^3)\left(-\frac{e^{-2x}}{2}\right)-(3x^2)\left(\frac{e^{-2x}}{4}\right)+(6x)\left(-\frac{e^{-2x}}{8}\right)-(6)\left(\frac{e^{-2x}}{16}\right)$

Hence, ${\displaystyle \int x^3{e}^{-2x}dx}=e^{-2x}\left[-\frac{x^3}{2}-\frac{3x^2}{4}-\frac{6x}{8}-\frac{6}{16}\right]$