Chapter 6.3, Problem 23E

To determine

To find: The anti derivative of the given integration.

Answer to Problem 23E

The anti derivative is: $-\frac{1}{8}{e}^{-2x}\left(4x^3+6x^2+6x+3\right)+C$

Explanation of Solution

Given information:

The integration is: $\int x^3{e}^{-2x}dx$

Calculation:

The given integration is: $\int x^3{e}^{-2x}dx$

Use tabular integration with

  $\begin{array}{c}f(x)=x^3\\ g(x)=e^{-2x}\end{array}$

Then, discover the anti derivatives of $e^{-2x}$ and the derivatives of $x^3$ until finally reach 0. Then, draw arrows connecting each first-column expression (except from the final expression of 0) to the expression in the second-column row below. Start with a positive sign and alternate between positive and negative signs as write the signs above the arrows.

  

Utilize the operation signals located above each arrow to combine the results of the functions that are connected by arrows. Remember to add the $+C$ :

  $\begin{array}{c}\int x^3{e}^{-2x}dx=x^3\left(-\frac{1}{2}{e}^{-2x}\right)-3x^2\left(\frac{1}{4}{e}^{-2x}\right)+6x\left(-\frac{1}{8}{e}^{-2x}\right)-6\left(\frac{1}{16}{e}^{-2x}\right)+C\\ =-\frac{1}{2}{x}^3{e}^{-2x}-\frac{3}{4}{x}^2{e}^{-2x}-\frac{3}{4}x{e}^{-2x}-\frac{3}{8}{e}^{-2x}+C\\ =-\frac{1}{8}{e}^{-2x}\left(4x^3+6x^2+6x+3\right)+C\end{array}$

Therefore the required anti derivative of the given integration $\int x^3{e}^{-2x}dx$ is

  $-\frac{1}{8}{e}^{-2x}\left(4x^3+6x^2+6x+3\right)+C$ .