Chapter 6, Problem 22RE

To determine

To find: The value of integral.

Answer to Problem 22RE

The answer is: $\left(-\frac{x^2}{3}-\frac{2x}{9}-\frac{2}{27}\right)e^{-3x}+C$

Explanation of Solution

Given information:

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

Calculation:

The function $x^2{e}^{-3x}$ is a product so use tabular integration with $f(x)=x^2$ and $g(x)=e^{-3x}$ Make a table of the derivatives of $f(x)$ and the anti derivatives of $g(x)$ .

Then, beginning with $+$ , write alternate signs for each row. Draw an arrow from the left column to the expression one row lower in the right column for all rows except the last. Draw a straight line from the right column to the left column for the final row.

  

Utilize the operation indicators located above each arrow to combine the results of the functions connected by arrows. Then, simplify.

  $\begin{array}{c}\int x^2{e}^{-3x}dx\\ =x^2\left(-\frac{1}{3}{e}^{-3x}\right)-2x\left(\frac{1}{9}{e}^{-3x}\right)+2\left(-\frac{1}{27}{e}^{-3x}\right)+C\\ =-\frac{1}{3}{x}^2{e}^{-3x}-\frac{2}{9}x{e}^{-3x}-\frac{2}{27}{e}^{-3x}+C\\ =\left(-\frac{x^2}{3}-\frac{2x}{9}-\frac{2}{27}\right)e^{-3x}+C\end{array}$

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

  $\left(-\frac{x^2}{3}-\frac{2x}{9}-\frac{2}{27}\right)e^{-3x}+C$