Chapter 7, Problem 22RE

To determine

To evaluate the integral.

Answer to Problem 22RE

The required solution is, ${\displaystyle \int x^2{e}^{-3x}dx}=\left(-\frac{x^2}{3}-\frac{2x}{9}-\frac{2}{27}\right)e^{-3x}+C$

Explanation of Solution

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

Concept used: Tabular integration.

We have seen that integrals of the form ${\displaystyle \int f\left(x\right)}g\left(x\right)dx$ in which f can be differentiated repeatedly to become zero and g can be integrated repeatedly without difficulty, are natural candidates for integration by parts. However, if many repetitions are required, the calculation

Can be cumbersome.

Calculation : Using Tabular intergration, the required solution is obtained as,

with $f\left(x\right)=x^2$ and $g\left(x\right)=e^{-3x}$ , we list

  

We combine the products of the functions connected by the arrows according to the

Operation signs above the arrows to obtain,

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

Hence, the required solution is ${\displaystyle \int x^2{e}^{-3x}dx}=\left(-\frac{x^2}{3}-\frac{2x}{9}-\frac{2}{27}\right)e^{-3x}+C$