Chapter 7.3, Problem 3QQ

To determine

To find which of the following is correct.

Answer to Problem 3QQ

  $\text{The correct answer is A, }\frac{1}{2}x{e}^{2x}-\frac{1}{4}{e}^{2x}+C\text{.}$

Explanation of Solution

Given information:

  $\begin{array}{l}\text{The equation is, }{\displaystyle \int x{e}^{2x}dx}=\\ \left(\text{A}\right)\frac{x{e}^{2x}}{2}-\frac{e^{2x}}{4}\text{+ C }\left(\text{B}\right)\frac{x{e}^{2x}}{2}-\frac{e^{2x}}{2}\text{+ C}\\ \left(C\right)\frac{x{e}^{2x}}{2}+\frac{e^{2x}}{4}\text{+ C }\left(\text{D}\right)\frac{x{e}^{2x}}{2}+\frac{e^{2x}}{2}\text{+ C}\\ \left(\text{E}\right)\frac{x^2{e}^{2x}}{4}\text{+ C }\end{array}$

Calculation : The required solution is obtained as,

  $\begin{array}{l}{\displaystyle \int x{e}^{2x}dx}\\ \text{ Let }u=x,du=dx,dv=e^{2x}dx,v=\frac{1}{2}{e}^{2x}\\ \frac{1}{2}x{e}^{2x}-{\displaystyle \int \frac{1}{2}{e}^{2x}dx}\\ \frac{1}{2}x{e}^{2x}-\frac{1}{4}{e}^{2x}+C\\ \text{ Hence, the correct answer is A, }\frac{1}{2}x{e}^{2x}-\frac{1}{4}{e}^{2x}+C\text{.}\end{array}$