Chapter 5.6, Problem 26E

To determine

Tofind: the value of integral.

Answer to Problem 26E

The value of integral is $-\frac{1}{2}{e}^{-t^2}\left[t^2+1\right]$ .

Explanation of Solution

Given information:

The integral is ${\displaystyle \int t^3{e}^{-t^2}dt}$ .

Formula used: integral by parts,

  ${\displaystyle \int u·vdx=u{\displaystyle \int vdx}}-{\displaystyle \int u\text{'}}{\displaystyle \int vdx}dx$

Calculation: to find the value of integral, first substitute and then use integral by parts and then simplify,

  $\begin{array}{l}I={\displaystyle \int t^3{e}^{-t^2}dt}\\ I={\displaystyle \int \frac{x}{2}{e}^{-x}dx}\text{ [substitute }{t}^2=x\Rightarrow 2tdt=dx]\\ I=\frac{1}{2}\left[x{\displaystyle \int e^{-x}dx}-{\displaystyle \int \left(1\right)\left(-e^{-x}\right)dx}\right]\text{ [use integral by parts and simplify]}\\ I=\frac{1}{2}\left[-x{e}^{-x}-e^{-x}\right]+c\text{ [integrate]}\\ I=-\frac{1}{2}{e}^{-t^2}\left[t^2+1\right]\text{ [substitute]}\end{array}$

Thus, the value of integral is $-\frac{1}{2}{e}^{-t^2}\left[t^2+1\right]$ .