Chapter 5.10, Problem 1E

a.

To determine

To find whether the integral ${\displaystyle \underset{1}{\overset{\infty }{\int }}{x}^4{e}^{-x^4}dx}$ is improper or proper

Answer to Problem 1E

The given integral ${\displaystyle \underset{1}{\overset{\infty }{\int }}{x}^4{e}^{-x^4}dx}$ is improper

Explanation of Solution

Given information:

The given integral is ${\displaystyle \underset{1}{\overset{\infty }{\int }}{x}^4{e}^{-x^4}dx}$

Given

  $\begin{array}{l}{\displaystyle \underset{1}{\overset{\infty }{\int }}{x}^4{e}^{-x^4}dx}\\ =\underset{t\to \infty }{{\displaystyle \lim }}{\displaystyle \underset{1}{\overset{t}{\int }}{x}^4{e}^{-x^4}dx}\end{array}$

Integrating by parts, the integral becomes

  $\begin{array}{l}\underset{t\to \infty }{{\displaystyle \lim }}\left[{\left(x^4.\frac{e^{-x^4}}{(-4x^3)}\right)}_1^t-{\displaystyle \underset{1}{\overset{t}{\int }}4x^3\frac{e^{-x^4}}{(-4x^3)}dx}\right]\\ =\underset{t\to \infty }{{\displaystyle \lim }}\left[{\left(x.\frac{e^{-x^4}}{(-4)}\right)}_1^t+{\displaystyle \underset{1}{\overset{t}{\int }}{x}^3\frac{e^{-x^4}}{(-4x^3)}dx}\right]\\ =\underset{t\to \infty }{{\displaystyle \lim }}\left[\frac{e^{-1}}{4}+{\displaystyle \underset{1}{\overset{t}{\int }}{x}^3\frac{e^{-x^4}}{(-4x^3)}dx}\right]\end{array}$

Since the given integral has infinite interval of integration.

Hence, the given integral is improper

b.

To determine

To find whether the integral ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\sec xdx}$ is improper or proper

Answer to Problem 1E

The given integral ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\sec xdx}$ is improper

Explanation of Solution

Given information:

The given integral is ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\sec xdx}$

Given

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\sec xdx}\\ =\underset{t\to {(\pi /2)}^{-}}{{\displaystyle \lim }}{\displaystyle \underset{0}{\overset{t}{\int }}secxdx}\\ \text{As t}\to \frac{\pi }{\text{2}}\text{,}\text{}\sec x=\infty \end{array}$

Therefore, the given integral has infinite discontinuities.

Hence, the given integral is improper

(c) 

To determine

To find whether the integral ${\displaystyle \underset{0}{\overset{2}{\int }}\frac{x}{x^2-5x+6}dx}$ is improper or proper

Answer to Problem 1E

The given integral ${\displaystyle \underset{0}{\overset{2}{\int }}\frac{x}{x^2-5x+6}dx}$ is improper

Explanation of Solution

Given,

  ${\displaystyle \underset{0}{\overset{2}{\int }}\frac{x}{x^2-5x+6}dx}$

Rewrite the integral,

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{2}{\int }}\frac{x}{(x-2)(x-3)}dx}\\ \text{=}\underset{\text{t}\to {\text{2}}^{\text{-}}}{\text{lim}}{\displaystyle \underset{0}{\overset{t}{\int }}\frac{x}{(x-2)(x-3)}dx}\\ =\underset{\text{t}\to {\text{2}}^{\text{-}}}{\text{lim}}{\displaystyle \underset{0}{\overset{t}{\int }}\frac{x-2+2}{(x-2)(x-3)}dx}\\ =\underset{\text{t}\to {\text{2}}^{\text{-}}}{\text{lim}}{\displaystyle \underset{0}{\overset{t}{\int }}\left(\frac{1}{(x-3)}+\frac{2}{(x-2)(x-3)}\right)dx}\end{array}$

Since, the above expression has infinite discontinuity at $x=2$ .

Hence, the given integral is improper

(d) 

To determine

To find whether the integral ${\displaystyle \underset{-\infty }{\overset{0}{\int }}\frac{1}{x^2+5}dx}$ is improper or proper

Answer to Problem 1E

The given integral ${\displaystyle \underset{-\infty }{\overset{0}{\int }}\frac{1}{x^2+5}dx}$ is improper

Explanation of Solution

Given,

  $\begin{array}{l}{\displaystyle \underset{-\infty }{\overset{0}{\int }}\frac{1}{x^2+5}dx}\\ \text{=}\underset{\text{t}\to -\infty }{\text{lim}}{\displaystyle \underset{t}{\overset{0}{\int }}\frac{1}{x^2+5}dx}\end{array}$

Since, the above expression has infinite intervals.

Hence, the given integral is improper