Chapter 9.4, Problem 4QQ

a.

To determine

Tofind: the area as an improper integral.

Answer to Problem 4QQ

The area as an improper integral is $R={\displaystyle {\int }_1^{\infty }\frac{2\ln x}{x^2}dx}$ .

Explanation of Solution

Given information:

The curve is $y=\frac{2\ln x}{x^2}$ .

Calculation: to find the area as an improper integral, integrate the curve from $1\text{ to }\infty$ ,

  $R={\displaystyle {\int }_1^{\infty }\frac{2\ln x}{x^2}dx}$

b.

To determine

To find: the area as a limit of a definite integral.

Answer to Problem 4QQ

The area as a limit of a definite integral is $R=\underset{a\to \infty }{\lim }{\displaystyle {\int }_1^a\frac{2\ln x}{x^2}}dx$ .

Explanation of Solution

Given information:

The curve is $y=\frac{2\ln x}{x^2}$ .

Calculation: to find the area as an improper integral, integrate the curve from $1\text{ to }\infty$ ,

  $R={\displaystyle {\int }_1^{\infty }\frac{2\ln x}{x^2}dx}$

As a limit of a definite integral,

  $\begin{array}{l}R={\displaystyle {\int }_1^{\infty }\frac{2\ln x}{x^2}dx}\\ R=\underset{a\to \infty }{\lim }{\displaystyle {\int }_1^a\frac{2\ln x}{x^2}}dx\end{array}$

Thus, the area a limit of a definite integral is $R=\underset{a\to \infty }{\lim }{\displaystyle {\int }_1^a\frac{2\ln x}{x^2}}dx$ .

c.

To determine

To find: the area.

Answer to Problem 4QQ

The area is $1$ .

Explanation of Solution

Given information:

The curve is $y=\frac{2\ln x}{x^2}$ .

Calculation: to find the area as an improper integral, integrate the curve from $1\text{ to }\infty$ ,

  $R={\displaystyle {\int }_1^{\infty }\frac{2\ln x}{x^2}dx}$

As a limit of a definite integral,

  $\begin{array}{l}R={\displaystyle {\int }_1^{\infty }\frac{2\ln x}{x^2}dx}\\ R=\underset{a\to \infty }{\lim }{\displaystyle {\int }_1^a\frac{2\ln x}{x^2}}dx\end{array}$

Thus, the area a limit of a definite integral is $R=\underset{a\to \infty }{\lim }{\displaystyle {\int }_1^a\frac{2\ln x}{x^2}}dx$ .

To find the area first integrate it and simplify it,

  $\begin{array}{l}R=\underset{a\to \infty }{\lim }{\displaystyle {\int }_1^a\frac{2\ln x}{x^2}}dx\\ R=\underset{a\to \infty }{\lim }{\left[-\frac{1}{x}\left(\ln x+1\right)\right]}_1^a\text{ [}\ln x=t\Rightarrow \frac{1}{x}dx=dt\Rightarrow {\displaystyle \int uvdx=u{\displaystyle \int vdx}-{\displaystyle \int \frac{du}{dx}}{\displaystyle \int vdxdx}}]\\ R=\underset{a\to \infty }{\lim }\left[-\frac{1}{a}\left(\ln a+1\right)+\frac{1}{1}\left(\ln 1+1\right)\right]\\ R=0+1\left(0+1\right)\text{ [simplify]}\\ R=1\end{array}$

Thus, the area is $1$ .