Chapter 9.4, Problem 43E

To determine

Tofind: the area of the region in the first quadrant that lies under the curve.

Answer to Problem 43E

The area of the region in the first quadrant that lies under the curve is $1$ .

Explanation of Solution

Given information:

Thecurve equation is $y=\frac{\ln x}{x^2}$ .

Calculation: the curve intercept the x -axis at the point $x=1$ , after that it’s increasing.

So the area of the region in the first quadrant can be find by find the value of integral from $1\text{ to }\infty$ ,

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

Thus, the area of the region is $1$ .