Chapter 5, Problem 18P

To determine

To calculate:

The minimum value of the area of the region

Answer to Problem 18P

The minimum value of the region under the curve $y=\frac{x+1}{x}$ is 3.26

Explanation of Solution

Given information:

The area of the region is under curve $y=\frac{x+1}{x}$ from $x=a$ to $x=a+1.5$ , for all $a>0$

Calculation:

Evaluate the integral,

  $\begin{array}{l}{\displaystyle {\int }_a^{a+1.5}\left(x+\frac{1}{x}\right)dx}={\displaystyle {\int }_a^{a+1.5}dx}+{\displaystyle {\int }_a^{a+1.5}\frac{1}{x}dx}\\ ={\left[\frac{x^2}{2}\right]}_a^{a+1.5}+{\left[\ln x\right]}_a^{a+1.5}\\ =\frac{1}{2}\left[{\left(a+1.5\right)}^2-a^2\right]+\ln \left(a+1.5\right)+\ln a\end{array}$

Differentiate obtained result by $a$ and equalize with zero

  $\frac{d}{da}\left[\frac{1}{2}\left[{\left(a+1.5\right)}^2-a^2\right]+\ln \left(a+1.5\right)+\ln a\right]=a+1.5-a+\frac{1}{a+1.5}-\frac{1}{a}=0$

Simplify

  $\frac{3a^2+4.5a-3}{2a^2+3a}=0$

This expression from the left side of the equation is zero only when the numerator is equal to zero. Therefore solve this quadratic equation

  $\begin{array}{l}3a^2+4.5a-3=0\\ a=-2\\ a=0.5\end{array}$

The two values of $a$ are obtained. But it is given that $a>0$ , therefore accept $a=0.5$

Find the second derivative of $a+1.5-a+\frac{1}{a+1.5}-\frac{1}{a}$

  $\frac{d}{da}\left(a+1.5-a+\frac{1}{a+1.5}-\frac{1}{a}\right)=-\frac{1}{{\left(a+1.5\right)}^2}+\frac{1}{a^2}$

Substitute $a=0.5$ in last result, then second derivative is obtained as,

  $-\frac{1}{4}+\frac{1}{{0.5}^2}>0$

Notice that the second derivative is positive for $a=0.5$ and it is the point of minimum of area of the region under the curve $y=x+\frac{1}{x}$ , therefore

  ${\displaystyle {\int }_a^{a+1.5}\left(x+\frac{1}{x}\right)dx}=\frac{1}{2}\left[{\left(0.5+1.5\right)}^2-{0.5}^2\right]+\ln \left(0.5+1.5\right)+\ln \left(0.5\right)\approx 3.26$

Therefore, the minimum value of the region under the curve $y=\frac{x+1}{x}$ is 3.26