Chapter 8.2, Problem 42E

To determine

To Find:

To find the area of the region in the first quadrant bounded on the left by y axis, below by the line $y=\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$ ,above left by the curve $y=1+\sqrt{x}$ and above right by the curve $y=\frac{2}{\sqrt{x}}$ .

Answer to Problem 42E

The area of the region in the first quadrant bounded by curves is $\frac{11}{3}$ .

Explanation of Solution

Given information:

The given functions are $y=\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$ , $y=1+\sqrt{x}$ , $y=\frac{2}{\sqrt{x}}$ and bounded by y-axis.

The red line is the graph of $y=\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$ , the blue curve is the graph of $y=1+\sqrt{x}$ and the green curve is the graph of $y=\frac{2}{\sqrt{x}}$ .

  

The functions $y=\frac{2}{\sqrt{x}}$ and $y=1+\sqrt{x}$ intersects when $x=1,y=\frac{x}{4}$ and $y=\frac{2}{\sqrt{x}}$ intersect when $x=4$ .

Therefore, the area enclosed by these curves is given by the sum of the integrals,

  $\begin{array}{l}A={\displaystyle \underset{0}{\overset{1}{\int }}\left(1+\sqrt{x}-\frac{x}{4}\right)dx}+{\displaystyle \underset{1}{\overset{4}{\int }}\left(\frac{2}{\sqrt{x}}-\frac{x}{4}\right)}dx\\ A={\left[x+\frac{2x^{\frac{3}{2}}}{3}-\frac{x^2}{8}\right]}_0^1+{\left[\frac{4}{\sqrt{x}}-\frac{x^2}{8}\right]}_1^4\\ A=\left[1+\frac{2}{3}-\frac{1}{8}\right]+\left[4-\frac{16}{8}-\frac{1}{8}\right]\\ A=\left[\frac{37}{24}\right]+\left[4-\frac{15}{8}\right]\\ A=\left[\frac{37}{24}\right]+\left[\frac{17}{8}\right]\\ A=\frac{37+51}{24}\\ A=\frac{88}{24}\\ A=\frac{11}{3}\end{array}$

Therefore, The area of the region in the first quadrant bounded by curves is $\frac{11}{3}$ .