Chapter 7.2, Problem 42E

To determine

To find: The area of the region in the first quadrant bounded on the left by the $y$ -axis.

Answer to Problem 42E

The area: $A=\frac{11}{3}$

Explanation of Solution

Given information:

The line $y=x/4$ and above left by the curve $y=1+\sqrt{x}$ and above right by the curve $y=2/\sqrt{x}$ .

Calculation:

The first one is bounded by $y=1+\sqrt{x}$ and

Below by $y=\frac{x}{4}$ for each $x\in [0,1]$ .

Therefore, for each $x\in [0,1]$ and get:

  $1+\sqrt{x}\ge \frac{x}{4}$

The second area from $x=1$

To

  $x=4$ is bounded above by $\frac{2}{\sqrt{x}}$ and below by $\frac{x}{4}$ .

Therefore, for each $x\in [1,4]$ and get:

  $\frac{2}{\sqrt{x}}\ge \frac{x}{4}$

Finally to use the definition and evaluate the shaded area:

  

  $\begin{array}{c}={\left.\left[x+\frac{2}{3}\cdot \sqrt{x^3}-\frac{1}{4}\cdot \frac{x^2}{2}\right]\right|}_0^1+{\left.\left[2\cdot 2\sqrt{x}-\frac{1}{4}\cdot \frac{x^2}{2}\right]\right|}_1^4\\ =\left[1+\frac{2}{3}-\frac{1}{8}-0\right]+\left[4\cdot 2-\frac{16}{8}-\left(4-\frac{1}{8}\right)\right]\\ =\frac{37}{24}+\frac{51}{24}\\ =\frac{88}{24}\\ =\frac{11}{3}\end{array}$

The graph is shown below:

  

Therefore, the required area of the region in the first quadrant bounded on the left by the $y$ -axis is $\frac{11}{3}$ .