The area of the region is 1st quadrant bounded by the y-axis, `y=(x)/(4), y=1 +sqrt(x), and y=(2)/(sqrt(x))` is

Options

A. `2//3` sq. units
B. `8//3` sq. units
C. `11//3` sq. units
D. `13//6` sq. units

Explanation

To find the area of the region in the first quadrant bounded by the curves \( y = \frac{x}{4} \), \( y = 1 + \sqrt{x} \), and \( y = \frac{2}{\sqrt{x}} \), we will follow these steps:

### Step 1: Identify the Points of Intersection
First, we need to find the points where the curves intersect to determine the limits of integration.

1. **Intersection of \( y = 1 + \sqrt{x} \) and \( y = \frac{x}{4} \)**:
\[ 1 + \sqrt{x} = \frac{x}{4} \]
Rearranging gives:
\[ \sqrt{x} = \frac{x}{4} - 1 \]
Squaring both sides:
\[ x = \left(\frac{x}{4} - 1\right)^2 \]
Expanding and simplifying leads to:
\[ 4x = x^2 - 8x + 16 \implies x^2 - 12x + 16 = 0 \]
Using the quadratic formula:
\[ x = \frac{12 \pm \sqrt{144 - 64}}{2} = \frac{12 \pm \sqrt{80}}{2} = 6 \pm 2\sqrt{5} \]
Thus, the points of intersection are \( x = 6 - 2\sqrt{5} \) and \( x = 6 + 2\sqrt{5} \).

2. **Intersection of \( y = 1 + \sqrt{x} \) and \( y = \frac{2}{\sqrt{x}} \)**:
\[ 1 + \sqrt{x} = \frac{2}{\sqrt{x}} \]
Rearranging gives:
\[ \sqrt{x} + 1 = \frac{2}{\sqrt{x}} \implies \sqrt{x} \cdot \sqrt{x} + \sqrt{x} = 2 \]
This leads to:
\[ x + \sqrt{x} - 2 = 0 \]
Let \( u = \sqrt{x} \):
\[ u^2 + u - 2 = 0 \]
Factoring gives:
\[ (u - 1)(u + 2) = 0 \implies u = 1 \quad (\text{since } u = -2 \text{ is not valid}) \]
Thus, \( x = 1 \).

3. **Intersection of \( y = \frac{x}{4} \) and \( y = \frac{2}{\sqrt{x}} \)**:
\[ \frac{x}{4} = \frac{2}{\sqrt{x}} \]
Cross-multiplying gives:
\[ x^{3/2} = 8 \implies x = 8^{2/3} = 4 \]

### Step 2: Set Up the Integrals
Now that we have the points of intersection, we can set up our integrals.

1. **Area \( A_1 \) from \( x = 0 \) to \( x = 1 \)**:
\[ A_1 = \int_0^1 \left( (1 + \sqrt{x}) - \left(\frac{x}{4}\right) \right) dx \]

2. **Area \( A_2 \) from \( x = 1 \) to \( x = 4 \)**:
\[ A_2 = \int_1^4 \left( \left(\frac{2}{\sqrt{x}}\right) - \left(\frac{x}{4}\right) \right) dx \]

### Step 3: Calculate the Integrals
1. **Calculating \( A_1 \)**:
\[ A_1 = \int_0^1 \left( 1 + \sqrt{x} - \frac{x}{4} \right) dx \]
\[ = \int_0^1 1 \, dx + \int_0^1 \sqrt{x} \, dx - \int_0^1 \frac{x}{4} \, dx \]
Evaluating each integral:
\[ = [x]_0^1 + \left[\frac{2}{3} x^{3/2}\right]_0^1 - \left[\frac{x^2}{8}\right]_0^1 \]
\[ = 1 + \frac{2}{3} - \frac{1}{8} = 1 + \frac{16}{24} - \frac{3}{24} = 1 + \frac{13}{24} = \frac{37}{24} \]

2. **Calculating \( A_2 \)**:
\[ A_2 = \int_1^4 \left( \frac{2}{\sqrt{x}} - \frac{x}{4} \right) dx \]
\[ = \int_1^4 2x^{-1/2} \, dx - \int_1^4 \frac{x}{4} \, dx \]
Evaluating each integral:
\[ = \left[ 4\sqrt{x} \right]_1^4 - \left[ \frac{x^2}{8} \right]_1^4 \]
\[ = 4(2) - 4 + \frac{1}{8} = 8 - 2 - \frac{1}{8} = 6 - \frac{1}{8} = \frac{48}{8} - \frac{1}{8} = \frac{47}{8} \]

### Step 4: Total Area
Now we can find the total area:
\[ A = A_1 + A_2 = \frac{37}{24} + \frac{47}{8} \]
To add these, we need a common denominator (which is 24):
\[ = \frac{37}{24} + \frac{141}{24} = \frac{178}{24} = \frac{89}{12} \]

### Final Answer
The area of the region is:
\[ \boxed{\frac{89}{12}} \text{ square units} \]