Sketch the graph of `y=sqrt(x)+1 "in"[0,4]` and determine the area of the region enclosed by the curve, the axis of X and the lines `x=0,x=4`.

Answer

To solve the problem of finding the area of the region enclosed by the curve \( y = \sqrt{x} + 1 \), the x-axis, and the lines \( x = 0 \) and \( x = 4 \), we will follow these steps:

### Step 1: Sketch the Graph
First, we need to sketch the graph of the function \( y = \sqrt{x} + 1 \) over the interval \([0, 4]\).

- At \( x = 0 \):
\[ y = \sqrt{0} + 1 = 1 \]
- At \( x = 4 \):
\[ y = \sqrt{4} + 1 = 2 + 1 = 3 \]

The points we have are \( (0, 1) \) and \( (4, 3) \). The graph will start at \( (0, 1) \) and end at \( (4, 3) \), and it will be a curve that rises as \( x \) increases.

### Step 2: Set Up the Area Calculation
The area we want to calculate is bounded by the curve, the x-axis, and the vertical lines \( x = 0 \) and \( x = 4 \).

To find the area, we will use integration. The area \( A \) can be expressed as:
\[ A = \int_{0}^{4} (\sqrt{x} + 1) \, dx \]

### Step 3: Calculate the Integral
Now we will compute the integral:
\[ A = \int_{0}^{4} (\sqrt{x} + 1) \, dx = \int_{0}^{4} \sqrt{x} \, dx + \int_{0}^{4} 1 \, dx \]

Calculating each part:
1. For \( \int_{0}^{4} \sqrt{x} \, dx \):
\[ \int \sqrt{x} \, dx = \frac{2}{3} x^{3/2} \]
Evaluating from 0 to 4:
\[ \left[ \frac{2}{3} x^{3/2} \right]_{0}^{4} = \frac{2}{3} (4^{3/2}) - \frac{2}{3} (0^{3/2}) = \frac{2}{3} (8) = \frac{16}{3} \]

2. For \( \int_{0}^{4} 1 \, dx \):
\[ \int 1 \, dx = x \]
Evaluating from 0 to 4:
\[ [x]_{0}^{4} = 4 - 0 = 4 \]

Combining both results:
\[ A = \frac{16}{3} + 4 = \frac{16}{3} + \frac{12}{3} = \frac{28}{3} \]

### Step 4: Conclusion
Thus, the area of the region enclosed by the curve \( y = \sqrt{x} + 1 \), the x-axis, and the lines \( x = 0 \) and \( x = 4 \) is:
\[ \boxed{\frac{28}{3}} \]