Draw a rough sketch of the curve `y=sqrt(x-1)` in the interval [1, 5] and find the area under the given curve and between the lines x = 1 and x = 5.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Sketch the Curve
The curve given is \( y = \sqrt{x - 1} \). This is a square root function that starts from the point \( (1, 0) \) because \( \sqrt{0} = 0 \) when \( x = 1 \). As \( x \) increases, \( y \) will also increase. The interval we are interested in is from \( x = 1 \) to \( x = 5 \).

### Step 2: Identify Key Points
- At \( x = 1 \):
\[ y = \sqrt{1 - 1} = \sqrt{0} = 0 \quad \text{(Point: (1, 0))} \]
- At \( x = 5 \):
\[ y = \sqrt{5 - 1} = \sqrt{4} = 2 \quad \text{(Point: (5, 2))} \]

### Step 3: Draw the Curve
Plot the points \( (1, 0) \) and \( (5, 2) \) on the coordinate plane. The curve will start at \( (1, 0) \) and rise to \( (5, 2) \), resembling the right half of a parabola.

### Step 4: Set Up the Integral for Area
To find the area under the curve from \( x = 1 \) to \( x = 5 \), we set up the integral:
\[ \text{Area} = \int_{1}^{5} \sqrt{x - 1} \, dx \]

### Step 5: Evaluate the Integral
To evaluate the integral, we first find the antiderivative of \( \sqrt{x - 1} \):
\[ \int \sqrt{x - 1} \, dx = \int (x - 1)^{1/2} \, dx \]
Using the power rule:
\[ = \frac{(x - 1)^{3/2}}{3/2} = \frac{2}{3}(x - 1)^{3/2} \]

Now, we evaluate it from 1 to 5:
\[ \text{Area} = \left[ \frac{2}{3}(x - 1)^{3/2} \right]_{1}^{5} \]

Calculating the upper limit:
\[ = \frac{2}{3}(5 - 1)^{3/2} = \frac{2}{3}(4)^{3/2} = \frac{2}{3}(8) = \frac{16}{3} \]

Calculating the lower limit:
\[ = \frac{2}{3}(1 - 1)^{3/2} = \frac{2}{3}(0) = 0 \]

Thus, the area is:
\[ \text{Area} = \frac{16}{3} - 0 = \frac{16}{3} \]

### Final Answer
The area under the curve \( y = \sqrt{x - 1} \) from \( x = 1 \) to \( x = 5 \) is \( \frac{16}{3} \) square units.

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