Using integration, find the area of the region bounded by the line `2y=5x+7` , X-axis and the line x = 2 and x = 8.

Explanation

To find the area of the region bounded by the line \(2y = 5x + 7\), the x-axis, and the vertical lines \(x = 2\) and \(x = 8\), we can follow these steps:

### Step 1: Rewrite the equation of the line
First, we rewrite the equation of the line in terms of \(y\):
\[ 2y = 5x + 7 \implies y = \frac{5}{2}x + \frac{7}{2} \]

### Step 2: Identify the boundaries
The area we want to find is bounded by:
- The line \(y = \frac{5}{2}x + \frac{7}{2}\)
- The x-axis (where \(y = 0\))
- The vertical lines \(x = 2\) and \(x = 8\)

### Step 3: Set up the integral
The area \(A\) can be calculated using the integral of the upper curve minus the lower curve from \(x = 2\) to \(x = 8\):
\[ A = \int_{2}^{8} \left(\frac{5}{2}x + \frac{7}{2} - 0\right) \, dx \]
This simplifies to:
\[ A = \int_{2}^{8} \left(\frac{5}{2}x + \frac{7}{2}\right) \, dx \]

### Step 4: Calculate the integral
Now we compute the integral:
\[ A = \int_{2}^{8} \left(\frac{5}{2}x + \frac{7}{2}\right) \, dx \]
We can split the integral:
\[ A = \frac{5}{2} \int_{2}^{8} x \, dx + \frac{7}{2} \int_{2}^{8} 1 \, dx \]

Calculating each integral separately:
1. For \(\int x \, dx\):
\[ \int x \, dx = \frac{x^2}{2} \]
So,
\[ \int_{2}^{8} x \, dx = \left[\frac{x^2}{2}\right]_{2}^{8} = \frac{8^2}{2} - \frac{2^2}{2} = \frac{64}{2} - \frac{4}{2} = 32 - 2 = 30 \]

2. For \(\int 1 \, dx\):
\[ \int 1 \, dx = x \]
So,
\[ \int_{2}^{8} 1 \, dx = [x]_{2}^{8} = 8 - 2 = 6 \]

Now substituting back into the area formula:
\[ A = \frac{5}{2} \cdot 30 + \frac{7}{2} \cdot 6 \]
Calculating each term:
\[ A = 75 + 21 = 96 \]

### Final Answer
Thus, the area of the region bounded by the given lines is:
\[ \boxed{96} \text{ square units} \]