What is the area of the region bounded by the line `3x-5y = 15`. `x =1, x = 3` and x-axis in sq units ?

Options

A. `(36)/(5)`
B. `(18)/(5)`
C. `(9)/(5)`
D. `(3)/(5)`

Explanation

To find the area of the region bounded by the line \(3x - 5y = 15\), the lines \(x = 1\) and \(x = 3\), and the x-axis, we can follow these steps:

### Step 1: Rearrange the Line Equation
We start with the line equation:
\[ 3x - 5y = 15 \]
We need to express \(y\) in terms of \(x\):
\[ 5y = 3x - 15 \quad \Rightarrow \quad y = \frac{3}{5}x - 3 \]

### Step 2: Set Up the Integral
The area \(A\) under the curve from \(x = 1\) to \(x = 3\) can be found using the definite integral:
\[ A = \int_{1}^{3} y \, dx = \int_{1}^{3} \left(\frac{3}{5}x - 3\right) \, dx \]

### Step 3: Calculate the Integral
Now we calculate the integral:
\[ A = \int_{1}^{3} \left(\frac{3}{5}x - 3\right) \, dx \]
We can break this into two separate integrals:
\[ A = \int_{1}^{3} \frac{3}{5}x \, dx - \int_{1}^{3} 3 \, dx \]

Calculating the first integral:
\[ \int \frac{3}{5}x \, dx = \frac{3}{5} \cdot \frac{x^2}{2} = \frac{3}{10}x^2 \]
Evaluating from 1 to 3:
\[ \left[\frac{3}{10}x^2\right]_{1}^{3} = \frac{3}{10}(3^2) - \frac{3}{10}(1^2) = \frac{3}{10}(9) - \frac{3}{10}(1) = \frac{27}{10} - \frac{3}{10} = \frac{24}{10} = \frac{12}{5} \]

Calculating the second integral:
\[ \int 3 \, dx = 3x \]
Evaluating from 1 to 3:
\[ \left[3x\right]_{1}^{3} = 3(3) - 3(1) = 9 - 3 = 6 \]

### Step 4: Combine the Results
Now we combine the results of both integrals:
\[ A = \frac{12}{5} - 6 = \frac{12}{5} - \frac{30}{5} = \frac{12 - 30}{5} = \frac{-18}{5} \]
Since area cannot be negative, we take the absolute value:
\[ A = \frac{18}{5} \]

### Final Answer
Thus, the area of the region bounded by the line, the x-axis, and the vertical lines \(x = 1\) and \(x = 3\) is:
\[ \boxed{\frac{18}{5}} \text{ square units} \]