The area of the reggion bounded by the lines `y=2x+1,y=3x+1" and "x=4` is

Options

A. 16 sq. units
B. `(121)/(3)` sq. units
C. `(121)/(6)` sq. units
D. 8 sq. units

Answer

To find the area of the region bounded by the lines \(y = 2x + 1\), \(y = 3x + 1\), and \(x = 4\), we can follow these steps:

### Step 1: Identify the equations of the lines
We have the following equations:
1. \(y = 2x + 1\)
2. \(y = 3x + 1\)
3. \(x = 4\)

### Step 2: Find the points of intersection
To find the area, we first need to determine the points where the lines intersect.

- Set \(y = 2x + 1\) equal to \(y = 3x + 1\):
\[ 2x + 1 = 3x + 1 \]
Simplifying gives:
\[ 2x = 3x \implies 0 = x \implies x = 0 \]
Substitute \(x = 0\) into either equation:
\[ y = 2(0) + 1 = 1 \implies (0, 1) \]

- The second intersection point is where \(x = 4\):
- For \(y = 2x + 1\):
\[ y = 2(4) + 1 = 8 + 1 = 9 \implies (4, 9) \]
- For \(y = 3x + 1\):
\[ y = 3(4) + 1 = 12 + 1 = 13 \implies (4, 13) \]

### Step 3: Determine the area between the curves
The area between the curves from \(x = 0\) to \(x = 4\) can be calculated using integration. The upper curve is \(y = 3x + 1\) and the lower curve is \(y = 2x + 1\).

### Step 4: Set up the integral
The area \(A\) can be expressed as:
\[ A = \int_{0}^{4} \left( (3x + 1) - (2x + 1) \right) dx \]
This simplifies to:
\[ A = \int_{0}^{4} (3x + 1 - 2x - 1) dx = \int_{0}^{4} x \, dx \]

### Step 5: Calculate the integral
Now, we compute the integral:
\[ A = \int_{0}^{4} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{4} \]
Calculating the limits:
\[ A = \left( \frac{4^2}{2} - \frac{0^2}{2} \right) = \left( \frac{16}{2} - 0 \right) = 8 \]

### Final Answer
The area of the region bounded by the lines is \(8\) square units.

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