Question

Find area bounded by region, y=3x+1, y=4x+1 and x=3?

Answer 1

The area of triangle ABC, bounded by the lines \( y = 3x + 1 \), \( y = 2x + 1 \), and \( x = 0 \), can be calculated as follows:

Step 1: Understanding the Problem
The area of triangle ABC can be determined by first calculating the area of the regions formed by the lines and the x-axis, then subtracting the area of one triangle from another.

Step 2: Define the areas of triangles ABED and ACED
The area of triangle ABED is the integral of the function \( (3x + 1) \) minus 0 with respect to \( x \), evaluated from \( x = 0 \) to \( x = 4 \). Similarly, the area of triangle ACED is the integral of the function \( (2x + 1) \) minus 0 with respect to \( x \), also evaluated from \( x = 0 \) to \( x = 4 \).

Step 3: Set up the integrals
The area of triangle ABED is: \[ \text{Area(ABED)} = \int_0^4 (3x + 1) \, dx \] The area of triangle ACED is: \[ \text{Area(ACED)} = \int_0^4 (2x + 1) \, dx \]

Step 4: Simplify the integrals
Expanding the integrals: \[ \text{Area(ABED)} = \int_0^4 3x \, dx + \int_0^4 1 \, dx \] \[ \text{Area(ACED)} = \int_0^4 2x \, dx + \int_0^4 1 \, dx \]

Step 5: Evaluate the integrals
First, evaluate the integral for \( \text{Area(ABED)} \): \[ \int_0^4 3x \, dx = \left[ \frac{3x^2}{2} \right]_0^4 = \frac{3(4^2)}{2} - 0 = \frac{48}{2} = 24 \] \[ \int_0^4 1 \, dx = \left[ x \right]_0^4 = 4 - 0 = 4 \] So: \[ \text{Area(ABED)} = 24 + 4 = 28 \] Next, evaluate the integral for \( \text{Area(ACED)} \): \[ \int_0^4 2x \, dx = \left[ \frac{2x^2}{2} \right]_0^4 = \frac{2(4^2)}{2} - 0 = 32/2 = 16 \] \[ \int_0^4 1 \, dx = 4 \] So: \[ \text{Area(ACED)} = 16 + 4 = 20 \]

Step 6: Find the area of triangle ABC
The area of triangle ABC is the difference between the areas of triangle ABED and triangle ACED: \[ \text{Area(ABC)} = \text{Area(ABED)} - \text{Area(ACED)} = 28 - 20 = 8 \]

Final Answer:
Therefore, the area of triangle ABC is \( \boxed{8} \) square units.