Question

Find the area of the region bounded by the curve \( y^2 = 4x \,\,\, \text{and the line} \,\,\, x = 3. \)

Hint:

When dealing with equations of the form \( y^2 = 4x \), express \( y \) as \( \pm 2\sqrt{x} \) and integrate accordingly.

Answer 1

Step 1: Set up the equation.
The area of the region is bounded by the curve \( y^2 = 4x \) and the line \( x = 3 \). To find the limits of integration, we need to express \( y \) in terms of \( x \) from the equation \( y^2 = 4x \): \[ y = \pm 2\sqrt{x}. \] We are given the line \( x = 3 \), so we will integrate from \( x = 0 \) to \( x = 3 \) to find the area.

Step 2: Find the area using integration.
The area between the curve and the line is given by: \[ \text{Area} = \int_{0}^{3} 2\sqrt{x} \, dx \] Since the function \( y = 2\sqrt{x} \) is symmetric, we only need to integrate the positive half.

Step 3: Solve the integral.
To solve \( \int 2\sqrt{x} \, dx \), we use the power rule: \[ \int 2\sqrt{x} \, dx = \int 2x^{1/2} \, dx = \frac{4x^{3/2}}{3}. \] Thus, the area is: \[ \text{Area} = \frac{4(3)^{3/2}}{3} = \frac{4 \times 3\sqrt{3}}{3} = 4\sqrt{3}. \]

Step 4: Conclusion.
The area of the region bounded by the curve \( y^2 = 4x \) and the line \( x = 3 \) is \( 4\sqrt{3} \) square units.