Question

Find the area of the region bounded by the curve \( y^2 = 4x \), the X-axis and the lines \( x = 1 \), \( x = 4 \) for \( y \geq 0 \).

Hint:

When finding the area under a curve, express the curve as a function of \( x \) or \( y \) and integrate over the given limits.

Answer 1

The given equation \( y^2 = 4x \) is a parabola. To find the area, we first express \( y \) as: \[ y = \sqrt{4x} = 2\sqrt{x} \] The area is given by the integral of \( y \) from \( x = 1 \) to \( x = 4 \): \[ \text{Area} = \int_1^4 2\sqrt{x} \, dx \] Now, compute the integral: \[ \int 2\sqrt{x} \, dx = 2 \cdot \frac{2}{3} x^{3/2} = \frac{4}{3} x^{3/2} \] Substitute the limits: \[ \text{Area} = \frac{4}{3} \left[ x^{3/2} \right]_1^4 = \frac{4}{3} \left( 4^{3/2} - 1^{3/2} \right) \] Now calculate the values: \[ 4^{3/2} = 8, \quad 1^{3/2} = 1 \] Thus, the area is: \[ \text{Area} = \frac{4}{3} \left( 8 - 1 \right) = \frac{4}{3} \times 7 = \frac{28}{3} \] The area of the region is: \[ \boxed{\frac{28}{3}} \]