Question

The area (in sq. units) of the region bounded by the parabola y2 = 4x and the line x = 1 is

• A. $\frac{1}{3}$
• B. $\frac{4}{3}$
• C. $\frac{5}{3}$
• D. $\frac{8}{3}$

Hint:

Always sketch the curves to visualize the region. For curves symmetric about an axis (like \(y^2 = 4ax\)), calculating the area of one half and doubling it is often simpler and less prone to errors.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We need to find the area of a region enclosed by a curve and a line. This is a classic application of definite integration. The curve \(y^2 = 4x\) is a parabola opening to the right with its vertex at the origin.
Step 2: Key Formula or Approach:
The area of a region bounded by a curve \(y = f(x)\), the x-axis, and the lines \(x = a\) and \(x = b\) is given by \(\int_{a}^{b} f(x) dx\). Since the parabola is symmetric about the x-axis, we can find the area in the first quadrant and multiply it by 2.
Step 3: Detailed Explanation:
The given parabola is \(y^2 = 4x\). For the upper half of the parabola (in the first quadrant), we have \(y = \sqrt{4x} = 2\sqrt{x}\).
The region is bounded by \(x = 0\) (the y-axis, where the parabola starts) and \(x = 1\).
The area in the first quadrant is given by the integral:
\[ A_1 = \int_{0}^{1} y \, dx = \int_{0}^{1} 2\sqrt{x} \, dx \] \[ A_1 = 2 \int_{0}^{1} x^{1/2} \, dx \] Using the power rule for integration:
\[ A_1 = 2 \left[ \frac{x^{1/2 + 1}}{1/2 + 1} \right]_{0}^{1} = 2 \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{1} = 2 \left[ \frac{2}{3}x^{3/2} \right]_{0}^{1} \] \[ A_1 = \frac{4}{3} [x^{3/2}]_{0}^{1} = \frac{4}{3} (1^{3/2} - 0^{3/2}) = \frac{4}{3}(1 - 0) = \frac{4}{3} \] This is the area of the region above the x-axis. Due to symmetry, the area of the region below the x-axis is also \(\frac{4}{3}\).
The total area is:
\[ A_{\text{total}} = 2 \times A_1 = 2 \times \frac{4}{3} = \frac{8}{3} \] Step 4: Final Answer:
The total area of the region is $\frac{8{3}$} sq. units.