Question:medium

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

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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.
Updated On: Apr 6, 2026
  • $\frac{1}{3}$
  • $\frac{4}{3}$
  • $\frac{5}{3}$
  • $\frac{8}{3}$
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The Correct Option is D

Solution and Explanation

Step 1: Concept Identification:
Determine the area of a region bounded by a curve and a line. This problem is a direct application of definite integration. The curve \(y^2 = 4x\) represents a parabola opening rightward with its vertex at the origin.
Step 2: Methodology:
The area bounded by \(y = f(x)\), the x-axis, and vertical lines \(x = a\) and \(x = b\) is calculated using \(\int_{a}^{b} f(x) dx\). Given the parabola's symmetry about the x-axis, we can compute the area in the first quadrant and then double it.
Step 3: Calculation:
The parabola is defined by \(y^2 = 4x\). For the upper half (first quadrant), \(y = \sqrt{4x} = 2\sqrt{x}\).
The boundaries for the region are \(x = 0\) (the y-axis) and \(x = 1\).
The area in the first quadrant is:
\[ A_1 = \int_{0}^{1} y \, dx = \int_{0}^{1} 2\sqrt{x} \, dx \]\[ A_1 = 2 \int_{0}^{1} x^{1/2} \, dx \]Applying the power rule for integration:
\[ A_1 = 2 \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{1} = 2 \left[ \frac{2}{3}x^{3/2} \right]_{0}^{1} \]\[ A_1 = \frac{4}{3} [x^{3/2}]_{0}^{1} = \frac{4}{3} (1^{3/2} - 0^{3/2}) = \frac{4}{3}(1 - 0) = \frac{4}{3} \]This value represents the area above the x-axis. Due to symmetry, the area below the x-axis is also \(\frac{4}{3}\).
The total area is calculated as:
\[ A_{\text{total}} = 2 \times A_1 = 2 \times \frac{4}{3} = \frac{8}{3} \]Step 4: Conclusion:
The total area of the region is $\frac{8{3}$} square units.
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