Question

The area of the region bounded by the curve \(y^2 = 4x\), y -axis and the line\(y = 2\)  is

• A. \(\frac23\) sq. units
• B. \(\frac43\) sq. units
• C. \(\frac32\) sq. units
• D. \(\frac34\) sq. units

Answer 1

The Correct Option is A

The problem requires us to find the area of the region bounded by the parabola \(y^2 = 4x\), the y-axis, and the line \(y = 2\).

To solve this, we follow these steps:

1. Note that the equation \(y^2 = 4x\) describes a parabola that opens to the right. Rewriting, we have \(x = \frac{y^2}{4}\).
2. Identify the limits for \(y\). The vertical line \(y = 2\) and the parabola intersect the y-axis at \(y = 0\). Hence, the limits for \(y\) are from 0 to 2.
3. Compute the area using the definite integral of \(x\) with respect to \(y\) between these limits:

\(\int_{0}^{2} \frac{y^2}{4} \, dy\)

4. Evaluate the integral:

\(\frac{1}{4} \int_{0}^{2} y^2 \, dy = \frac{1}{4} \left[ \frac{y^3}{3} \right]_{0}^{2}\)

Substitute the limits:

\(\frac{1}{4} \left[ \frac{2^3}{3} - \frac{0^3}{3} \right] = \frac{1}{4} \times \frac{8}{3} = \frac{8}{12} = \frac{2}{3}\) sq. units

Thus, the area of the region bounded by the curve, the y-axis, and the line \(y = 2\) is \(\frac{2}{3}\) square units.