The area of the region (in square units) bounded by \(x=1, x=2\) and the curve \(y^2 = 4x\) in the first quadrant is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{4}{3}[2\sqrt{2}-1]\)
In the first quadrant \(y \ge 0\), so the curve \(y^2 = 4x\) is \(y = 2\sqrt{x}\). The area between this curve and the x-axis from x = 1 to x = 2 is
\( \displaystyle \text{Area} = \int_{1}^{2} 2\sqrt{x}\,dx \)
\( \displaystyle = 2 \cdot \frac{2}{3}\Big[x^{3/2}\Big]_{1}^{2} \)
\( \displaystyle = \frac{4}{3}\left(2^{3/2} - 1\right) \)
Since \(2^{3/2} = 2\sqrt{2}\), the area is \(\frac{4}{3}[2\sqrt{2} - 1]\) square units.