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) → $\frac{4}{3}[2\sqrt{2}-1]$
$y^2 = 4x \Rightarrow y = 2\sqrt{x}$ (first quadrant)
Area $= \displaystyle \int_{1}^{2} y\,dx = \int_{1}^{2} 2\sqrt{x}\,dx$
$= 2 \int_{1}^{2} x^{1/2}\,dx$
$= 2\left[\frac{2}{3}x^{3/2}\right]_{1}^{2}$
$= \frac{4}{3}\left(2^{3/2} - 1^{3/2}\right)$
$= \frac{4}{3}(2\sqrt{2} - 1)$
$\frac{4}{3}(2\sqrt{2} - 1)$