The area (in sq.units) of region bounded by \(y^2 = 9x, x=2, x =4\) and the x-axis in the first quadrant is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(4[4 - \sqrt{2}]\)
In the first quadrant y ≥ 0, so the curve is \(y = 3\sqrt{x}\). The area between the curve and the x-axis from x = 2 to x = 4 is
\( \displaystyle A = \int_{2}^{4}3\sqrt{x}\,dx = 3\left[\frac{2}{3}x^{3/2}\right]_{2}^{4} \)
\( \displaystyle = 2\left(4^{3/2} - 2^{3/2}\right) \)
\( \displaystyle = 2\left(8 - 2\sqrt{2}\right) \)
\(= 16 - 4\sqrt{2} = 4[4 - \sqrt{2}]\)