For k > 0, let A(k) denote the area (in sq. units) of the region bounded by the parabola \(y^{2}=4x\) and the line x = k. Which of the following statements are correct?
(A) \(A(k) = \displaystyle\int_{0}^{k}2\sqrt{x}\,dx\)
(B) \(A(1) = \displaystyle\frac{8}{3}\)
(C) A(4) = 4A(1)
(D) A(k) is two-thirds of the area of the rectangle with sides k and \(4\sqrt{k}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (B) and (D) only
The line x = k meets the parabola where \(y^{2} = 4k\), that is \(y = \pm2\sqrt{k}\). The parabola is symmetric about the x-axis, so the area is twice the part above the axis, under \(y = 2\sqrt{x}\):
\( \displaystyle A(k) = 2\int_{0}^{k}2\sqrt{x}\,dx \)
\( \displaystyle = 4\left[\frac{2}{3}\,x^{3/2}\right]_{0}^{k} = \frac{8}{3}\,k^{3/2} \)
The single integral \(\displaystyle\int_{0}^{k}2\sqrt{x}\,dx\) is only the upper half, \(\displaystyle\frac{4}{3}\,k^{3/2}\). The values at k = 1 and k = 4 are
\( \displaystyle A(1) = \frac{8}{3},\ \ A(4) = \frac{8}{3}\times8 = \frac{64}{3} \)
The rectangle with sides k and \(4\sqrt{k}\) has area \(4k^{3/2}\), and
\( \displaystyle \frac{2}{3}\times4k^{3/2} = \frac{8}{3}\,k^{3/2} \)
So \(\displaystyle A(k) = \frac{8}{3}\,k^{3/2}\), giving \(\displaystyle A(1) = \frac{8}{3}\) and \(\displaystyle A(4) = \frac{64}{3} = 8A(1)\), while the integral in (A) equals \(\displaystyle\frac{4}{3}\,k^{3/2}\).
Checking the statements
(A) \(A(k) = \displaystyle\int_{0}^{k}2\sqrt{x}\,dx\) — Incorrect. This integral is only the half above the x-axis, \(\displaystyle\frac{4}{3}\,k^{3/2}\).
(B) \(A(1) = \displaystyle\frac{8}{3}\) — Correct. \(\displaystyle\frac{8}{3}\times1^{3/2} = \frac{8}{3}\).
(C) A(4) = 4A(1) — Incorrect. \(\displaystyle A(4) = \frac{64}{3}\), which is 8A(1).
(D) A(k) is two-thirds of the area of the rectangle with sides k and \(4\sqrt{k}\) — Correct. \(\displaystyle\frac{2}{3}\times4k^{3/2} = \frac{8}{3}\,k^{3/2}\).