Let R be the region bounded by the curve \(y = 2x^{2}\), the x-axis and the line x = 3. Which of the following statements are correct?
(A) The area of R is \(\int_{0}^{3}\displaystyle 2x^{2}\,dx\)
(B) The area of R is 18 sq. units
(C) The area of R is \(\int_{0}^{18}\displaystyle \sqrt{\frac{y}{2}}\,dy\)
(D) The area of R is one-third of the area of the rectangle with vertices (0, 0), (3, 0), (3, 18) and (0, 18)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (B) and (D) only
The curve touches the x-axis at the origin, so R lies under the curve from x = 0 to x = 3:
\( \displaystyle \int_{0}^{3}2x^{2}\,dx = \left[\frac{2x^{3}}{3}\right]_{0}^{3} = 18 \)
The integral in (C) uses \(x = \sqrt{\frac{y}{2}}\) from y = 0 to y = 18 (the height at x = 3), so it measures the area between the curve and the y-axis:
\( \displaystyle \int_{0}^{18}\sqrt{\frac{y}{2}}\,dy = \frac{1}{\sqrt{2}}\left[\frac{2}{3}y^{3/2}\right]_{0}^{18} \)
\( \displaystyle = \frac{1}{\sqrt{2}}\times\frac{2}{3}\times54\sqrt{2} = 36 \)
The rectangle in (D) has area
\(3\times18 = 54\)
So the area of R is \(\int_{0}^{3}2x^{2}\,dx = 18\), the integral in (C) is 36, and the rectangle has area 54.
Checking the statements
(A) The area of R is \(\int_{0}^{3}2x^{2}\,dx\) — Correct. R lies under the curve from x = 0 to x = 3.
(B) The area of R is 18 sq. units — Correct. \(\frac{2\times27}{3} = 18\).
(C) The area of R is \(\int_{0}^{18}\sqrt{\frac{y}{2}}\,dy\) — Incorrect. This integral is 36, the area between the curve and the y-axis.
(D) The area of R is one-third of the rectangle with vertices (0, 0), (3, 0), (3, 18), (0, 18) — Correct. \(\frac{54}{3} = 18\).