Match List-I with List-II
| List-I (Region) | List-II (Area in sq. units) | ||
| (A) | Bounded by y = 4x + 3, the x-axis and the lines x = 1 and x = 2 | (I) | \(\displaystyle\frac{16}{3}\) |
| (B) | Bounded by \(y = \sqrt{x}\), the x-axis and the line x = 4 | (II) | 4 |
| (C) | Bounded by \(y = x^{3}\), the x-axis and the line x = 2 | (III) | \(\pi\) |
| (D) | Enclosed by the circle \(x^{2}+y^{2}=4\) in the first quadrant | (IV) | 9 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
In (A), (B) and (C) the curve lies above the x-axis over the given interval, so each area is the integral of y over that interval.
(A) Bounded by y = 4x + 3, the x-axis and the lines x = 1 and x = 2
This matches (IV) 9.
\( \displaystyle \int_{1}^{2} (4x+3)\,dx = \left[2x^{2}+3x\right]_{1}^{2} \)
\(= (8+6)-(2+3) = 9\)
(B) Bounded by \(y = \sqrt{x}\), the x-axis and the line x = 4
This matches (I) \(\frac{16}{3}\).
\( \displaystyle \int_{0}^{4} \sqrt{x}\,dx = \left[\frac{2}{3}x^{3/2}\right]_{0}^{4} = \frac{2}{3}\times 8 = \frac{16}{3} \)
(C) Bounded by \(y = x^{3}\), the x-axis and the line x = 2
This matches (II) 4.
\( \displaystyle \int_{0}^{2} x^{3}\,dx = \left[\frac{x^{4}}{4}\right]_{0}^{2} = 4 \)
(D) Enclosed by the circle \(x^{2}+y^{2}=4\) in the first quadrant
This matches (III) \(\pi\).
The circle has radius 2, and the first quadrant holds one quarter of it:
\( \displaystyle \frac{1}{4}\pi\times 2^{2} = \pi \)