Match the List-I with List-II
| List-I | List-II | ||
| Definite integral | Value of definite integral | ||
| (A) | \( \int_{4}^{9}\displaystyle\frac{1}{\sqrt{x}}\,dx \) | (I) | \( \displaystyle\frac{2}{3} \) |
| (B) | \( \int_{-1}^{1}\displaystyle x^{3}\,dx \) | (II) | 1 |
| (C) | \( \int_{-1}^{1}\displaystyle x^{2}\,dx \) | (III) | 0 |
| (D) | \( \int_{1}^{2}\displaystyle [x]\,dx \), where \( [\bullet] \) denotes greatest integer function | (IV) | 2 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) - (IV), (B) - (III), (C) - (I), (D) - (II)
(A) \( \int_{4}^{9}\frac{1}{\sqrt{x}}\,dx \)
This matches (IV) 2.
\( \displaystyle \int_{4}^{9}x^{-1/2}\,dx=\left[2\sqrt{x}\right]_{4}^{9} \)
\( \displaystyle =2(3)-2(2) \)
\( \displaystyle =2 \)
(B) \( \int_{-1}^{1}x^{3}\,dx \)
This matches (III) 0.
\( x^{3} \) is an odd function, so its integral over the symmetric interval [−1, 1] is 0.
\( \displaystyle \int_{-1}^{1}x^{3}\,dx=\left[\frac{x^{4}}{4}\right]_{-1}^{1} \)
\( \displaystyle =\frac{1}{4}-\frac{1}{4} \)
\( \displaystyle =0 \)
(C) \( \int_{-1}^{1}x^{2}\,dx \)
This matches (I) \( \frac{2}{3} \).
\( x^{2} \) is an even function, so
\( \displaystyle \int_{-1}^{1}x^{2}\,dx=2\int_{0}^{1}x^{2}\,dx \)
\( \displaystyle =2\left[\frac{x^{3}}{3}\right]_{0}^{1} \)
\( \displaystyle =\frac{2}{3} \)
(D) \( \int_{1}^{2}[x]\,dx \)
This matches (II) 1.
For 1 ≤ x < 2, [x] = 1 (the single point x = 2 does not change the integral).
\( \displaystyle \int_{1}^{2}[x]\,dx=\int_{1}^{2}1\,dx \)
\( \displaystyle =2-1=1 \)
So the matching is (A) - (IV), (B) - (III), (C) - (I), (D) - (II).