Match the LIST-I with LIST-II
| LIST-I (Definite integrals) | LIST-II (Value) | ||
| A. | \( \int_{-6}^{6}\displaystyle|x|\,dx \) | I. | 1 |
| B. | \( \int_{-\pi/2}^{\pi/2}\displaystyle\sin^{2017}x\,dx \) | II. | 36 |
| C. | \( \int_{0}^{\pi/4}\displaystyle\cos x\,dx \) | III. | 0 |
| D. | \( \int_{0}^{2}\displaystyle\frac{\sqrt{x}}{\sqrt{x}+\sqrt{2-x}}\,dx \) | IV. | \( \displaystyle\frac{1}{\sqrt{2}} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A-II, B-III, C-IV, D-I
(A) \( \int_{-6}^{6}|x|\,dx \)
This matches (II) 36. |x| is an even function, so
\( \displaystyle \int_{-6}^{6}|x|\,dx=2\int_{0}^{6}x\,dx \)
\( \displaystyle =2\left[\frac{x^{2}}{2}\right]_{0}^{6} \)
\( \displaystyle =36 \)
(B) \( \int_{-\pi/2}^{\pi/2}\sin^{2017}x\,dx \)
This matches (III) 0. \( \sin^{2017}(-x)=-\sin^{2017}x \), an odd function, and the interval is symmetric about 0.
(C) \( \int_{0}^{\pi/4}\cos x\,dx \)
This matches (IV) \( \frac{1}{\sqrt{2}} \).
\( \displaystyle \int_{0}^{\pi/4}\cos x\,dx=\Big[\sin x\Big]_{0}^{\pi/4} \)
\( \displaystyle =\frac{1}{\sqrt{2}}-0 \)
\( \displaystyle =\frac{1}{\sqrt{2}} \)
(D) \( \int_{0}^{2}\frac{\sqrt{x}}{\sqrt{x}+\sqrt{2-x}}\,dx \)
This matches (I) 1. Call it I and use \( \int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx \) with a = 2:
\( \displaystyle I=\int_{0}^{2}\frac{\sqrt{2-x}}{\sqrt{2-x}+\sqrt{x}}\,dx \)
Adding the two forms of I:
\( \displaystyle 2I=\int_{0}^{2}\frac{\sqrt{x}+\sqrt{2-x}}{\sqrt{x}+\sqrt{2-x}}\,dx \)
\( \displaystyle =\int_{0}^{2}1\,dx \)
\( \displaystyle =2 \)
\( \displaystyle I=1 \)
So the matching is (A)-(II), (B)-(III), (C)-(IV), (D)-(I).
Tip: Remember the result \( \int_{0}^{a}\frac{f(x)}{f(x)+f(a-x)}\,dx=\frac{a}{2} \): (D) is this with a = 2.