Match the LIST-I with LIST-II (where [x] denotes greatest integer function), LIST-I Integral, LIST-II Values, A., _0^2 [x] dx, I., 0, B., _0^1.5 [x^2] dx, II., 1, C., _(-/4)^(/4) x dx, III., √2, D., _(-/4)^(/4) x dx, IV., 2-√2, Choose the correct answer from the options given below: Option 1: A-II, B-IV, C-I, D-III Option 2: A-IV, B-II, C-I, D-III Option 3: A-II, B-IV, C-III, D-I Option 4: A-IV, B-II, C-III, D-I

Answer: A-II, B-IV, C-I, D-III. Solution: _0^2[x]dx = 0+1 = 1. For [x^2], the values 0,1,2 occur on [0,1), [1,√2), [√2,1.5], giving (√2-1)+2(1.5-√2) = 2-√2. Being odd, x integrates to 0, while x = 2()/4 = √2.

Calculus

Integrals

Hard

core

Match the LIST-I with LIST-II (where $[x]$ denotes greatest integer function)

LIST-I
Integral LIST-II
Values
A. $\displaystyle\int_{0}^{2} [x]\, dx$ I. $0$
B. $\displaystyle\int_{0}^{1.5} [x^2]\, dx$ II. $1$
C. $\displaystyle\int_{-\pi/4}^{\pi/4} \sin x \, dx$ III. $\sqrt{2}$
D. $\displaystyle\int_{-\pi/4}^{\pi/4} \cos x \, dx$ IV. $2-\sqrt{2}$

Choose the correct answer from the options given below:

	LIST-I Integral		LIST-II Values
A.	$\displaystyle\int_{0}^{2} [x]\, dx$	I.	$0$
B.	$\displaystyle\int_{0}^{1.5} [x^2]\, dx$	II.	$1$
C.	$\displaystyle\int_{-\pi/4}^{\pi/4} \sin x \, dx$	III.	$\sqrt{2}$
D.	$\displaystyle\int_{-\pi/4}^{\pi/4} \cos x \, dx$	IV.	$2-\sqrt{2}$

A-II, B-IV, C-I, D-III

A-IV, B-II, C-I, D-III

A-II, B-IV, C-III, D-I

A-IV, B-II, C-III, D-I

✅ Correct Option: 1

$\int_0^2[x]dx = 0+1 = 1$ . For $[x^2]$ , the values $0,1,2$ occur on $[0,1)$ , $[1,\sqrt{2})$ , $[\sqrt{2},1.5]$ , giving $(\sqrt{2}-1)+2(1.5-\sqrt{2}) = 2-\sqrt{2}$ . Being odd, $\sin x$ integrates to $0$ , while $\int\cos x = 2\sin\frac{\pi}{4} = \sqrt{2}$ .

2026: 22 May Shift 1

$\int \frac{\sqrt{9+(\log x)^2}}{x} dx$ is equal to: (where $C$ is an arbitrary constant and consider $\log_e x = \log x$ )

2026: 18 May Shift 1

Match List-I with List-II

List-I	List-II
(A) $\displaystyle\int_{\pi/6}^{\pi/3} \cos x \, dx$	(I) $0$
(B) $\displaystyle\int_{-\pi}^{\pi} (\sin^{61}x + x^{123}) \, dx$	(II) $4$
(C) $\displaystyle\int_{0}^{\pi/2} \dfrac{1}{1+\tan^3 x} \, dx$	(III) $\dfrac{\sqrt{3}-1}{2}$
(D) $\displaystyle\int_{-2}^{2} (2 - \left\vert x \right\vert) \, dx$	(IV) $\dfrac{\pi}{4}$

Choose the correct answer from the options given below:

2025: 14 May Shift 2

The value of the definite integral $I = \int_{1}^{2} \frac{1}{x(1 + x^2)}dx$ is:

2026: 29 May Shift 1

Match List-I with List-II

List-I	List-II
(A) $\displaystyle\int_{-2}^{2}\cos^4x\sin^5x\,dx$	(I) $\pi$
(B) $\displaystyle\int_{-1}^{1}\lvert x \rvert dx$	(II) $\dfrac{\pi}{4}$
(C) $\displaystyle\int_0^{\pi/2}\cos^2x\,dx$	(III) 0
(D) $\displaystyle\int_{-\pi/2}^{\pi/2}(x^3+x\cos x+\tan^5x+1)dx$	(IV) 1

Choose the correct answer from the options given below:

2025: 30 May Shift 1

$\int \left(\frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha}\right) dx =$

(Given that $c$ is an arbitrary constant)

2025: 21 May Shift 2

The value of $\int_1^3 \frac{x^2}{x^3+1}dx$

Solution

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