CUET Mathematics - Calculus - Integrals | Free Questions - Page 7

Q1:

2025: 21 May Shift 2

Calculus

>

Integrals

Medium

Core

Value of $\int_2^3 \frac{\sqrt{x}}{\sqrt{x} + \sqrt{5-x}}dx$ is

0

$\frac{1}{2}$

1

5

Correct Answer

Option 2

Correct Answer

Explanation

Q2:

2025: 21 May Shift 2

Calculus

>

Integrals

Medium

Core

$\int_0^{\pi/2} \sqrt{1 - \sin 2x}\,dx$ is equal to:

$2(\sqrt{2} - 1)$

$2(\sqrt{2} + 1)$

2

$2\sqrt{2}$

Correct Answer

Option 1

Correct Answer

Explanation

Q3:

2025: 21 May Shift 2

Calculus

>

Integrals

Medium

Core

$\int \frac{\sin x - x\cos x}{x(x + \sin x)}dx =$ (where C is an arbitrary constant)

$\log\left|\frac{x}{x - \sin x}\right| + C$

$\log\left|\frac{x}{x + x\sin x}\right| + C$

$\log\left|\frac{x}{x + \sin x}\right| + C$

$\log\left|\frac{x + \sin x}{x}\right| + C$

Correct Answer

Option 3

Correct Answer

Explanation

Q4:

2025: 21 May Shift 2

Calculus

>

Integrals

Medium

Core

If $\int \frac{x^4}{x-2}dx = px + qx^2 + rx^3 + sx^4 + t\log |x - 2| + C$ , where C is an arbitrary constant and p, q, r, s, t are real numbers, then the correct arrangement of p, q, r, s, t is:

$p > q > r > s > t$

$q > r > p > t > s$

$r > t > s > p > q$

$t > p > q > r > s$

Correct Answer

Option 4

Correct Answer

Explanation

Q5:

2025: 21 May Shift 2

Calculus

>

Integrals

Hard

Applied

The value of the definite integral $\int_0^1 e^x \frac{(1-x)^2}{(1+x^2)^2}dx$ is:

$\frac{e}{2}$

$\frac{e}{2} - 1$

$\frac{e}{2} + 1$

$2e + 1$

Correct Answer

Option 2

Correct Answer

Explanation

Q6:

2025: 21 May Shift 1

Calculus

>

Integrals

Medium

Common

For $x > e$ , $\int \frac{dx}{x - \sqrt{x}}$ is equal to

$-2 \log_e |\sqrt{x} - 1| + C$ ; C is an arbitrary constant

$4 \log_e |\sqrt{x} - 1| + C$ ; C is an arbitrary constant

$\log_e |\sqrt{x} - 1| + C$ ; C is an arbitrary constant

$2 \log_e |\sqrt{x} - 1| + C$ ; C is an arbitrary constant

Correct Answer

Option 4

Correct Answer

Explanation

Q7:

2025: 21 May Shift 1

Calculus

>

Integrals

Medium

Common

Match List-I with List-II

List-I	List-II
Definite integral	Value
---	---
(A) $\int_{1}^{e} \frac{\log x}{x} dx$	(I) 4
(B) $\int_{-2}^{2} x^3(1 - x^2) dx$	(II) $\frac{1}{2}$
(C) $\int_{1}^{2} x \, dx$	(III) 0
(D) $\int_{-2}^{2} \vert x\vert dx$	(IV) $\frac{3}{2}$

Choose the correct answer from the options given below:

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

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

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

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

Correct Answer

Option 1

Correct Answer

Explanation

Q8:

2025: 21 May Shift 1

Calculus

>

Integrals

Medium

Core

The value of $\int \left\{ \frac{1}{\log_e x} - \frac{1}{(\log_e x)^2} \right\} dx$ is

$x \log_e x + c$ ; where c is an arbitrary constant

$e^x \log_e x + c$ ; where c is an arbitrary constant

$\frac{e^x}{\log_e x} + c$ ; where c is an arbitrary constant

$\frac{x}{\log_e x} + c$ ; where c is an arbitrary constant

Correct Answer

Option 4

Correct Answer

Explanation

Q9:

2025: 21 May Shift 1

Calculus

>

Integrals

Medium

Core

The value of $\int_0^{\pi/2} \log_e \left(\frac{5 + 2 \sin x}{5 + 2 \cos x}\right) dx$ is

0

1

2

$\pi$

Correct Answer

Option 1

Correct Answer

Explanation

Q10:

2025: 21 May Shift 1

Calculus

>

Integrals

Easy

Core

Match List-I with List-II (Given that $c$ is an arbitrary constant)

List-I	List-II
(A) $\int \frac{dx}{\sqrt{a^2 - x^2}} =$	(I) $\log_e \vert x + \sqrt{x^2 - a^2}\vert + c$
(B) $\int \sqrt{a^2 - x^2} dx =$	(II) $\sin^{-1} \frac{x}{a} + c$
(C) $\int \sqrt{x^2 - a^2} dx =$	(III) $\frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1} \frac{x}{a} + c$
(D) $\int \frac{dx}{\sqrt{x^2 - a^2}} =$	(IV) $\frac{x}{2} \sqrt{x^2 - a^2} - \frac{a^2}{2} \log_e \vert x + \sqrt{x^2 - a^2}\vert + c$

Choose the correct answer from the options given below:

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

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

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

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

Correct Answer

Option 4

Correct Answer

Explanation

Q11:

2025: 19 May Shift 1

Calculus

>

Integrals

Medium

Common

The value of $\int_0^1 x e^x dx$ is:

$\frac{1}{e}$

$e$

$1$

$-1$

Correct Answer

Option 3

Correct Answer

Explanation

Q12:

2025: 19 May Shift 1

Calculus

>

Integrals

Medium

Common

Value of $\int \frac{2}{(x-3)\sqrt{x+1}} dx$ is: (Here C is an arbitrary constant)

$\log\left|\frac{x-3}{x+1}\right| + C$

$\log\left|\frac{\sqrt{x-1}-2}{\sqrt{x+1}}\right| + C$

$\frac{1}{2}\log\left|\frac{\sqrt{x+1}-1}{\sqrt{x+1}+1}\right| + C$

$\log\left|\frac{\sqrt{x+1}-2}{\sqrt{x+1}+2}\right| + C$

Correct Answer

Option 4

Correct Answer

Explanation

Q13:

2025: 19 May Shift 1

Calculus

>

Integrals

Hard

Core

If $2f(x) + f\left(\frac{1}{x}\right) = x^2 + 1$ , then $\int f(x) dx$ is: (Here C is an arbitrary constant)

$\frac{1}{3}\left(\frac{2}{3}x^3 + \frac{1}{x} + x\right) + C$

$\frac{2}{3}x^3 - \frac{1}{x} + x + C$

$\frac{x^3}{3} + x + C$

$\frac{1}{3}\left(\frac{2}{3}x^3 - x\right) + C$

Correct Answer

Option 1

Correct Answer

Explanation

Q14:

2025: 19 May Shift 1

Calculus

>

Integrals

Medium

Core

Match List-I with List-II

List-I	List-II
(A) $\int_{-a}^a f(x) dx = 0$	(I) 0
(B) $\int_0^{2a} f(x) dx = 2\int_0^a f(x) dx$	(II) 1
(C) $\int_{-\pi}^{\pi} \cos x dx$	(III) $f$ is an odd function
(D) $\int_{-1}^1 x^{101} dx + 1$	(IV) $f(2a-x) = f(x)$

Choose the correct answer from the options given below:

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

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

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

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

Correct Answer

Option 1

Correct Answer

Explanation

Q15:

2025: 19 May Shift 1

Calculus

>

Integrals

Hard

Core

$\int \tan^{-1}\sqrt{x}$ $dx$ equals to: (Here C is an arbitrary constant)

$(x+1) \tan^{-1}\sqrt{x} - \sqrt{x} + C$

$x \tan^{-1}\sqrt{x} - \sqrt{x} + C$

$\sqrt{x} - x \tan^{-1}\sqrt{x} + C$

$\sqrt{x} - (x+1) \tan^{-1}\sqrt{x} + C$

Correct Answer

Option 1

Correct Answer

Explanation

Q16:

2025: 19 May Shift 1

Calculus

>

Integrals

Medium

Applied

The value of the definite integral $I = \int_0^2 x\sqrt{2-x} dx$ is:

$\frac{16}{15\sqrt{2}}$

$\frac{16\sqrt{3}}{15}$

$\frac{16\sqrt{2}}{15}$

$\frac{5\sqrt{2}}{7}$

Correct Answer

Option 3

Correct Answer

Explanation

Q17:

2025: 16 May Shift 1

Calculus

>

Integrals

Medium

Common

The value of $\int_0^1 [\log x - \log(1-x)] dx$ is

$\frac{1}{2}$

2

1

0

Correct Answer

Option 4

Correct Answer

Explanation

Q18:

2025: 16 May Shift 1

Calculus

>

Integrals

Medium

Common

$\int \frac{(x-3)e^x}{(x-1)^3} dx$ is equal to

$\frac{e^x}{(x-1)} + C$ , C is an arbitary constant

$\frac{e^x}{(x-3)^2} + C$ , C is an arbitary constant

$\frac{e^x}{(x-1)^2} + C$ , C is an arbitary constant

$\frac{e^x}{(x-1)^3} + C$ , C is an arbitary constant

Correct Answer

Option 3

Correct Answer

Explanation

Q19:

2025: 16 May Shift 1

Calculus

>

Integrals

Hard

Core

Match List-I with List-II

List-I	List-II
Functions	Integrals
---	---
(A) $\int \frac{dx}{x^2 - 4}, x \neq \pm 2$	(I) $\log \vert x + \sqrt{4 + x^2}\vert + C$ , where $C$ is an arbitrary constant
(B) $\int \frac{1}{\sqrt{16 - x^2}} dx; \vert x\vert < 4$	(II) $\sin^{-1} \left( \frac{x}{4} \right) + C$ , where $C$ is an arbitrary constant
(C) $\int \frac{1}{16 + x^2} dx$	(III) $\frac{1}{4} \log \left\vert \frac{x-2}{x+2} \right\vert + C$ , where $C$ is an arbitrary constant
(D) $\int \frac{1}{\sqrt{4 + x^2}} dx$	(IV) $\frac{1}{4} \tan^{-1} \left( \frac{x}{4} \right) + C$ , where $C$ is an arbitrary constant

Choose the correct answer from the options given below:

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

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

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

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

Correct Answer

Option 2

Correct Answer

Explanation

Q20:

2025: 16 May Shift 1

Calculus

>

Integrals

Medium

Core

$\int \frac{\cos 2x - \cos 2α}{\cos x - \cos α} dx$ is equal to

$2 \sin x + 2x \cos α + C$ , where C is an arbitary constant

$2 \sin x + 2 \sin α + C$ , where C is an arbitary constant

$\frac{\sin 2x}{2} + x \cos 2α + C$ , where C is an arbitary constant

$\frac{\sin 2x}{2} + \frac{\sin 2α}{2} + C$ , where C is an arbitary constant

Correct Answer

Option 1

Correct Answer

Explanation

Q21:

2025: 16 May Shift 1

Calculus

>

Integrals

Medium

Core

$\int_{3\pi/8}^{\pi/8} \frac{ \tan^{2025} x}{ \tan^{2025} x + \cot^{2025} x} dx$ is equal to

$\frac{\pi}{8}$

$-\frac{\pi}{8}$

$\frac{\pi}{4}$

$-\frac{\pi}{4}$

Correct Answer

Option 2

Correct Answer

Explanation

Q22:

2025: 16 May Shift 1

Calculus

>

Integrals

Medium

Applied

If the integral $I = \int \frac{x^2}{\sqrt{1+x}} dx = \frac{1}{\alpha} (1+x)^\alpha - \frac{8\alpha}{15} (1+x)^{\alpha-1} + 2(1+x)^{\alpha-2} + C$ , $C$ is constant of integration, then the value of $\alpha$ is:

1

1/2

3/2

5/2

Correct Answer

Option 4

Correct Answer

Explanation

Q23:

2025: 15 May Shift 2

Calculus

>

Integrals

Medium

Common

$\int \frac{f'(x)}{f(x) \log_e[f(x)]} dx$ is equal to

$f(x) \cdot log_e[f(x)] + C$ ; $C$ is a constant of integration

$\frac{log_e[f(x)]}{f(x)} + C$ ; $C$ is a constant of integration

$log_e(log_e[f(x)]) + C$ ; $C$ is a constant of integration

$\frac{log_e(log_e[f(x)])}{f(x)} + C$ ; $C$ is a constant of integration

Correct Answer

Option 3

Correct Answer

Explanation

Q24:

2025: 15 May Shift 2

Calculus

>

Integrals

Hard

Common

$\int \frac{(x^4 - x)^{1/4}}{x^5} dx$ is equal to

$\frac{4}{15}\left(1 - \frac{1}{x^3}\right)^{5/4} + C$ ; $C$ is a constant of integration

$\frac{4}{3}\left(1 - \frac{1}{x^3}\right)^{5/4} + C$ ; $C$ is a constant of integration

$\frac{4}{15}\left(1 - \frac{1}{x^3}\right)^{3/4} + C$ ; $C$ is a constant of integration

$\frac{4}{15}\left(1 - \frac{1}{x^3}\right)^{1/5} + C$ ; $C$ is a constant of integration

Correct Answer

Option 1

Correct Answer

Explanation

Q25:

2025: 15 May Shift 2

Calculus

>

Integrals

Medium

Core

The value of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\sin|x| + \cos|x|) dx$ is

0

2

4

$\frac{5}{2}$

Correct Answer

Option 3

Correct Answer

Explanation

Q26:

2025: 15 May Shift 2

Calculus

>

Integrals

Medium

Core

Match List-I with List-II

[.] denotes the greatest integer function.

List-I	List-II
(A) $\int_0^3 [x]dx$	(I) $\frac{1}{2}$
(B) $\int_0^1 [2x]dx$	(II) 1
(C) $\int_0^1 [3x]dx$	(III) $\frac{3}{2}$
(D) $\int_0^1 [4x]dx$	(IV) 3

Choose the correct answer from the options given below:

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

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

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

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

Correct Answer

Option 2

Correct Answer

Explanation

Q27:

2025: 15 May Shift 2

Calculus

>

Integrals

Medium

Core

For $x > 1$ , $\int \frac{e^{7\log x} - e^{5\log x}}{e^{5\log x} - e^{4\log x}} dx$ equals.

$\frac{x^3}{3} + \frac{x^2}{2} + C$ ; $C$ is a constant of integration

$\frac{x^3}{3} - \frac{x^2}{2} + C$ ; $C$ is a constant of integration

$\frac{x^3}{6} + \frac{x^2}{4} + C$ ; $C$ is a constant of integration

$\frac{x^3}{6} - \frac{x^2}{4} + C$ ; $C$ is a constant of integration

Correct Answer

Option 1

Correct Answer

Explanation

Q28:

2025: 15 May Shift 2

Calculus

>

Integrals

Medium

Applied

The value of the definite integral $I = \int_{-1}^{1} \frac{1}{1 + \sqrt{e^x}} dx$ is:

0

1

2

3

Correct Answer

Option 2

Correct Answer

Explanation

Q29:

2025: 15 May Shift 2

Calculus

>

Integrals

Medium

Applied

If $I = \int \frac{x}{x - \sqrt{x^2 - 4}} dx = \alpha x^3 + \beta(x^2 - 4)^{\frac{3}{2}} + \gamma$ , where $\gamma$ is constant of integration, then

$\alpha = 2\beta$

$2\alpha = \beta$

$\alpha = \beta$

$\alpha = 3\beta$

Correct Answer

Option 3

Correct Answer

Explanation

Q30:

2025: 15 May Shift 1

Calculus

>

Integrals

Medium

Common

Match List-I with List-II

List-I	List-II
(A) The value of $\int_{0}^{4} \vert x\vert \, dx$ is	(I) 3
(B) The value of $\int_{-2}^{2} \vert x\vert \, dx$ is	(II) -1
(C) The value of $\int_{0}^{3} [x] \, dx$ is	(III) 8
(D) The value of $\int_{-1}^{1} [x] \, dx$ is	(IV) 4

Choose the correct answer from the options given below:

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

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

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

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

Correct Answer

Option 4

Correct Answer

Explanation

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