$\mathop \smallint \nolimits_0^{\pi /2} \,\,\log \,\,\sin \,\,x{\rm{ }}\,dx$ is equal to

$ - \,\left( {\frac{{\rm{\pi }}}{2}} \right)\,\,\log \,\,2$

${\rm{\pi }}\,\,\log \,\,\frac{1}{2}$

$ - \,{\rm{\pi }}\,\,\log \,\,\frac{1}{2}$

$\frac{{\rm{\pi }}}{2}\,\,\log \,\,2$

If $f(x)$ is a periodic function, with period $T$, then

$\int_\limits a^b f(x) d x=\int_\limits a^{b+T} f(x) d x$

$\int_\limits a^b f(x) d x=\int_\limits {a+T}^b f(x) d x$

$\int_\limits a^b f(x) d x=\int_\limits{a+T}^{b+2 T} f(x) d x$

$\smallint \,\,\frac{1}{{1\,\, + \,\,\cos \,\,ax}}\,\,\;dx$ is equal to

$\cot \,\,\frac{{ax}}{2}\,\, + \,\,c$

$\frac{1}{a}\;\,\,\left( {{\rm{cosec}}\,\,\;ax\,\, - \,\,{\rm{cot}}\,\,\;ax} \right)\,\, + \,\,c$

$\frac{1}{a}\,\,({\rm{cosec}}\;\,\,ax\,\, + \,\,\cot \,\,ax)\,\, + \,\,c$

If $f\left( y \right) = {e^y},\,\,\;g\left( y \right) = y;\;\;\,\;y > 0$ and $F\left( t \right) = \mathop \smallint \nolimits_0^t \,\,f\left( {t - y} \right)\,\,g\,\,\left( y \right)\;\,\,dy$, then

$F\,\,\left( t \right) = 1 - {e^{ - t}}\,\,\left( {1 + t} \right)$

$F\,\,\left( t \right) = t{e^t}$

$F\,\,\left( t \right) = t{e^{ - t}}$

Back to DEFINITE INTEGRATION

Your Progress

0 / 791 Answered

Correct

Incorrect

Accuracy

JEE • MATHEMATICS • DEFINITE INTEGRATION

Definite integration practice · Easy

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

Take a chapter quiz →

Showing 541-570 of 791 questions

Preparing practice and saved progress…

541 Easy

541 of 791

Application +4 | -1

\int\limits_{0}^1|\sin 2 \pi x| \mathrm{d} x

is equal to

View answer and explanation

Correct option: D

2 / \pi

Option D (

2 / \pi

) is correct because the integral of the absolute value of sin(2πx) from 0 to 1 is calculated by splitting the integral into two parts and evaluating each part separately.

542 Easy

542 of 791

Application +4 | -1

\int\limits_{0}^a \frac{1}{1+4 x^2} d x=\frac{\pi}{8}

, then

a

equals

View answer and explanation

Correct option: B

\frac{\pi}{4}

Option D is incorrect. The correct answer is 1/2. The integral is evaluated using the arctangent function, and solving for 'a' gives 1/2.

543 Easy

543 of 791

Analyzing +4 | -1

\int \nolimits_0^{\pi /2} \,\,\log \,\,\sin \,\,x\mathrm{ }\,dx

is equal to

View answer and explanation

Correct option: A

- \,\left( {\frac{\mathrm{\pi }}{2}} \right)\,\,\log \,\,2

Option A is correct because by using properties of definite integrals and trigonometric identities, the integral can be simplified and evaluated to

- \left( {\frac{\mathrm{\pi }}{2}} \right)\,\log \,2

544 Easy

544 of 791

Application +4 | -1

\int\limits_{{\pi / 4}}^{3 \pi / 4} \frac{d x}{1+\cos x}

is equal to

View answer and explanation

Correct option: A

Option A is correct because the integral is solved using trigonometric identities and properties of definite integrals.

545 Easy

545 of 791

Application +4 | -1

The value of the integral

\int \limits_0^2 \,x\,\,\left[ x \right]\,\,dx

, is

View answer and explanation

Correct option: B

\frac{3}{2}

Option B (

\frac{3}{2}

) is correct because the integral of

x[x]

from 0 to 2, where

[x]

is the greatest integer function, is

\frac{3}{2}

546 Easy

546 of 791

Application +4 | -1

The whole area of the curves

x=a \cos ^3 t, y=b \sin ^3 t

is given by

View answer and explanation

Correct option: A

\frac{3}{8} \pi a b

Option A (

\frac{3}{8} \pi a b

) is correct because the area enclosed by the parametric curve is

\frac{3}{8} \pi a b

547 Easy

547 of 791

Application +4 | -1

The integral

\int_1^2 {{e^x} \cdot {x^x}\left( {2 + {{\log }_e}x} \right)dx}

equals

View answer and explanation

Correct option: C

e\left( {4e - 1} \right)

Option C (

e(4e - 1)

) is correct because it is the result of evaluating the definite integral after applying the substitution method.

548 Easy

548 of 791

Application +4 | -1

\phi(x)=\int\limits_{{1 / x}}^{\sqrt{x}} \sin \left(t^2\right) \, d t

, then

\phi^{\prime}(1)

is equal to

View answer and explanation

Correct option: C

(3 / 2) \sin 1

Option C (

(3 / 2) \sin 1

) is correct because it is the result of applying Leibniz's rule for differentiating an integral with variable limits.

549 Easy

549 of 791

Understanding +4 | -1

Statement-I :

\int e^x \sin x d x=\frac{e^x}{2}(\sin x-\cos x)+c

.

Statement-II :

\int e^x\left(f(x)+f^{\prime}(x)\right) d x=e^x f(x)+c

View answer and explanation

Correct option: A

If both Statement - I and Statement - II are true, and Statement - II is the correct explanation of Statement- I.

Option A is correct because both statements are true, and Statement II correctly explains Statement I, as Statement I is a specific case of Statement II.

550 Easy

550 of 791

Application +4 | -1

Let

g(x)=\int_0^x f(t) d t

, where

f

is such that

\frac{1}{2} \leq f(t) \leq 1

for

t \in[0,1]

and,

0 \leq f(t) \leq \frac{1}{2}

for

t \in[1,2]

Then,

g(2)

satisfies the inequality,

View answer and explanation

Correct option: B

0 \le g(2) < 2

Option B (

0 \le g(2) < 2

) is correct because it is the range that

g(2)

satisfies based on the given inequalities for

f(t)

551 Easy

551 of 791

Understanding +4 | -1

f(x)

is a periodic function, with period

T

, then

View answer and explanation

Correct option: C

\int\limits_{a}^b f(x) d x=\int\limits_{{a+T}}^{b+T} f(x) d x

Option C is correct because shifting the integration limits by the period T of a periodic function does not change the value of the definite integral.

552 Easy

552 of 791

Application +4 | -1

The value of

\int \nolimits_2^4 \,\,\left\{ {\left| {x - 2} \right|\,\, + \,\,\left| {x - 3} \right|} \right\}dx\,\,\mathrm{is}

View answer and explanation

Correct option: C

\mathrm{3}

Option C (3) is correct because the integral is evaluated by splitting the interval based on the absolute value functions and then integrating.

553 Easy

553 of 791

Application +4 | -1

\int \,\,\frac{1}{{1\,\, + \,\,\cos \,\,ax}}\,\,\;dx

is equal to

View answer and explanation

Correct option: B

\frac{1}{a}\,\,\tan \,\,\frac{{ax}}{2}\,\, + \,\,c

Option B,

\frac{1}{a}\,\,\tan \,\,\frac{{ax}}{2}\,\, + \,\,c

, is correct because using trigonometric identities, the integral can be simplified and solved, resulting in this expression.

554 Easy

554 of 791

Application +4 | -1

f\left( y \right) = {e^y},\,\,\;g\left( y \right) = y;\;\;\,\;y > 0

and

F\left( t \right) = \int \nolimits_0^t \,\,f\left( {t - y} \right)\,\,g\,\,\left( y \right)\;\,\,dy

, then

View answer and explanation

Correct option: B

F\,\,\left( t \right) = {e^t} - \,\,\left( {1 + t} \right)

Option B (

F\,\left( t \right) = {e^t} - \,\left( {1 + t} \right)

) is correct because it is the result of evaluating the convolution integral using integration by parts.

555 Easy

555 of 791

Application +4 | -1

\frac{d}{d x} G(x)=\frac{e^{\tan x}}{x}, x \in(0, \pi / 2)

, then

\int\limits_{{1 / 4}}^{1 / 2} \frac{2}{x} \cdot e^{\tan \left(\pi x^2\right)} d x

is equal to

View answer and explanation

Correct option: A

G(\pi / 4)-G(\pi / 16)

Option A (

G(\pi / 4)-G(\pi / 16)

) is correct because after substituting and changing variables, the integral can be expressed in terms of G(x) and evaluated at the limits.

Explore Other Chapters in MATHEMATICS

View All →

SEQUENCE OF SERIES 1764 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs DIFFERENTIAL EQUATIONS 977 MCQs

556 Easy

556 of 791

Application +4 | -1

\int \nolimits_0^{1.5} \,\,\left[ {{x^2}} \right]\,\,dx

View answer and explanation

Correct option: C

2\,\, - \,\,\sqrt 2

Option C (

2\,\, - \,\,\sqrt 2

) is correct because the integral of the greatest integer function of x² from 0 to 1.5 is calculated by splitting the integral into intervals where the greatest integer function has a constant value.

557 Easy

557 of 791

Application +4 | -1

\int\limits_{0}^2|x-1| d x=

View answer and explanation

Correct option: D

Option D (1) is correct because the integral of the absolute value function |x-1| from 0 to 2 is equal to 1.

558 Easy

558 of 791

Analyzing +4 | -1

{I_1} = \int \limits_0^{3\pi } \,\,f\left( {{{\cos }^2}\,\,x} \right)\,dx

and

{I_2} = \int \limits_0^\pi \,\,f\left( {{{\cos }^2}\,\,x} \right)\,\,dx

then

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because

{I_1} = 3{I_2}

, which is not represented by options A, B, or C.

559 Easy

559 of 791

Analyzing +4 | -1

\lim \limits_{{x \rightarrow \pi / 4}} \frac{\int\limits_{2}^{\sec ^2 x} f(t) d t}{x^2-\frac{\pi^2}{16}}

, equals

View answer and explanation

Correct option: A

\frac{8}{\pi} f(2)

Option A (

\frac{{8}}{{\pi}} f(2)

) is correct because applying L'Hopital's rule and the Fundamental Theorem of Calculus yields this result.

560 Easy

560 of 791

Analyzing +4 | -1

Let

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {{\text{In}}\left( {\left\{ x \right\}} \right)}&;&{x \notin I} \\ 0&;&{x \in I} \end{array}} \right.

; (where {x} is fractional part function), then the area enclosed by curves

y = f\left( x \right),y = f\left( {\left| x \right|} \right),x \in \left[ { - 5,0} \right]

and

y = 0

, is

View answer and explanation

Correct option: A

5\,{\text{In}}\left( {\frac{e}{2}} \right)

Option D (3) is correct because the equation f(x) - g(x) = 0 has 3 real solutions based on the intersection of the graphs.

561 Easy

561 of 791

Application +4 | -1

\int \nolimits_{ - 2}^2 \,\,\left( {x\,\, - \,\,\left| {\,x\,} \right|\,} \right)\,\,dx

is equal to

View answer and explanation

Correct option: D

- \,4

Option D (

\ - \,4

) is correct because the integral evaluates to -4 after considering the definition of the absolute value function.

562 Easy

562 of 791

Application +4 | -1

F(x)=\int\limits_{3}^x\left(2+\frac{d}{\mathrm{~d} t} \cos t\right) \mathrm{d} t

then

F^{\prime}\left(\frac{\pi}{6}\right)

is equal to

View answer and explanation

Correct option: D

3 / 2

Option D (3/2) is correct because the derivative of the integral is found using the Fundamental Theorem of Calculus and evaluating at x = π/6.

563 Easy

563 of 791

Application +4 | -1

The value of

\int \nolimits_1^4 \,\,\left| {\,x - 3\,} \right|dx

is equal to

View answer and explanation

Correct option: B

\frac{5}{2}

Option B is correct because the absolute value function requires splitting the integral at x=3, and then integrating each part separately, resulting in 5/2.

564 Easy

564 of 791

Application +4 | -1

\int\limits_{0}^{\pi / 4} \log \left(1+\tan ^2 \theta+2 \tan \theta\right) d \theta=

View answer and explanation

Correct option: C

(\pi \log 2) / 4

Option C is correct because the integral is solved using properties of logarithms and definite integrals, specifically using substitution and the property of integrals with limits 0 to a.

565 Easy

565 of 791

Application +4 | -1

The value of

\int \limits_0^1 \,\frac{{{x^3}}}{{1 + {x^8}}}\,\,dx

, is

View answer and explanation

Correct option: C

\frac{\pi }{16}

Option C (

\frac{\pi }{16}

) is correct because it is the result of substituting

x^4 = t

and evaluating the resulting integral.

566 Easy

566 of 791

Application +4 | -1

The value of

I=\int\limits_{{-\pi / 2}}^{\pi / 2} \sqrt{\cos x-\cos ^3 x} \mathrm{~d} x

View answer and explanation

Correct option: C

4 / 3

Option C (4/3) is correct because the integral simplifies to

2 \int_0^{\pi/2} \sqrt{{\cos x}} \sin x dx

, which can be easily evaluated using substitution.

567 Easy

567 of 791

Application +4 | -1

Let

g(x)=\left\{\begin{array}{cl}1, & 0 \leq x < 1 \\ x^3, & 1 \leq x < 4 \\ \sqrt{x}, & 4 \leq x < 9\end{array}\right.

then

\int\limits_{0}^9 g(x) \mathrm{d} x

View answer and explanation

Correct option: C

929 / 12

Option C is correct because the integral can be solved by splitting the integral based on the piecewise function definition.

568 Easy

568 of 791

Analyzing +4 | -1

I_n=\int\limits_{0}^{\pi / 4} \tan ^n x \, d x, n \in N

, then

I_{n+2}+I_n

equals

View answer and explanation

Correct option: C

\frac{1}{n+1}

Option C (

\frac{1}{n+1}

) is correct because the sum of the integrals can be simplified using trigonometric identities and then evaluated.

569 Easy

569 of 791

Application +4 | -1

\int \limits_0^{\pi /2} \,\,\frac{{{{\sin }^2}\,\,x}}{{\sin \,\,x\,\, + \,\,\cos \,\,x}}\,\,dx

is equal to

View answer and explanation

Correct option: C

\frac{1}{{\sqrt 2 }}\,\,\log \,\,\left( {\sqrt 2 + 1} \right)

Option C (

\frac{1}{{\sqrt 2 }}\,\,\log \,\,\left( {\sqrt 2 + 1} \right)

) is correct because using the property of definite integrals and simplifying leads to the given answer.

570 Easy

570 of 791

Application +4 | -1

For

x > 0

, let

f(x)=\int\limits_{1}^x \frac{\log t}{1+\mathrm{t}} \mathrm{dt}

. Then

f(x)+f\left(\frac{1}{x}\right)

is equal to

View answer and explanation

Correct option: C

\frac{1}{2}(\log x)^2

Option C (

\frac{1}{2}(\log x)^2

) is correct because it is the result obtained after simplifying the expression

f(x) + f(\frac{1}{x})

using a suitable substitution and evaluating the resulting integral.

Revision notes and key takeaways

Review DEFINITE INTEGRATION in your textbook or class notes, then use this set to check your understanding.

Showing 541-570 of 791 questions

Your Progress

Definite integration practice · Easy

Explore Other Chapters in MATHEMATICS

Continue Your Practice

Your practice