Let $I = \mathop \smallint \nolimits_0^1 \,\,\frac{{\sin \,\,x}}{{\sqrt x }}\,\,dx\,\,{\rm{and}}\,\,J = \mathop \smallint \nolimits_0^1 \,\,\frac{{\cos \,\,x}}{{\sqrt x }}\,\,dx$. Then, which one of the following is true?

$I < \frac{2}{3}\,\,{\rm{and}}\,\,J < 2$

$I\,\, > \,\,\frac{2}{3}\,\,{\rm{and}}\,\,J < \,\,2$

$I > \frac{2}{3}\,\,{\rm{and}}\,\,J > 2$

$I\,\,\left\langle {\,\frac{2}{3}\,\,{\rm{and}}\,\,J\,} \right\rangle \,\,2$

The integral $\mathop \smallint \limits_0^{\pi /2} \,\,f\,\,\left( {\sin \,\,2x} \right)\,\,\sin \,\,x\,\,dx$ is equal to

$\mathop \smallint \limits_0^{\pi /2} \,\,f\left( {\sin \,\,2x} \right)\,\,\cos \,\,x\,\,dx = \sqrt 2 \,\,\mathop \smallint \limits_0^{\pi /4} \,\,f\left( {\cos \,\,\,2x} \right)\cos \,\,x\,\,dx$

$\mathop \smallint \limits_0^{\pi /2} \,\,f\,\,\left( {\cos \,\,2x} \right)\,\,\sin \,\,x\,\,dx = \sqrt 2 \,\mathop \smallint \limits_0^{\pi /2} \,\,f\,\,\left( {\cos \,\,\,2x} \right)\sin \,\,x\,\,dx$

$\mathop \smallint \limits_0^{\pi /2} \,\,f\left( {\cos \,\,2x} \right)\cos \,\,x\,\,dx = \sqrt 2 \,\,\mathop \smallint \limits_0^{\pi /2} \,\,f\left( {\cos \,\,2x} \right)\,\,\cos \,\,x\,\,dx$

$\mathop \smallint \limits_0^{0\pi /2} \,\,f\left( {\sin \,\,2x} \right)\,\,\cos \,\,x\,\,dx = \sqrt 2 \,\,\mathop \smallint \limits_0^{\pi /2} \,\,f\left( {\cos \,\,\,2x} \right)\,\,\cos \,\,x\,\,dx$

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JEE • MATHEMATICS • DEFINITE INTEGRATION

Definite integration practice · Mixed difficulty

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661 Easy

661 of 995

Analyzing +4 | -1

The value of the integral

\int \nolimits_{1/n}^{\left( {an - 1} \right)/n} \,\,\frac{{\sqrt x }}{{\sqrt {a - x} \,\, + \,\,\sqrt x }}\;\,\,dx

〗 is

View answer and explanation

Correct option: C

\frac{{na\,\, - \,\,2}}{{2n}}

Option C (

\frac{{na\, - \,2}}{{2n}}

) is correct because after applying the property of definite integrals and adding the two forms of the integral, the result simplifies to the given expression.

662 Easy

662 of 995

Application +4 | -1

The area of the region bounced by

{y^2} = x\;\mathrm{and\;}y = \left| x \right|

View answer and explanation

Correct option: B

\frac{1}{6}\;\mathrm{sq\;unit}

Option B (

\frac{1}{6}\,\mathrm{sq unit}

) is correct because it represents the definite integral of the difference between

\sqrt{x}

and

x

from 0 to 1.

663 Medium

663 of 995

Application +4 | -1

f\left(\frac{3 x-4}{3 x+4}\right)=x+2, x \neq-\frac{4}{3}

, and

\int f(x) d x=\mathrm{A} \log |1-x|+\mathrm{B} x+\mathrm{C}

, then the ordered pair

(A, B)

is equal to :
(where

C

is a constant of integration)

View answer and explanation

Correct option: B

\left(-\frac{8}{3}, \frac{2}{3}\right)

Option B

\left(-\frac{8}{3}, \frac{2}{3}\right)

is correct because after finding f(x) and integrating, the coefficients A and B match these values.

664 Easy

664 of 995

Understanding +4 | -1

\int \nolimits_0^{2\pi } {\sin ^6}\,\,x\,{\cos ^5}\,\,x\,dx

is equal to

View answer and explanation

Correct option: C

\mathrm{0}

Option C (0) is correct because the integral of an odd function over a symmetric interval is zero.

665 Easy

665 of 995

Application +4 | -1

The integral

\int\limits_{1}^2 e^x x^x\left(2+\log _e x\right) d x

equals

View answer and explanation

Correct option: D

e(4 e-1)

Option D is correct. The integral is solved by recognizing the derivative of (ex)^x and evaluating the definite integral.

666 Easy

666 of 995

Understanding +4 | -1

The value of

\int\limits_{{-\pi / 2}}^{\pi / 2} \log \left(\frac{2-\sin \theta}{2+\sin \theta}\right) \mathrm{d} \theta

View answer and explanation

Correct option: A

Option A (0) is correct because the integrand is an odd function, and the integral of an odd function over a symmetric interval is zero.

667 Medium

667 of 995

Application +4 | -1

Let

f

be a continuous function satisfying

\int\limits_{0}^{t^2}\left(f(x)+x^2\right) d x=\frac{4}{3} t^3, \forall t>0

. Then

f\left(\frac{\pi^2}{4}\right)

is equal to

View answer and explanation

Correct option: C

\pi\left(1-\frac{\pi^3}{16}\right)

Option C (

\pi(1-\frac{\pi^3}{16})

) is correct because it is the value of the function

f(x)

evaluated at

x = \frac{\pi^2}{4}

after solving the given integral equation.

668 Medium

668 of 995

Application +4 | -1

\int\left\{\frac{(\log x-1)}{1+(\log x)^2}\right\}^2 d x

is equal to

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because the integral can be solved by recognizing the form of the integrand after substitution.

669 Easy

669 of 995

Analyzing +4 | -1

I(m, n)=\int\limits_{0}^1 t^m(1-t)^n d t

, then the expression for

I(m, n)

in terms of

I(m+1, n-1)

, is

View answer and explanation

Correct option: B

\frac{n}{m+1} I(m+1, n-1)

The correct answer is B (

\frac{n}{m+1} I(m+1, n-1)

) because integration by parts allows expressing

I(m, n)

in terms of

I(m+1, n-1)

670 Hard

670 of 995

Application +4 | -1

\int_{\pi / 2}^x \sqrt{\left(3-2 \sin ^2 t\right)} d t+\int_0^y \cos t d t=0

, then

\left(\frac{d y}{d x}\right)

x=\pi

and

y=\pi

View answer and explanation

Correct option: A

\sqrt{3}

Option A is correct because differentiating both sides of the given equation and then substituting the given values of x and y leads to the correct derivative value.

671 Easy

671 of 995

Application +4 | -1

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

is equal to

View answer and explanation

Correct option: C

\mathrm{0}

Option C,

\mathrm{0}

, is correct because using the property of definite integrals and adding the original integral to the transformed integral results in zero.

672 Easy

672 of 995

Understanding +4 | -1

The value of

\int\limits_{0}^x d[t],

View answer and explanation

Correct option: D

[x]

Option D is correct because the integral of the differential of the greatest integer function [t] from 0 to x is simply [x].

673 Easy

673 of 995

Analyzing +4 | -1

Let

I = \int \nolimits_0^1 \,\,\frac{{\sin \,\,x}}{{\sqrt x }}\,\,dx\,\,\mathrm{and}\,\,J = \int \nolimits_0^1 \,\,\frac{{\cos \,\,x}}{{\sqrt x }}\,\,dx

. Then, which one of the following is true?

View answer and explanation

Correct option: C

I < \frac{2}{3}\,\,\mathrm{and}\,\,J < 2

Option C is correct because, for x in (0,1), sin(x) < x and cos(x) < 1, allowing us to bound the integrals I and J, resulting in I < 2/3 and J < 2.

674 Easy

674 of 995

Application +4 | -1

\int_a^b\{f(x)-3 x\} d x=a^2-b^2

, then the value of

f\left(\frac{\pi}{6}\right)

, is

View answer and explanation

Correct option: A

\frac{\pi}{6}

The correct answer is A (

\frac{\pi}{6}

) because after evaluating the integral and differentiating, we find that

f(x) = x

, thus

f(\frac{\pi}{6}) = \frac{\pi}{6}

675 Medium

675 of 995

Application +4 | -1

h\left( a \right) = h\left( b \right)

, the value of the integral

\int \limits_a^b \,\,{\left[ {f\,\,\left( {g\;\left( {h\,\,\left( x \right)\,} \right)\,} \right)} \right]^{ - 1}}\,\,f'\,\,\left( {\,g\,\,\left( {h\,\,\left( x \right)\,} \right)\,} \right)\,\,g'\,\,\left( {h\,\left( x \right)\,} \right)\,\,h'\,\,\left( x \right)\,\,dx

is equal to

View answer and explanation

Correct option: A

\mathrm{0}

Option A (

\mathrm{0}

) is correct because the limits of integration become equal, making the integral zero.

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

676 Easy

676 of 995

Application +4 | -1

\lim \limits_\limits {n \rightarrow \infty} \frac{1^p+2^p+3^p+\ldots .+n^p}{n^{p+1}}

View answer and explanation

Correct option: A

\frac{1}{p+1}

Option A (

\frac{1}{p+1}

) is correct because the limit of the sum is evaluated as a definite integral of

x^p

from 0 to 1.

677 Easy

677 of 995

Application +4 | -1

\int\limits_{0}^{\pi / 2}|\sin x-\cos x| d x=

View answer and explanation

Correct option: B

2(\sqrt{2}-1)

Option B (

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

) is correct because it is the result of evaluating the definite integral of the absolute difference between sine and cosine functions over the given interval.

678 Easy

678 of 995

Analyzing +4 | -1

Let

I=\int_{\pi / 4}^{\pi / 3}\left(\frac{8 \sin x-\sin 2 x}{x}\right) d x

. Then

View answer and explanation

Correct option: C

I > \sqrt 3 - \frac{\pi }{6}

The correct answer is C (

I > \sqrt 3 - \frac{\pi }{6}

) because the integral I is bounded below by the given expression, which is derived by using the minimum value of

\frac{{\sin x}}{x}

in the interval.

679 Easy

679 of 995

Application +4 | -1

\lim\limits_{n \to \infty } \,\,\frac{{{1^{99}} + \,\,{2^{99}}\,\, + \,\, \ldots + \,\,{n^{99}}}}{{{n^{100}}}} =

View answer and explanation

Correct option: B

\frac{{1}}{{100}}

Option B (

\frac{{1}}{{100}}

) is correct because the limit of the sum can be expressed as a definite integral, which evaluates to

\frac{1}{100}

680 Easy

680 of 995

Application +4 | -1

The integral

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

, is equal to

View answer and explanation

Correct option: D

Option D (2) is correct because by using the property

\int_a^b f(x) dx = \int_a^b f(a+b-x) dx

and trigonometric identities, the integral simplifies to 2.

681 Medium

681 of 995

Application +4 | -1

The value of definite integral

\int\limits_{ - \pi }^\pi {\frac{{2x\left( {1 + \sin x} \right)}}{{1 + {{\cos }^2}x}}} \,dx

is:

View answer and explanation

Correct option: C

{\pi ^2}

Option D (

{\pi ^2}

) is correct because it is the value of the definite integral after applying properties of definite integrals and trigonometric substitutions.

682 Easy

682 of 995

Application +4 | -1

Let

f(x)

be a function satisfying

f^{\prime}(x)=f(x)

with

f(0)=1

and

g(x)

be a function which satisfies

f(x)+g(x)=x^2

. Then

\int\limits_{0}^1 f(x) g(x) d x=

View answer and explanation

Correct option: D

e-\frac{e^2}{2}-\frac{3}{2}

Option D (

e-\frac{e^2}{2}-\frac{3}{2}

) is correct because it is the result of integrating the product of f(x) and g(x) from 0 to 1, where f(x) = e^x and g(x) = x² - e^x.

683 Easy

683 of 995

Application +4 | -1

\int \nolimits_{\left[ x \right]}^{\left[ x \right] + 1} \,\,f\left( t \right)\,\,dt = \left[ x \right]

,〗 then the value of

\int \nolimits_{ - 2}^4 \,\,f\left( x \right)\,\,dx

is equal to

View answer and explanation

Correct option: D

\mathrm{3}

Option D (3) is correct because the integral from [x] to [x]+1 of f(t) dt equals [x], allowing us to sum the integrals from -2 to 4 to get 3.

684 Easy

684 of 995

Application +4 | -1

\int \frac{1}{\left(e^x+e^{-x}\right)^2} d x=

View answer and explanation

Correct option: A

-\frac{1}{2\left(e^{2 x}+1\right)}+c

Option A (-1/(2(e^(2x)+1))+c) is correct because it is the result of integrating the given function using substitution.

685 Medium

685 of 995

Application +4 | -1

\phi '\,\,\left( x \right) = \frac{{{{\log }_e}\,\,\left| {\,\sin \,\,x\,} \right|}}{x},\;\,x \ne n\;\pi ,\,\;n \in Z

and

\int \limits_1^3 \,\,\frac{{3\,\,{{\log }_e}\,\left| {\,\sin \,\,{x^3}\,} \right|}}{x}\,\,dx = \phi \,\,\left( 1 \right)

, then the possible value of

k

View answer and explanation

Correct option: A

\mathrm{27}

Option A (27) is correct because the integral is evaluated using substitution and the properties of definite integrals, resulting in k = 27.

686 Medium

686 of 995

Analyzing +4 | -1

For any

n \in N

, the value of the integral

\int_0^\pi \frac{\sin 2 n x}{\sin x} d x

, is

View answer and explanation

Correct option: D

none of these

Option D (none of these) is correct because the value of the integral is 0 for all n in N, which is not among the provided options.

687 Easy

687 of 995

Application +4 | -1

Let

f

be an odd function then

\int\limits_{{-1}}^1(|x|+f(x) \cos x) \mathrm{d} x

is equal to

View answer and explanation

Correct option: B

Option B is correct because the integral simplifies to integrating the absolute value function, as the other term is an odd function integrated over a symmetric interval.

688 Medium

688 of 995

Analyzing +4 | -1

\left| {\mathop {\,\int }\limits_{10}^{19} \,\,\frac{{\sin \,\,x}}{{1 + {x^8}}}\,\,dx\,} \right|

is less than

View answer and explanation

Correct option: C

{10^{ - 7}}

Option C is correct because the absolute value of the integral is less than

{10^{ - 7}}

after bounding the integrand.

689 Easy

689 of 995

Analyzing +4 | -1

The integral

\int \limits_0^{\pi /2} \,\,f\,\,\left( {\sin \,\,2x} \right)\,\,\sin \,\,x\,\,dx

is equal to

View answer and explanation

Correct option: B

\int \limits_0^{\pi /2} \,\,f\left( {\sin \,\,2x} \right)\,\,\cos \,\,x\,\,dx = \sqrt 2 \,\,\int \limits_0^{\pi /4} \,\,f\left( {\cos \,\,\,2x} \right)\cos \,\,x\,\,dx

Option B is correct because after applying the property of definite integrals and substitution, the given integral is equal to

\sqrt 2 \,\,\int \limits_0^{\pi /4} \,\,f\left( {\cos \,\,\,2x} \right)\cos \,\,x\,\,dx

690 Easy

690 of 995

Application +4 | -1

The value of

\int\limits_{0}^{[x]}(-x-[x]) d x

, is

View answer and explanation

Correct option: C

(1 / 2)[x]

Option C (

(1 / 2)[x]

) is correct because the integral simplifies using the periodic property of the function

x-[x]

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