If ${\rm{g}}\left( x \right) = \mathop \smallint \nolimits_0^x \,\,{\cos ^4}\,\,t\,dt$, then ${\rm{g}}\left( {x\,\, + \,\,\pi } \right)$ equals

${\rm{g}}\left( x \right)\,\, + \,\,{\rm{g}}\left( \pi \right)$

${\rm{g}}\left( x \right)\,\, - \,\,{\rm{g}}\left( \pi \right)$

${\rm{g}}\left( x \right)\,\,{\rm{g}}\left( \pi \right)$

$\frac{{{\rm{g}}\left( x \right)}}{{{\rm{g}}\left( \pi \right)}}$

The value of $\mathop \smallint \limits_0^{\pi /2} \,\,\frac{{\sin \,\,8x\,\,\log \,\,\cot \,\,x}}{{\cos \,\,2x}}\,\,dx$, is

$\frac{{{\rm{5}}\,{\rm{\pi }}}}{{\rm{2}}}$

$\frac{{{\rm{3}}\,{\rm{\pi }}}}{{\rm{2}}}$

If $I\left( {m,\;\,n} \right) = \mathop \smallint \limits_0^1 \,{x^{m - 1}}{\left( {1 - x} \right)^{n - 1}}\,dx$, then

$I\left( {m,\;\,n} \right) = \mathop \smallint \limits_0^\infty \,\,\frac{{{x^{m - 1}}}}{{{{\left( {1 + x} \right)}^{m + n}}}}\,\,dx = \mathop \smallint \limits_0^\infty \,\,\frac{{{x^{n - 1}}}}{{{{\left( {1 + x} \right)}^{m + n}}}}\,\,dx$

$I\left( {m,\,\;n} \right) = \mathop \smallint \limits_0^\infty \,\,\frac{{{x^m}}}{{{{\left( {1 + x} \right)}^{m + n}}}}\,\,dx = \mathop \smallint \limits_0^\infty \,\,\frac{{{x^n}}}{{{{\left( {1 + x} \right)}^{m + n}}}}\,\,dx$

$I\left( {m,\;n} \right) = \mathop \smallint \limits_0^\infty \,\,\frac{{{x^n}}}{{{{\left( {1 + x} \right)}^{m + n - 1}}}}\,\,dx = \mathop \smallint \limits_0^\infty \,\,\frac{{{x^m}}}{{{{\left( {1 + x} \right)}^{m + n - 1}}}}\,\,dx$

If $F\left( x \right) = \mathop \smallint \nolimits_{{x^2}}^{{x^3}} \,\,\log \,\,t\,dt,\;\,\left( {x > 0} \right)$, then $F'\,\left( x \right)$ is equal to

$\left( {4x\,\, - \,\,9{x^2}} \right)\log \,\,x$

$\frac{d}{{dx}}\,\,\mathop \smallint \nolimits_{f\left( x \right)}^{{\rm{g}}\left( x \right)} \,\,{\rm{h}}\,\,\left( t \right)\,\,dt$ is equal to

$g'\left( x \right)\,\,h\,\,\left[ {{\rm{g}}\,\,\left( x \right)} \right]\,\, - \,\,f'\left( x \right)\,\,h\,\,\left[ {f\left( x \right)} \right]$

$g'\left( x \right)\,\,h\,\,\left[ {{\rm{g}}\,\,\left( x \right)} \right]$

$h\,\,\left[ {\,{\rm{g}}\,\,\left( x \right)} \right]\,\, - \,\,h\left[ {\,f\,\,\left( x \right)} \right]$

If x satisfies the equation ${x^2}\,\left( {\mathop \smallint \limits_0^1 \,\,\frac{{dt}}{{{t^2} + \,\,2t\,\,\cos \,\,\alpha \,\, + \,\,1}}} \right)\,\, - \,\,x\left( {\mathop \smallint \limits_{ - 3}^3 \frac{{{t^2}\,\,\sin \,\,2t}}{{{t^2} + \,\,1}}\,\,dt} \right) - 2 = 0$ $(0 < \,\,\alpha \,\, < \pi )$, then the value of $x$ is

$ \pm \,\,\sqrt {\frac{{\sin \,\,\alpha }}{\alpha }} $

$ \pm 4\,\,\sqrt {\frac{{\sin \,\,\alpha }}{\alpha }} $

If $\mathop \smallint \limits_{\pi /2}^x \,\,\sqrt {3\, - \,2\,\,{{\sin }^2}\,\,u} \,du\,\, + \,\,\mathop \smallint \nolimits_0^y \,\,\cos \,\,t\,dt = 0$ then, $\frac{{dy}}{{dx}}$ is equal to

$\frac{{\sqrt {4 - 3\,\,{{\sin }^2}\,\,x} }}{{\cos \,\,y}}$

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JEE • MATHEMATICS • DEFINITE INTEGRATION

Definite integration practice · Mixed difficulty

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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.

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Showing 151-180 of 995 questions

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151 Easy

151 of 995

Application +4 | -1

\int \nolimits_{ - \pi /2}^{\pi /2} \,\,\frac{{dx}}{{1 + \cos \,\,x}}\,\,\mathrm{is}

is equal to

View answer and explanation

Correct option: C

\mathrm{2}

Option C (2) is correct because the integral simplifies to

\int \nolimits_0^{\pi /2} {\sec ^2}\,\frac{x}{2}\,dx

which evaluates to 2.

152 Easy

152 of 995

Application +4 | -1

The value of

I = \int \nolimits_0^1 \,\,x\,\,\left| {\,x\,\, - \,\,\frac{1}{2}\,} \right|\,\,dx

View answer and explanation

Correct option: C

\mathrm{1}\,\mathrm{/}\,\mathrm{8}

Option C (

\mathrm{1}\,\mathrm{/}\,\mathrm{8}

) is correct because the definite integral is evaluated by splitting the integral at x=1/2 due to the absolute value function.

153 Medium

153 of 995

Application +4 | -1

\int \nolimits_0^{\pi /4} \,\,(\cos \,\,x\,\, - \,\,\sin \,\,x)\,\,dx\,\, + \,\,\int \nolimits_{\pi /4}^{5\pi /4} \,\,(\sin \,\,x\,\, - \,\,\cos \,\,x)\,\,dx\,\, + \,\,\int \nolimits_{2\pi }^{\pi /4} \,\,(\cos \,\,x\,\, - \,\,\sin \,\,x)\,\,dx

is equal to

View answer and explanation

Correct option: D

4\,\,\sqrt 2 - 2

Option D (4√2 - 2) is correct because the integral is evaluated by computing the definite integrals and summing the results, leading to the correct value.

154 Medium

154 of 995

Application +4 | -1

\mathrm{g}\left( x \right) = \int \nolimits_0^x \,\,{\cos ^4}\,\,t\,dt

, then

\mathrm{g}\left( {x\,\, + \,\,\pi } \right)

equals

View answer and explanation

Correct option: A

\mathrm{g}\left( x \right)\,\, + \,\,\mathrm{g}\left( \pi \right)

Option A (g(x) + g(π)) is correct because substituting t = π + y allows us to separate the integral into g(x) + g(π).

155 Easy

155 of 995

Application +4 | -1

f(x)

and

g(x)

are two continuous functions defined on

[-a, a]

, then the value of

\int_{ - a}^a {\{ f(} x) + f( - x)\} \{ g(x) - g( - x)\} \,\,dx\mathrm{, }\,\,\mathrm{is }

View answer and explanation

Correct option: D

none of these

Option D (none of these) is correct because the integrand is an odd function, making the integral over a symmetric interval equal to zero.

156 Easy

156 of 995

Analyzing +4 | -1

The value of

\int \limits_0^{\pi /2} \,\,\frac{{\sin \,\,8x\,\,\log \,\,\cot \,\,x}}{{\cos \,\,2x}}\,\,dx

, is

View answer and explanation

Correct option: A

0

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.

157 Medium

157 of 995

Application +4 | -1

The value of

\mathop {\int \,\,}\limits_0^3 x\,\,\sqrt {1 + x} \,\,dx

, is

View answer and explanation

Correct option: D

None of these

The correct answer is D (None of these) because the value of the integral is

\frac{116}{15}

, which is not among the options.

158 Easy

158 of 995

Application +4 | -1

The ratio of the area bounded by the curves

y=\cos x

and

y=\cos 2 x

between

x=0, x=\pi / 3

and

x

-axis, is

View answer and explanation

Correct option: D

2: 1

Option D (2:1) is correct because the ratio of the areas is calculated by integrating the given functions over the specified interval and then dividing the results.

159 Medium

159 of 995

Application +4 | -1

\{x\}

denotes the fractional part of

x

, then

\int\limits_{0}^2\{x\} d x

is equal to

View answer and explanation

Correct option: A

Option A (1) is correct because the integral of the fractional part of x from 0 to 2 is equal to 1.

160 Medium

160 of 995

Application +4 | -1

The value of the integral

\int \limits_0^1 \,\,{\cot ^{ - 1}}\,\,\left( {1 - \,\,x + \,\,{x^2}} \right)\,\,dx

, is

View answer and explanation

Correct option: B

\frac{\pi }{2}\,\, - \,\,\log \,\,2

Option B (

\frac{\pi }{2}\\ - \\,\log \\,2

) is correct because the integral is solved by using the property

\cot^{-1}(x) = \tan^{-1}(1/x)

and simplifying the integrand using the tangent addition formula.

161 Medium

161 of 995

Application +4 | -1

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

View answer and explanation

Correct option: C

\sqrt 2 - 1

Option C (√2 - 1) is correct because the integral of the greatest integer function [x²] from 0 to √2 can be split into two integrals, one from 0 to 1 where [x²] = 0 and another from 1 to √2 where [x²] = 1.

162 Medium

162 of 995

Analyzing +4 | -1

Let

\Delta \,\,\left( y \right) = \left| {\,\begin{array}{cccccccccccccccccccc} {y\,\, + \,\,a}&{y\,\, + \,\,b}&{y\,\, + \,\,a\,\, - \,\,c}\\ {y\,\, + \,\,b}&{y\,\, + \,\,c}&{y\,\, - \,\,1}\\ {y\,\, + \,\,c}&{y\,\, + \,\,d}&{y\,\, - \,\,b\,\, + \,\,d} \end{array}\,} \right|

and,

\int \limits_0^2 \,\Delta \,\,\left( y \right)\,\,dy = - 16

, where

a,\;\,b,\,\,c,\,\,d

are in A.P., then the common difference of the A.P. is equal to

View answer and explanation

Correct option: B

\pm \,2

Option B (

\pm \,2

) is correct because the common difference of the A.P. is found by evaluating the integral and solving the resulting equation.

163 Easy

163 of 995

Application +4 | -1

I\left( {m,\;\,n} \right) = \int \limits_0^1 \,{x^{m - 1}}{\left( {1 - x} \right)^{n - 1}}\,dx

, then

View answer and explanation

Correct option: A

I\left( {m,\;\,n} \right) = \int \limits_0^\infty \,\,\frac{{{x^{m - 1}}}}{{{{\left( {1 + x} \right)}^{m + n}}}}\,\,dx = \int \limits_0^\infty \,\,\frac{{{x^{n - 1}}}}{{{{\left( {1 + x} \right)}^{m + n}}}}\,\,dx

Option A is correct because it represents the Beta function in terms of an integral from 0 to infinity, derived through substitution.

164 Easy

164 of 995

Application +4 | -1

F\left( x \right) = \int \nolimits_{{x^2}}^{{x^3}} \,\,\log \,\,t\,dt,\;\,\left( {x > 0} \right)

, then

F'\,\left( x \right)

is equal to

View answer and explanation

Correct option: A

\left( {9{x^2} - \,\,4x} \right)\,\,\log \,\,x

Option A (

\left( {9{x^2} - \,4x} \right)\,\,\log \,x

) is correct because it is the result of applying Leibniz's rule for differentiation under the integral sign.

165 Easy

165 of 995

Application +4 | -1

\frac{d}{{dx}}\,\,\int \nolimits_{f\left( x \right)}^{\mathrm{g}\left( x \right)} \,\,\mathrm{h}\,\,\left( t \right)\,\,dt

is equal to

View answer and explanation

Correct option: C

g'\left( x \right)\,\,h\,\,\left[ {\mathrm{g}\,\,\left( x \right)} \right]\,\, - \,\,f'\left( x \right)\,\,h\,\,\left[ {f\left( x \right)} \right]

Option C (

g'\left( x \right)\\,,h\\,\left[ {\mathrm{g}\\,,\left( x \right)} \right]\\, - \\,f'\left( x \right)\\,,h\\,\left[ {f\left( x \right)} \right]

) is correct because it represents the Leibniz rule for differentiating an integral with variable limits.

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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

166 Medium

166 of 995

Analyzing +4 | -1

If x satisfies the equation

{x^2}\,\left( {\int \limits_0^1 \,\,\frac{{dt}}{{{t^2} + \,\,2t\,\,\cos \,\,\alpha \,\, + \,\,1}}} \right)\,\, - \,\,x\left( {\int \limits_{ - 3}^3 \frac{{{t^2}\,\,\sin \,\,2t}}{{{t^2} + \,\,1}}\,\,dt} \right) - 2 = 0

(0 < \,\,\alpha \,\, < \pi )

, then the value of

x

View answer and explanation

Correct option: A

\pm 2\,\,\sqrt {\frac{{\sin \,\,\alpha }}{\alpha }}

Option A (

\pm 2\\,\sqrt {\frac{{\sin \\,\alpha }}{\alpha }}

) is correct because after simplifying the equation using properties of odd functions and evaluating the integral, the value of x can be found.

167 Easy

167 of 995

Application +4 | -1

f(x)=\left\{\begin{array}{lll}4 x+3, & \text { if } & 1 \leq x \leq 2 \\ 3 x+5, & \text { if } & 2 < x \leq 4\end{array}\right.

then

\int\limits_{1}^4 f(x) d x=

View answer and explanation

Correct option: D

Option D (37) is correct because it is the result of evaluating the piecewise defined function over the given interval by splitting the integral at the point where the function definition changes.

168 Medium

168 of 995

Application +4 | -1

The integral

\int \frac{x d x}{2-x^2+\sqrt{2-x^2}}

equals :

View answer and explanation

Correct option: B

-\log \left|1+\sqrt{2-x^2}\right|+c

Option B (

-\log \left|1+\sqrt{{2-x^2}}\right|+c

) is correct because the integral can be solved by substituting

t = \sqrt{{2-x^2}}

169 Easy

169 of 995

Application +4 | -1

The value of

\int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{{{\sin }^2}x}}{{1 + {2^x}}}dx}

View answer and explanation

Correct option: D

\frac{\pi }{4}

Option D (

\frac{\pi }{4}

) is correct because the integral is solved by using the property of definite integrals and combining the integral with a modified form to simplify and evaluate.

170 Easy

170 of 995

Application +4 | -1

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

is equal to

View answer and explanation

Correct option: A

Option A is correct. The integral is evaluated by using trigonometric identities to simplify the integrand and then integrating.

171 Hard

171 of 995

Application +4 | -1

The value of

\int\limits_{{-20 \pi}}^{20 \pi}[\sin x+\cos x] d x

is: (where [.] denotes greatest integer function)

View answer and explanation

Correct option: B

-20 \pi

Option B (

-20 \pi

) is correct because the integral of the greatest integer function of

\sin x + \cos x

over the given interval evaluates to

-20\pi

172 Easy

172 of 995

Application +4 | -1

The value of

\int\limits_{0}^{\pi / 2} \frac{\mathrm{d} x}{1+\tan ^3 x}

View answer and explanation

Correct option: C

\pi / 4

Option C (π/4) is correct because the integral is evaluated using the property ∫₀ᵃ f(x) dx = ∫₀ᵃ f(a-x) dx and adding the two forms of the integral.

173 Medium

173 of 995

Application +4 | -1

The total area between

y=4 x-x^2

and

y=x

from

x=0

and

x=4

View answer and explanation

Correct option: B

37 / 6

Option B (37/6) is correct because it is the total area calculated by splitting the integral into two parts and summing them up.

174 Medium

174 of 995

Application +4 | -1

\lim \limits_{{n \rightarrow \infty}} \frac{1}{n} \sum_{r=1}^{2 n} \frac{r}{\sqrt{n^2+r^2}}

equals

View answer and explanation

Correct option: B

-1+\sqrt{5}

Option B (-1+√5) is correct because the limit of the Riemann sum is converted to a definite integral and evaluated correctly.

175 Medium

175 of 995

Analyzing +4 | -1

{u_n} = \int \nolimits_0^{\pi /4} {\tan ^n}\,\,x\,dx,

, then

{u_n}\,\, + \,\,{u_{n - 2}}

is equal to

View answer and explanation

Correct option: A

\frac{1}{{n\,\, - \,\,1}}

Option A (

\frac{1}{{n\, - \,1}}

) is correct because by using integration by parts and simplifying, the expression

{u_n}\, + \,{u_{n - 2}}

simplifies to

\frac{1}{{n\, - \,1}}

176 Hard

176 of 995

Application +4 | -1

\int \limits_{\pi /2}^x \,\,\sqrt {3\, - \,2\,\,{{\sin }^2}\,\,u} \,du\,\, + \,\,\int \nolimits_0^y \,\,\cos \,\,t\,dt = 0

then,

\frac{{dy}}{{dx}}

is equal to

View answer and explanation

Correct option: B

- \,\,\frac{{\sqrt {3\,\, - \,\,2\,\,{{\sin }^2}\,\,x} }}{{\cos \,\,y}}

Option B is correct because differentiating the given equation with respect to x and solving for dy/dx yields the stated result.

177 Medium

177 of 995

Application +4 | -1

The value of the parameter '

a

' such that the area bounded by

y=a^2 x^2+a x+1

, coordinate axes and the line

x=1

attains its least value, is equal to

View answer and explanation

Correct option: C

-\frac{3}{4}

Option C (-

\frac{{3}}{{4}}

) is correct because it is the value of 'a' that minimizes the area under the given curve between the specified limits, found by completing the square and identifying the vertex.

178 Easy

178 of 995

Application +4 | -1

\int\limits_{0}^{\pi / 4} \frac{x \sin x}{\cos ^3 x} \mathrm{~d} x=\frac{\pi}{4}+A

, then

A

is equal to

View answer and explanation

Correct option: D

-\frac{1}{2}

Option D (

-\frac{1}{2}

) is correct because integration by parts and trigonometric identities are used to evaluate the definite integral, resulting in the given value.

179 Easy

179 of 995

Application +4 | -1

The integral

\int\limits_{{-1 / 2}}^{1 / 2}\left([x]+2 \log \frac{1+x}{1-x}\right) {d} x \,

equals

View answer and explanation

Correct option: A

-1 / 2

Option A (-1/2) is correct because the integral simplifies to the integral of [x] from -1/2 to 1/2, which is -1/2.

180 Easy

180 of 995

Application +4 | -1

\int\limits_{{\pi / 2}}^x \sqrt{3-2 \sin ^2 z} d z+\int\limits_{0}^y \cos t d t=0

then

\frac{d y}{d x}

(\pi / 2, \pi)

View answer and explanation

Correct option: A

Option B (4) is correct because by setting up ratios of the combinations and solving the resulting equations, we can find the value of r.

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