The value of integral $\mathop \smallint \nolimits_0^\pi \,\,x\;\,\,f(\sin \,\,x)\;\,\,dx\,\,\;{\rm{is}}$

$\pi \,\mathop \smallint \nolimits_0^{\pi /2} \,\,f\;(\sin \,\,x)\;\,dx$

$\frac{\pi }{4}\,\,\mathop \smallint \nolimits_0^\pi \,\,f\;(\sin \,\,x)\;\,dx$

$\mathop \smallint \nolimits_{ - \pi /2}^{\pi /2} \,\,{\sin ^4}\,\,x\,\,{\cos ^6}\,\,x\,\,dx$ is equal to

$\frac{{3\,{\rm{\pi }}}}{{256}}$

$\frac{{3\,{\rm{\pi }}}}{{64}}$

$\frac{{3\,{\rm{\pi }}}}{{572}}$

$\frac{{3\,{\rm{\pi }}}}{{128}}$

If ${I_1} = \mathop \smallint \limits_0^\infty \,\,\frac{1}{{1 + {x^4}}}\,\,dx$ and ${I_2} = \mathop \smallint \limits_0^\infty \,\,\frac{{{x^2}}}{{1\,\, + \,\,{x^4}}}\,\,dx$. Then $\frac{{{I_1}}}{{{I_2}}} = $

$\begin{array}{*{20}{l}} {{\rm{1}}\,{\rm{/}}\,{\rm{2}}} \end{array}$

If $f\left( x \right) = f\left( {a - x} \right),\;\,\,{\rm{then}}\,\,\mathop \smallint \nolimits_0^a \,\,f\left( x \right)\,\,dx$ is equal to

$\frac{a}{2}\,\,\mathop \smallint \nolimits_0^a \,\,f\left( x \right)\,\,dx$

$\mathop \smallint \nolimits_0^a \,\,f\left( x \right)\,\,dx$

$\frac{{{a^2}}}{2}\,\,\mathop \smallint \nolimits_0^a \,\,f\left( x \right)dx$

$ - \frac{a}{2}\,\,\mathop \smallint \nolimits_0^a \,\,f\left( x \right)\,\,dx$

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JEE • MATHEMATICS • DEFINITE INTEGRATION

Definite integration practice · Medium

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181 Medium

181 of 596

Application +4 | -1

\int\limits_{0}^3|2-x| d x

equals

View answer and explanation

Correct option: B

5 / 2

Option B (

5 / 2

) is correct because the integral of

|2-x|

from 0 to 3 is equal to

5/2

182 Medium

182 of 596

Application +4 | -1

The value of

\int\limits_{0}^1 \tan ^{-1}\left(\frac{2 x-1}{1+x-x^2}\right) d x

View answer and explanation

Correct option: B

Option B (0) is correct because the integral simplifies to zero by using properties of inverse trigonometric functions and definite integrals.

183 Medium

183 of 596

Application +4 | -1

The value of integral

\int \nolimits_0^\pi \,\,x\;\,\,f(\sin \,\,x)\;\,\,dx\,\,\;\mathrm{is}

View answer and explanation

Correct option: B

\pi \,\int \nolimits_0^{\pi /2} \,\,f\;(\sin \,\,x)\;\,dx

Option B (

\pi \,\int \nolimits_0^{\pi /2} \,f\;(\sin \,x)\,dx

) is correct because it uses the property of definite integrals to simplify the expression.

184 Medium

184 of 596

Application +4 | -1

If area bounded by the curves

{y^2} = 4\;ax\,\mathrm{and}\,y = mx

{a^2}/3,

then the value of

m

View answer and explanation

Correct option: A

The correct answer is A (2) because it is the value of m that makes the area enclosed by the curves equal to a²/3, which is found by integrating the difference between the curves.

185 Medium

185 of 596

Application +4 | -1

\int\limits_{{-x / 2}}^{\pi / 2} \frac{\cos x}{1+e^x} d x=

View answer and explanation

Correct option: A

Option A (1) is correct because the integral evaluates to 1 after using properties of definite integrals and even/odd functions.

186 Medium

186 of 596

Analyzing +4 | -1

Let

f:[0,1] \rightarrow R

be a continuous function, then which of the following is the maximum value of

\int\limits_{0}^1 x^2 f(x) d x-\int\limits_{0}^1 x(f(x))^2 d x

for all such functions

View answer and explanation

Correct option: B

\frac{1}{16}

Option D (

\sqrt{22}

) is correct because it is the radius of the circle calculated using geometric properties and the power of a point.

187 Medium

187 of 596

Application +4 | -1

The integral

\int \frac{d x}{x^2\left(x^4+1\right)^{1 / 4}}

equals

View answer and explanation

Correct option: B

-\left(\frac{x^4+1}{x^4}\right)^{\frac{1}{4}}+c

Option B (

-\left(\frac{x^4+1}{x^4}\right)^{\frac{1}{4}}+c

) is correct because after substitution and integration, the integral evaluates to this expression.

188 Medium

188 of 596

Application +4 | -1

f(x)=\int\limits_{{x^2}}^{x^4} \sin \sqrt{t} \,d t

, then

f^{\prime}(x)

equals

View answer and explanation

Correct option: B

4 x^3 \sin x^2-2 x \sin x

Option B (

4 x^3 \sin x^2-2 x \sin x

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

189 Medium

189 of 596

Application +4 | -1

Suppose

f

is such that

f\,\,\left( { - x} \right) = - \,f\,\left( x \right)

for every real

x

and

\int \nolimits_0^1 \,\,f\left( x \right)\,\,dx = 5

, then

\int \nolimits_{ - 1}^0 \,\,\,\,f\left( t \right)\,\,dt

is equal to

View answer and explanation

Correct option: D

- \,5

Option D (-5) is correct because f(x) is an odd function, and the integral from -1 to 0 is the negative of the integral from 0 to 1.

190 Medium

190 of 596

Application +4 | -1

The value of

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

View answer and explanation

Correct option: D

\frac{\pi}{4}

Option D (

\frac{\pi}{4}

) is correct because, by using the property of definite integrals and adding the original integral to the transformed one, the integral simplifies to

\frac{\pi}{4}

191 Medium

191 of 596

Application +4 | -1

Area between the curve

y=\cos x

and

x

-axis when

0 \leq x

View answer and explanation

Correct option: B

Option B (4) is correct because the area is calculated by summing the absolute values of the integrals of

\cos x

over the specified intervals to account for regions below the x-axis.

192 Medium

192 of 596

Application +4 | -1

The integral

\int\limits_{{7 \pi / 4}}^{7 \pi / 3} \sqrt{\tan ^2 x} d x

is equal to :

View answer and explanation

Correct option: D

\log \sqrt{2}

Option D (

\log \sqrt{2}

) is correct because after evaluating the integral of

\sqrt{\tan^2 x}

from

\frac{7\pi}{4}

\frac{7\pi}{3}

, we obtain

\log \sqrt{2}

193 Medium

193 of 596

Application +4 | -1

\int \nolimits_{ - \pi /2}^{\pi /2} \,\,{\sin ^4}\,\,x\,\,{\cos ^6}\,\,x\,\,dx

is equal to

View answer and explanation

Correct option: C

\frac{{3\,\mathrm{\pi }}}{{256}}

Option C (

\frac{{3\\,\mathrm{\pi }}}{{256}}

) is correct because the integral of the even function is solved using the Gamma function.

194 Medium

194 of 596

Application +4 | -1

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

View answer and explanation

Correct option: B

\frac{2}{\pi}

Option B (

\frac{2}{\pi}

) is correct because after substitution and using the properties of the sine function, the integral evaluates to

\frac{2}{\pi}

195 Medium

195 of 596

Application +4 | -1

For which of the following values of

{m}

, is the area of the region bounded by the curve

y=x-x^2

and the line

y=m x

equals

\frac{9}{2}

View answer and explanation

Correct option: B

-2, 4

Option B (-2, 4) is correct because these are the values of m for which the area between the curve y=x-x² and the line y=mx equals 9/2.

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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

196 Medium

196 of 596

Analyzing +4 | -1

f\left( x \right) = \,\,\left| {\,\begin{array}{cccccccccccccccccccc} {\sec \,\,x}&{\cos \,\,x}&{{{\sec }^2}\,\,x\,\, + \,\,\mathrm{cosec}\,\,x\,\,\cot \,\,x}\\ {{{\cos }^2}\,\,x}&{{{\cos }^2}\,\,x}&{\mathrm{cose}{\mathrm{c}^2}\,\,x}\\ 1&{{{\cos }^2}\,\,x}&{{{\cos }^2}\,\,x} \end{array}\,} \right|

, then

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

equals

View answer and explanation

Correct option: C

- \,\,\left( {\frac{\pi }{4}\,\, + \,\,\frac{8}{{15}}} \right)

Option C (

- \left( {\frac{\pi }{{4}}\\ + \frac{{8}}{{15}}} \right)

) is correct because after evaluating the determinant and integrating the resulting trigonometric functions, we obtain this value.

197 Medium

197 of 596

Analyzing +4 | -1

The least value of the function

F(x)=\int_0^x(3 \sin u+4 \cos u) \mathrm{d} u

on the interval

(5 \pi / 4,4 \pi / 3]

View answer and explanation

Correct option: D

\frac{9-4 \sqrt{3}}{2}

Option D (

\frac{9-4 \sqrt{3}}{2}

) is correct because the minimum value of the function occurs at

x = 4\pi/3

within the given interval, and the function value there is

\frac{9-4 \sqrt{3}}{2}

198 Medium

198 of 596

Application +4 | -1

\int f(x) d x=\psi(x)

, then

\int x^5 f\left(x^3\right) d x

is equal to

View answer and explanation

Correct option: C

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

Option C (

\frac{1}{3} x^3 \psi(x^3)-\int x^2 \psi(x^3) d x+C

) is correct because integration by parts and substitution lead to this result.

199 Medium

199 of 596

Analyzing +4 | -1

{I_1} = \int \limits_0^\infty \,\,\frac{1}{{1 + {x^4}}}\,\,dx

and

{I_2} = \int \limits_0^\infty \,\,\frac{{{x^2}}}{{1\,\, + \,\,{x^4}}}\,\,dx

. Then

\frac{{{I_1}}}{{{I_2}}} =

View answer and explanation

Correct option: A

Option A (1) is correct because by substituting x = 1/t in I1, it transforms into I2, showing that I1 and I2 are equal.

200 Medium

200 of 596

Application +4 | -1

\int \limits_0^3 \,\,\left| {\,{x^3} + {x^2} + 3x\,} \right|\,\,\,dx

is equal to

View answer and explanation

Correct option: B

\frac{{171}}{4}

Option B (

\frac{{171}}{4}

) is correct because the integrand is positive over the interval [0, 3], and the integral evaluates to this value.

201 Medium

201 of 596

Application +4 | -1

The area bounded by

y=-x^2+2 x+3

and

y=0

View answer and explanation

Correct option: B

\frac{32}{3}

Option B (

\frac{32}{3}

) is correct because the area bounded by the curve

y = -x^2 + 2x + 3

and

y=0

is found by integrating the function from -1 to 3, which results in

\frac{32}{3}

202 Medium

202 of 596

Application +4 | -1

\int\limits_{0}^{\pi / 2} \frac{\sin ^{3 / 2} x d x}{\cos ^{3 / 2} x+\sin ^{3 / 2} x}=

View answer and explanation

Correct option: D

\pi / 4

Option D (

\pi / 4

) is correct because the integral simplifies to

\pi/4

after using the property of definite integrals and adding the original and transformed integrals.

203 Medium

203 of 596

Application +4 | -1

The area of the region bounded by the

\mathrm{x}

-axis and the curves defined by

y=\tan x,(-\pi / 3 \leq \mathrm{x} \leq \pi / 3)

View answer and explanation

Correct option: C

2 \log 2

Option C (

2 \log 2

) is correct because it represents the area of the region bounded by the x-axis and the curve y = tan(x) from -π/3 to π/3.

204 Medium

204 of 596

Analyzing +4 | -1

\int\limits_{0}^{\sin ^2 x}\left(\operatorname{Sin}^{-1} \sqrt{t}\right) d t+\int\limits_{0}^{\cos ^2 x}\left(\operatorname{Cos}^{-1} \sqrt{t}\right) d t=

View answer and explanation

Correct option: C

\frac{\pi}{4}

Option C (

\frac{{π}}{{4}}

) is correct because the derivative of the given expression is zero, implying it's a constant, and evaluating it at x = pi/4 gives pi/4.

205 Medium

205 of 596

Application +4 | -1

\int\limits_{0}^{r / 2} \frac{\sqrt{\cot x}}{\sqrt{\cot x}+\sqrt{\tan x}} d x=

View answer and explanation

Correct option: C

\frac{\pi}{4}

Option C (

\frac{\pi}{4}

) is correct because by using the property of definite integrals and adding the original integral to the transformed integral, we can simplify the expression and solve for the integral.

206 Medium

206 of 596

Analyzing +4 | -1

f(x)=\int\limits_{{0}}^{2} e^{|x-t|} d t

, then the minimum value of

f(x)

is equal to:

View answer and explanation

Correct option: A

2(e-1)

Option A (2(e-1)) is correct because the minimum value of the function f(x) is obtained within the interval [0, 2] by analyzing the function and its derivative.

207 Medium

207 of 596

Application +4 | -1

f\left( x \right) = f\left( {a - x} \right),\;\,\,\mathrm{then}\,\,\int \nolimits_0^a \,\,f\left( x \right)\,\,dx

is equal to

View answer and explanation

Correct option: C

\frac{a}{2}\,\,\int \nolimits_0^a \,\,f\left( x \right)\,\,dx

Option C (

\frac{a}{2}\,\,\int \nolimits_0^a \,\,f(x)\,\,\;dx

) is correct because using the property

f(x) = f(a - x)

simplifies the integral to

\frac{a}{2}\,\,\int \nolimits_0^a \,\,f(x)\,\,\;dx

208 Medium

208 of 596

Application +4 | -1

\int\limits_{{-\pi / 2}}^{\pi / 2} \sin ^2 x \cos ^2 x(\sin x+\cos x) d x=

View answer and explanation

Correct option: B

\frac{4}{15}

Option B (

\frac{4}{15}

) is correct because the integral simplifies to

2 \int_0^{\pi/2} \sin^2 x \cos^3 x dx

, which evaluates to

\frac{4}{15}

209 Medium

209 of 596

Application +4 | -1

a

and

b

are real numbers different from zero, then

\int\limits_{0}^{\pi / 2} \frac{|a b|}{a^2 \cos ^2 x+b^2 \sin ^2 x} d x=

View answer and explanation

Correct option: A

\frac{\pi}{2}

Option A (

\frac{\pi}{2}

) is correct because after simplification and using trigonometric substitutions, the integral evaluates to

\frac{\pi}{2}

210 Medium

210 of 596

Application +4 | -1

The value of

\int\limits_{0}^{11}[x]^3 . d x

, where

[*]

denotes the greatest integer function, is

View answer and explanation

Correct option: D

3025

Option D (3025) is correct because the integral is evaluated by summing the integrals over each integer interval from 0 to 10.

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