If ${I_1} = \mathop \smallint \nolimits_0^{\pi /2} \,\,x\,\,\sin \,\,x\,dx$ and ${I_2} = \mathop \smallint \nolimits_0^{\pi /2} \,x\,\,\cos \,\,x\,dx$, then which one of the following is true?

${I_2} - {I_1} = \frac{\pi }{2}$

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JEE • MATHEMATICS • DEFINITE INTEGRATION

Definite integration practice · Easy

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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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241 Easy

241 of 791

Understanding +4 | -1

\int\limits_{{-1}}^1 x^{17} \cos ^4 x d x=

View answer and explanation

Correct option: C

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

242 Easy

242 of 791

Application +4 | -1

The correct evaluation of

\int\limits_{0}^{\pi / 2}\left|\sin \left(x-\frac{\pi}{4}\right)\right| d x

View answer and explanation

Correct option: B

2-\sqrt{2}

Option B (

2-\sqrt{{2}}

) is correct because it is the result of evaluating the definite integral of the absolute value of a sine function over the given interval.

243 Easy

243 of 791

Analyzing +4 | -1

Given

\int\limits_{1}^2 e^{x^2} d x=a

, the value of

\int\limits_{e}^{e^4} \sqrt{\log _e x} \,d x

, is

View answer and explanation

Correct option: D

2 e^4-e-a

Option D (

2 e^4-e-a

) is correct because it is obtained by using substitution and integration by parts on the given integral.

244 Easy

244 of 791

Application +4 | -1

\int\limits_{{-1}}^1 \log \frac{2-x}{2+x} d x=

View answer and explanation

Correct option: D

Option D (0) is correct because the integrand is an odd function, and the integral is evaluated over a symmetric interval [-1, 1].

245 Easy

245 of 791

Application +4 | -1

\int \nolimits_0^\infty \,\,\left[ {\frac{{\left[ {x\,\, + \,\,1} \right]}}{{{e^x}}}} \right]\,\,dx

(where

\left[ {\,\, \cdot \,\,} \right]

denotes greatest integer function) is equal to

View answer and explanation

Correct option: A

\mathrm{0}

Option A (0) is correct because the expression inside the greatest integer function is always less than 1, making the greatest integer function equal to 0.

246 Easy

246 of 791

Analyzing +4 | -1

The area bounded by the curves

y^2=12 x

and

x^2=12 y

is divided by the line

x=3

in two parts. The area (in square units) of the larger part, is

View answer and explanation

Correct option: B

\frac{147}{4}

Option B (

\frac{147}{4}

) is correct because it represents the larger area obtained after subtracting the smaller area from the total area enclosed by the curves.

247 Easy

247 of 791

Application +4 | -1

Let

f: R \rightarrow R

be defined as

f(x)=a e^{2 x}+b e^x+c x

. If

f(0)=-1, f^{\prime}\left(\log _e 2\right)=21

and

\int\limits_{0}^{\log _e 4}(f(x)-c x) d x=\frac{39}{2}

, then the value of

|a+b+c|

equals

View answer and explanation

Correct option: D

Option D (8) is correct because it involves solving a system of equations derived from the given conditions on the function f(x) and its derivative.

248 Easy

248 of 791

Analyzing +4 | -1

The value of

\int\limits_{{-1}}^1 \{\sqrt{1+x+x^2}-\sqrt{1-x+x^2}\}\ d x

, is

View answer and explanation

Correct option: A

Option A (0) is correct because the integrand is an odd function integrated over a symmetric interval.

249 Easy

249 of 791

Application +4 | -1

Let

4 \alpha \int_{-1}^2 e^{-\alpha|x|} d x=5

. Then,

\alpha=

View answer and explanation

Correct option: A

\ln 2

The correct answer is A (

\ln 2

) because after evaluating the integral and solving for

\alpha

, the value of

\alpha

is found to be

\ln 2

250 Easy

250 of 791

Application +4 | -1

\int\limits_{{-a}}^a(|x|+|x-2|) d x=22

and

a>2

, then the value of

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

View answer and explanation

Correct option: A

Option A (3) is correct because the value of 'a' is found by solving the given integral equation, and then the second integral is evaluated using the periodic property.

251 Easy

251 of 791

Application +4 | -1

The value of

\int\limits_{0}^1 \frac{d x}{x+\sqrt{1-x^2}}

View answer and explanation

Correct option: D

\frac{\pi}{4}

Option D (

\frac{\pi}{4}

) is correct because the integral is evaluated by using a trigonometric substitution, which simplifies the integral to a known form.

252 Easy

252 of 791

Application +4 | -1

The area between the curve

y=\sin ^2 x, x

-axis and the ordinates

x=0

and

x=\frac{\pi}{2}

View answer and explanation

Correct option: B

\frac{\pi}{4}

Option B (

\frac{\pi}{4}

) is correct because the area under the curve

y = \sin^2 x

from 0 to

\frac{\pi}{2}

is found by integrating the function, which simplifies to

\frac{\pi}{4}

253 Easy

253 of 791

Analyzing +4 | -1

{I_1} = \int \nolimits_0^{3\pi } \,\,f\,\,({\cos ^2}\,\,x)\,\,dx

and

{I_2} = \int \nolimits_0^\pi \,\,f({\cos ^2}\,\,x)\,\,dx,\mathrm{then}

,〗 then

View answer and explanation

Correct option: C

{I_1} = 3\,{I_2}

Option C (

{I_1} = 3\,{I_2}

) is correct because the integral

{I_1}

can be split into three equal integrals, each equal to

{I_2}

254 Easy

254 of 791

Analyzing +4 | -1

The value of the definite integral

\int\limits_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {\frac{{dx}}{{(1 + \mathrm{ }{e^{x\,\cos x}})({{\sin }^4}x + {{\cos }^4}x)}}}

is equal to

View answer and explanation

Correct option: B

\frac{\pi }{{2\sqrt 2 }}

Option B (

\frac{\pi }{{2\sqrt 2 }}

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

255 Easy

255 of 791

Application +4 | -1

\int\limits_{0}^{\pi / 2} \frac{\cos x-\sin x}{1+\sin x \cos x} d x=

View answer and explanation

Correct option: C

Option C (0) is correct because the integral evaluates to zero due to the property of definite integrals where substituting

\frac{\pi}{2} - x

results in the negative of the original integrand.

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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

256 Easy

256 of 791

Application +4 | -1

The value of

\int\limits_0^{\frac{\pi }{2}} {\sqrt {\sin 2\theta } \sin \theta } \,d\theta

View answer and explanation

Correct option: D

\frac{\pi }{4}

Option D (

\frac{\pi }{4}

) is correct because the integral can be simplified using trigonometric identities and substitution to arrive at the final value.

257 Easy

257 of 791

Application +4 | -1

The value of integral,

: \int\limits_{3}^6 \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x

View answer and explanation

Correct option: B

\frac{3}{2}

Option B (

\frac{3}{2}

) is correct because using the property

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

simplifies the integral.

258 Easy

258 of 791

Application +4 | -1

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

View answer and explanation

Correct option: C

Option C (2) is correct because the integral of the absolute value function |1-x| over the interval [-1, 1] evaluates to 2.

259 Easy

259 of 791

Application +4 | -1

Area bounded by the curve

y=\ell n x+\tan ^{-1} x

and

x-a x i s

from

x=1

x=2

, is

View answer and explanation

Correct option: D

\frac{5}{2} \ln 2-\frac{1}{2} \ln 5+2 \tan ^{-1} 2-\frac{\pi}{4}-1

Option C is the correct answer because it represents the definite integral of the given function from x=1 to x=2, evaluated using integration by parts.

260 Easy

260 of 791

Analyzing +4 | -1

I_n=\int\limits_{1}^e(\log x)^n d x

, then

I_n+n I_{n-1}

is equal to

View answer and explanation

Correct option: B

e

Option B (

e

) is correct because integrating by parts simplifies the expression and leads to the result

I_n + nI_{n-1} = e

261 Easy

261 of 791

Application +4 | -1

\int \nolimits_0^{\pi /6} \frac{{\sin \,\,x}}{{{{\cos }^3}\,\,x}}\,\,dx

is equal to

View answer and explanation

Correct option: B

\frac{1}{6}

Option B (

\frac{1}{6}

) is correct because by substituting cos(x) = t, the integral simplifies and evaluates to 1/6.

262 Easy

262 of 791

Understanding +4 | -1

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

has the same value as the integral

View answer and explanation

Correct option: A

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

Option A (

\int_0^1 f(x) dx

) is correct because substituting 1-x = t transforms the integral into the same integral with variable x.

263 Easy

263 of 791

Application +4 | -1

Let

f\,\,\left( x \right) = x\,\, - \,\,\left[ x \right]

, for every real number

x

, where

\left[ x \right]

is the integral part of

x

. Then,

\int \nolimits_{ - 1}^1 \,\,f\,\,\left( x \right)\,\,dx

is equal to

View answer and explanation

Correct option: A

1

Option A (1) is correct because the integral of the fractional part of x over the given interval is equal to 1.

264 Easy

264 of 791

Understanding +4 | -1

The value of

\int\limits_{0}^\pi \cos ^{11} x \,d x

, is

View answer and explanation

Correct option: C

Option C is correct because the integral of cos¹¹(x) from 0 to pi is zero due to the symmetry of the cosine function around pi/2.

265 Easy

265 of 791

Application +4 | -1

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

View answer and explanation

Correct option: D

Option D (4) is correct because the integral of the absolute value of sin(x) from 0 to 2π is 4, calculated by splitting the integral into two parts.

266 Easy

266 of 791

Analyzing +4 | -1

\left[ {\,\sum \limits_{n = 1}^{10} \,\,\int \limits_{ - 2n - 1}^{ - 2n} \,\,{{\sin }^{27}}\,\,x\,dx} \right]\,\, + \,\,\left[ {\sum \limits_{n = 1}^{10} \,\,\int \limits_{2n}^{2n + 1} \,\,{{\sin }^{27}}\,\,x\,dx} \right]

equals

View answer and explanation

Correct option: D

\mathrm{0}

Option D (0) is correct because the sum of the integrals can be shown to cancel out by using the substitution x = -t and the properties of the sine function.

267 Easy

267 of 791

Application +4 | -1

f\left( x \right)

be continuous function such that the area bounded by the curve

y = f\left( x \right)

, the

x

-axis and the lines

x = a\mathrm{\;and}\;x = 0\;\mathrm{is}\frac{{{a^2}}}{2} + \frac{a}{2}\sin a + \frac{\mathrm{\pi }}{2}\cos a.\,\mathrm{Value\;of\;}f\left( {\frac{\mathrm{\pi }}{2}} \right)\;\mathrm{is}

View answer and explanation

Correct option: A

\frac{1}{2}

Option A (

\frac{1}{2}

) is correct because it is the result of differentiating the area function and substituting

\frac{\pi}{2}

for

a

268 Easy

268 of 791

Application +4 | -1

{I_1} = \int \nolimits_0^{\pi /2} \,\,x\,\,\sin \,\,x\,dx

and

{I_2} = \int \nolimits_0^{\pi /2} \,x\,\,\cos \,\,x\,dx

, then which one of the following is true?

View answer and explanation

Correct option: A

{I_1} + {I_2} = \frac{\pi }{2}

Option A (

{I_1} + {I_2} = \frac{\pi }{2}

) is correct because evaluating both integrals and summing them results in

\frac{\pi}{2}

269 Easy

269 of 791

Application +4 | -1

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

View answer and explanation

Correct option: B

Option B (0) is correct because the function

f(x) = x|x|

is an odd function, and the integral of an odd function over a symmetric interval [-a, a] is always zero.

270 Easy

270 of 791

Application +4 | -1

The area between the curve

y = 4 + 3x - {x^2}

and

x

-axis is

View answer and explanation

Correct option: A

125/6 sq unit

Option A (125/6 sq unit) is correct because it represents the definite integral of the quadratic function from -1 to 4.

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