If $f(x)=a+b x+c x^2$ where $a, b, c \in R$, then $\int_0^1 f(x) d x$ is equal to

$\frac{1}{3}\left\{f(1)+f(0)+2 f\left(\frac{1}{2}\right)\right\}$

Let $f\,\left( x \right)$ be a function satisfying $f'\left( x \right) = f\left( x \right)$ with $f\left( 0 \right) = 1$ and ${\rm{g}}\,\,\left( x \right)$ be a function that satisfies $f\left( x \right)\,\, + \,\,{\rm{g}}\left( x \right) = {x^2}$. Then, the value of the integral $\mathop \smallint \nolimits_0^1 \,\,f\left( x \right)\,\,{\rm{g}}\,\,\left( x \right)\,\,dx$, is

$e - \frac{{{e^2}}}{2} - \frac{5}{2}$

$e\,\, + \,\,\frac{{{e^2}}}{2}\,\, + \,\,\frac{5}{2}$

$\smallint \,\,\frac{{x\,\, + \,\,\sin \,\,x}}{{1\,\, + \,\,\cos \,\,x}}\,\,dx$ is equal to

$\log \,\,\left( {1\,\, + \,\,\cos \,\,x} \right)\,\, + \,\,c$

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

$\log \,\,\left( {x\,\, + \,\,\sin \,\,x} \right)\,\, + \,\,c$

Back to DEFINITE INTEGRATION

Your Progress

0 / 995 Answered

Correct

Incorrect

Accuracy

JEE • MATHEMATICS • DEFINITE INTEGRATION

Definite integration practice · Mixed difficulty

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 211-240 of 995 questions

Preparing practice and saved progress…

211 Easy

211 of 995

Application +4 | -1

The value of

\int_{ - \pi /2}^{\pi /2} {\frac{{{{\cos }^2}\,x}}{{1 + {3^x}}}\,dx}

View answer and explanation

Correct option: A

\frac{\pi }{4}

Option A (

\frac{\pi }{4}

) is correct because by using the property of definite integrals and adding the two forms of the integral, we can simplify the expression and evaluate it.

212 Easy

212 of 995

Application +4 | -1

\int \nolimits_{ - 1}^1 \,\,\left| {1\, - \,x\,} \right|\,\,dx

is equal to

View answer and explanation

Correct option: C

\mathrm{2}

Option C (

\mathrm{2}

) is correct because the integral of the absolute value function is evaluated by recognizing that |1-x| = 1-x on the interval [-1, 1].

213 Medium

213 of 995

Application +4 | -1

Area bounded by the curves

y^2=4 x, y^2=-4 x

and

y=4

is equal to

View answer and explanation

Correct option: C

\frac{32}{3}

sq. units

Option C (

\frac{32}{3}

sq. units) is correct because the area is calculated by integrating the difference between the curves from y = 0 to y = 4 and multiplying by 2 due to symmetry.

214 Easy

214 of 995

Application +4 | -1

The area bounded by

y = \log x,\;\;x

-axis and ordinates

x = 1,\;x = 2

View answer and explanation

Correct option: C

\log \left( {4/e} \right)

Option C,

\log \left( {4/e} \right)

, is the correct area calculated by integrating the function y = log x from x = 1 to x = 2 and simplifying the result.

215 Medium

215 of 995

Analyzing +4 | -1

The value of

\lim\limits_{n \to \infty } \,\,{\left\{ {\left( {1 + \frac{1}{{{n^2}}}} \right)\,\,\left( {1 + \frac{{{2^2}}}{{{n^2}}}} \right)\,\, \ldots \,\,\left( {1 + \frac{{{n^2}}}{{{n^2}}}} \right)} \right\}^{1/n}}

, is

View answer and explanation

Correct option: C

\frac{2}{{{e^2}}}\,\,{e^{\pi /2}}

Option C is correct because after converting the product to a summation and evaluating the limit using integration, the result is

\frac{2}{{{e^2}}}\,\,{e^{\pi /2}}

216 Easy

216 of 995

Application +4 | -1

Let

f(x)=\int\limits_{0}^x \sqrt{6-u^2} d u

. Then the real roots of the equation

x^2-f^{\prime}(x)=0

are

View answer and explanation

Correct option: C

x= \pm \sqrt{2}

Option C is correct because the real roots are found by solving the equation after finding the derivative of the integral and setting it equal to x squared.

217 Easy

217 of 995

Understanding +4 | -1

f(x)=a+b x+c x^2

where

a, b, c \in R

, then

\int_0^1 f(x) d x

is equal to

View answer and explanation

Correct option: B

\frac{1}{6}\left\{f(1)+f(0)+4 f\left(\frac{1}{2}\right)\right\}

Option B (

\frac{1}{6}\left{f(1)+f(0)+4 f\left(\frac{1}{2}\right)\right}

) is correct because by evaluating the integral

\int_0^1 f(x) dx

and comparing it with the given options, we find that the correct expression is

\frac{1}{6}\left{f(1)+f(0)+4 f\left(\frac{1}{2}\right)\right}

218 Easy

218 of 995

Application +4 | -1

\int \nolimits_{ - 2}^{10} \,\mathrm{sgn}\,(x - \,\,\left[ x \right]\,dx

Equals, where

\left[ {\,\mathrm{.}\,} \right]\,

denotes greatest integer function

View answer and explanation

Correct option: C

\mathrm{8}

Option C (8) is correct because the integral of the sign function simplifies to the difference between the lengths of the positive and negative intervals.

219 Easy

219 of 995

Application +4 | -1

\int\limits_{0}^1 \frac{\sin t}{1+t} d t=\alpha

, then the value of the integral

\int\limits_{{4 \pi-2}}^{4 \pi} \frac{\sin \frac{t}{2}}{4 \pi+2-t} d t

in terms of

\alpha

is given by

View answer and explanation

Correct option: D

-\alpha

Option D (

-\alpha

) is correct because after substitution and simplification, the integral becomes the negative of the given integral.

220 Medium

220 of 995

Application +4 | -1

\int \limits_0^\infty \,\,\frac{{dx}}{{{{\left[ {x\,\, + \,\,\sqrt {{x^2}\,\, + \,\,1} } \right]}^3}}} =

View answer and explanation

Correct option: A

\frac{3}{8}

Option A (

\frac{3}{8}

) is correct because the definite integral evaluates to 3/8 after using a trigonometric substitution.

221 Medium

221 of 995

Application +4 | -1

\int_0^2\left[x^2\right] d x

, where

[x]

is the greatest integer

\leq \mathrm{x}

View answer and explanation

Correct option: C

5-\sqrt{2}-\sqrt{3}

Option C (

5-\sqrt{2}-\sqrt{3}

) is correct because it is the result of evaluating the definite integral of the greatest integer function of x squared over the interval [0, 2].

222 Medium

222 of 995

Analyzing +4 | -1

Let

I_n(x)=\int_0^x \frac{1}{\left(t^2+5\right)^n} d t, n=1,2,3, \ldots

. Then

View answer and explanation

Correct option: A

50 I_6-9 I_5=x I_5^{\prime}

Option A (

50 I_6-9 I_5=x I_5^{\prime}

) is correct because it is derived from integration by parts and simplification of the given integral

I_n(x)

223 Medium

223 of 995

Application +4 | -1

The value of

\int_0^\pi|\cos x|^3 d x

View answer and explanation

Correct option: D

\frac{4}{3}

The correct answer is D (

\frac{4}{3}

) because after evaluating the integral of the absolute value of cosine cubed from 0 to pi, the result is

\frac{4}{3}

224 Medium

224 of 995

Analyzing +4 | -1

I_n=\int\limits_{0}^1\left(\cos ^{-1} x\right)^n \mathrm{~d} x

then

I_6-360 I_2

is given by

View answer and explanation

Correct option: B

6\left(\frac{\pi}{2}\right)^5-120\left(\frac{\pi}{2}\right)^3

Option B (

6(\frac{\pi}{2})^5-120(\frac{\pi}{2})^3

) is correct because after applying integration by parts twice, the expression

I_6 - 360I_2

simplifies to

6(\frac{\pi}{2})^5-120(\frac{\pi}{2})^3

225 Hard

225 of 995

Application +4 | -1

Let

f\,\left( x \right)

be a function satisfying

f'\left( x \right) = f\left( x \right)

with

f\left( 0 \right) = 1

and

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

be a function that satisfies

f\left( x \right)\,\, + \,\,\mathrm{g}\left( x \right) = {x^2}

. Then, the value of the integral

\int \nolimits_0^1 \,\,f\left( x \right)\,\,\mathrm{g}\,\,\left( x \right)\,\,dx

, is

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because it is the result of integrating the product of the two functions after finding f(x) and g(x).

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

226 Easy

226 of 995

Application +4 | -1

The value of

\int\limits_{{-\pi}}^\pi \frac{\cos ^2 x}{1+a^x} d x, a > 0

, is

View answer and explanation

Correct option: C

\pi / 2

Option C (

\pi / 2

) is correct because by using the property

\int_0^a f(x) dx = \int_0^a {f(x) + f(-x)} dx

, the integral simplifies to

\int_0^\pi \cos^2 x dx

, which evaluates to

\pi/2

227 Medium

227 of 995

Analyzing +4 | -1

The value of

\int \nolimits_{ - 2}^2 \,\,\left[ {p\,\,\log \,\,\left( {\frac{{1 + x}}{{1 - x}}} \right) + \,\,q\,\,\log \,\,{{\left( {\frac{{1 - x}}{{1 + x}}} \right)}^{ - 2}} + \,\,r} \right]\,\,dx

depends on

View answer and explanation

Correct option: C

The value of

r

Option C (The value of

r

) is correct because the integral depends only on the constant term

r

due to the odd function property of the logarithmic terms.

228 Medium

228 of 995

Application +4 | -1

Whenever

a < b

, the value of

\int\limits_{a}^b \frac{|x|}{x} d x

View answer and explanation

Correct option: C

|b|-|a|

Option C is correct because the integral evaluates to the difference in absolute values at the bounds, which is a direct consequence of the integrand being |x|/x.

229 Medium

229 of 995

Application +4 | -1

The definite integral

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

is equal to

View answer and explanation

Correct option: C

\frac{\pi}{2}-1

Option C (

\frac{\pi}{2}-1

) is correct because it is the result of integrating the given definite integral after rationalizing the denominator.

230 Easy

230 of 995

Analyzing +4 | -1

\int \nolimits_0^2 \,\,\sqrt {\frac{{2\,\, + \,\,x}}{{2\,\, - \,\,x}}} \,\,dx

is equal to

View answer and explanation

Correct option: A

\pi + 2

Option A is correct because after simplifying and using trigonometric substitution, the integral evaluates to π + 2.

231 Medium

231 of 995

Analyzing +4 | -1

Consider the following statements:
1.

\int \nolimits_{ - \pi /2}^{\pi /2} \,\,\sqrt {\cos \,\,x\,\, - \,\,{{\cos }^3}\,\,x} \,dx = \frac{3}{4}

\int \nolimits_0^4 \,\,\left( {\,\left| {\,x - 1\,} \right|\,\, + \,\,\left| {\,x - 3\,} \right|\,} \right)dx = 10

Which of these is/are correct?

View answer and explanation

Correct option: B

Only

\left( \mathrm{2} \right)

Option B (Only (2)) is correct because statement 2 is true after evaluating the integral, while statement 1 is false.

232 Easy

232 of 995

Understanding +4 | -1

Let

f\;:R \to R,\,\;g\;:R \to R

be continuous functions. Then the value of the integral

\int \limits_{ - \pi /2}^{\pi /2} \,\,\left[ {\,f\left( x \right)\,\, + \,\,f\left( { - x} \right)} \right]\,\,\left[ {\,\,g\,\,\left( x \right) - \,\,g\left( {\, - \,x} \right)} \right]\,\,dx

View answer and explanation

Correct option: D

\mathrm{0}

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

233 Easy

233 of 995

Application +4 | -1

\{x\}

denotes the fractional part of

x

, then the value of the integral

\int\limits_{2}^4\{2 x\} d x

View answer and explanation

Correct option: B

Option B (1) is correct because the integral simplifies due to the periodic nature of the fractional part function.

234 Medium

234 of 995

Application +4 | -1

The value of

\int\limits_{0}^2\left[x^2-x+1\right] d x

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

View answer and explanation

Correct option: C

\frac{5-\sqrt{5}}{2}

Option C (

\frac{{5-\sqrt{{5}}}}{{2}}

) is correct because the integral of the greatest integer function is evaluated by splitting the interval based on where the function takes integer values.

235 Easy

235 of 995

Application +4 | -1

\int\limits_{\sqrt{2}}^x \frac{d t}{t \sqrt{t^2-1}}=\frac{\pi}{12}

, then

x

equals

View answer and explanation

Correct option: C

Option C (2) is correct because after evaluating the integral and solving for x, the result is x=2.

236 Medium

236 of 995

Application +4 | -1

\int \nolimits_{\sin x}^1 \,\,{t^2}\,f\left( t \right)\,\,dt = 1 - \sin \,\,x,\;\,\,\forall \;x \in \,\,\left[ {0,\,\,\pi /2} \right]

, then

f\,\,\left( {\frac{1}{{\sqrt 3 }}} \right)

View answer and explanation

Correct option: A

Option A (3) is correct because differentiating both sides using Leibniz's rule and then substituting

x

such that

\sin x = \frac{1}{\sqrt{3}}

gives the value of

f(\frac{1}{\sqrt{3}})

237 Easy

237 of 995

Analyzing +4 | -1

The equation

\int \limits_{ - \pi /4}^{\pi /4} \,\,\left\{ {a\,\,\left| {\,\sin \,\,x\,} \right|\,\, + \,\,\frac{{b\,\,\sin \,\,x}}{{1\, + \,\,\cos \,\,x}}\,\, + \,\,c} \right\}dx = 0

, where

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

are constants, gives a relation between

View answer and explanation

Correct option: B

a

and

c

Option B (a and c) is correct because the integral of the term with 'b' vanishes due to it being an odd function, leaving a relation only between 'a' and 'c'.

238 Medium

238 of 995

Application +4 | -1

The tangent of the curve

y = f\left( x \right)

at the point with abscissa

x = 1

form an angle of

\pi /6

and at the point

x = 2

an angle of

\pi /3

and at the point

x = 3

an angle of

\pi /4

. If

f''\left( x \right)

is continuous, then the value of

\mathop {\int \,\,}\limits_1^3 f''\left( x \right)f'\left( x \right)\,\,dx + \mathop {\int \,}\limits_2^3 f''\left( x \right)\,\,dx

, is

View answer and explanation

Correct option: C

\frac{{4 - 3\,\,\sqrt 3 }}{3}

Option C is correct because the expression simplifies to

\frac{{4 - 3\,\,\sqrt 3 }}{3}

after evaluating the integrals using the given information about the derivatives.

239 Easy

239 of 995

Application +4 | -1

The value of

\int\limits_{0}^2 x d([x]-x)

, is

View answer and explanation

Correct option: B

Option B (1) is correct because the integral is evaluated using integration by parts and properties of the greatest integer function.

240 Medium

240 of 995

Analyzing +4 | -1

\int \,\,\frac{{x\,\, + \,\,\sin \,\,x}}{{1\,\, + \,\,\cos \,\,x}}\,\,dx

is equal to

View answer and explanation

Correct option: A

x\,\,\tan \,\,\frac{x}{2}\,\, + \,\,c

Option A (

x \tan \frac{x}{2} + c

) is correct because the integral can be solved by rewriting the integrand using trigonometric identities and then integrating by parts.

Revision notes and key takeaways

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

Showing 211-240 of 995 questions

Your Progress

Definite integration practice · Mixed difficulty

Explore Other Chapters in MATHEMATICS

Continue Your Practice

Your practice