$\mathop \smallint \nolimits_{ - \pi /3}^{\pi /3} \,\,\frac{{x\,\,\sin \,\,x}}{{{{\cos }^2}\,\,x}}\,\,dx$ is equal to

$\frac{1}{3}\,\,\left( {4\pi + 1} \right)$

$\frac{{4\pi }}{3}\,\, + \,\,\log \,\,\tan \,\,\frac{{5\pi }}{{12}}$

$\mathop \smallint \nolimits_0^{\pi /2} \,\,\log \,\,\sin \,\,x{\rm{ }}\,dx$ is equal to

$ - \,\left( {\frac{{\rm{\pi }}}{2}} \right)\,\,\log \,\,2$

${\rm{\pi }}\,\,\log \,\,\frac{1}{2}$

$ - \,{\rm{\pi }}\,\,\log \,\,\frac{1}{2}$

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

If $f(x)$ is a periodic function, with period $T$, then

$\int_\limits a^b f(x) d x=\int_\limits a^{b+T} f(x) d x$

$\int_\limits a^b f(x) d x=\int_\limits {a+T}^b f(x) d x$

$\int_\limits a^b f(x) d x=\int_\limits{a+T}^{b+2 T} f(x) d x$

Let $f\left( x \right)$ be a non-negative continuous function such that the area bounded by the curve $y = f\left( x \right),\;x$-axis and the ordinates $x = \frac{{\rm{\pi }}}{4}{\rm{\;}}$ and $x = {\rm{\beta }} > \frac{{\rm{\pi }}}{4}\,{\rm{is}}\,{\rm{\pi }}$ $\left( {{\rm{\beta }}\sin {\rm{\beta }} + \frac{{\rm{\pi }}}{4}\cos {\rm{\beta }} + \sqrt 2 {\rm{\beta }}} \right)\;{\rm{the\;}}f\left( {\frac{{\rm{\pi }}}{2}} \right){\rm{s}}$

$\left( {1 - \frac{{\rm{\pi }}}{4} + \sqrt 2 } \right)$

$\left( {1 - \frac{{\rm{\pi }}}{4} - \sqrt 2 } \right)$

$\left( {\frac{{\rm{\pi }}}{4} - \sqrt 2 + 1} \right)$

$\left( {\frac{{\rm{\pi }}}{4} + \sqrt 2 - 1} \right)$

$\smallint \,\,\frac{1}{{1\,\, + \,\,\cos \,\,ax}}\,\,\;dx$ is equal to

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

$\frac{1}{a}\;\,\,\left( {{\rm{cosec}}\,\,\;ax\,\, - \,\,{\rm{cot}}\,\,\;ax} \right)\,\, + \,\,c$

$\frac{1}{a}\,\,({\rm{cosec}}\;\,\,ax\,\, + \,\,\cot \,\,ax)\,\, + \,\,c$

If $f\left( y \right) = {e^y},\,\,\;g\left( y \right) = y;\;\;\,\;y > 0$ and $F\left( t \right) = \mathop \smallint \nolimits_0^t \,\,f\left( {t - y} \right)\,\,g\,\,\left( y \right)\;\,\,dy$, then

$F\,\,\left( t \right) = 1 - {e^{ - t}}\,\,\left( {1 + t} \right)$

$F\,\,\left( t \right) = t{e^t}$

$F\,\,\left( t \right) = t{e^{ - t}}$

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JEE • MATHEMATICS • DEFINITE INTEGRATION

Definite integration practice · Mixed difficulty

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691 Easy

691 of 995

Understanding +4 | -1

Let

f

be a continuous function

[a, b]

such that

f(x) > 0

for all

x \in[a, b]

. If

F(x)=\int\limits_{a}^x f(t) \mathrm{d} t

then

View answer and explanation

Correct option: B

F

is differentiable and increasing on

[a, b]

Option B (

F

is differentiable and increasing on

[a, b]

) is correct because

F'(x) = f(x) > 0

[a, b]

, which means

F

is increasing and differentiable.

692 Medium

692 of 995

Analyzing +4 | -1

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

is equal to

View answer and explanation

Correct option: B

\frac{{4\pi }}{3}\,\, - \,\,2\,\,\log \,\,\tan \,\,\frac{{5\pi }}{{12}}

Option B (

\frac{{4\pi }}{3}\, - \,2\,\log \,\tan \,\frac{{5\pi }}{{12}}

) is correct because after applying integration by parts and simplifying, the result matches this option.

693 Easy

693 of 995

Analyzing +4 | -1

The value of

I=\int\limits_{0}^{\pi / 2} \frac{(\sin x+\cos x)^2}{\sqrt{1+\sin 2 x}} d x

, is

View answer and explanation

Correct option: C

2

The correct answer is C (

2

) because the integral simplifies to

\int_0^{\pi/2} (\sin x + \cos x) dx

, which evaluates to 2.

694 Easy

694 of 995

Application +4 | -1

\int \nolimits_0^1 \,\,\frac{{x\,\,dx}}{{\left[ {x + \sqrt {1 - {x^2}} } \right]\,\,\sqrt {1 - {x^2}} }}

is equal to

View answer and explanation

Correct option: C

\mathrm{\pi }/4

Option C (

\frac{\mathrm{\pi }}{4}

) is correct because after substituting

x = \sin \theta

and using properties of definite integrals, the integral simplifies to this value.

695 Medium

695 of 995

Application +4 | -1

f:R \to R

is given by

f(x) = x + 1

, then the value of

\lim\limits_{n \to \infty } \frac{1}{n}\left[ {f(0) + f\left( {\frac{5}{n}} \right) + f\left( {\frac{{10}}{n}} \right) + ..... + f\left( {\frac{{5(n - 1)}}{n}} \right)} \right]

View answer and explanation

Correct option: D

\frac{7}{2}

Option D (7/2) is correct because the limit can be expressed as a definite integral, which evaluates to 7/2.

696 Medium

696 of 995

Application +4 | -1

The value of

\int \limits_{ - 1}^2 \,\,\frac{{\left| {\,x\,} \right|}}{x}\,\,dx

, is

View answer and explanation

Correct option: B

\mathrm{1}

Option B (1) is correct because the integral of |x|/x from -1 to 2 can be split into two integrals, one from -1 to 0 and another from 0 to 2, and then evaluated.

697 Easy

697 of 995

Application +4 | -1

\int \nolimits_0^1 \,\,{x^{3/2}}\,\,\sqrt {1 - x\,\,} dx

is equal to

View answer and explanation

Correct option: D

\frac{\pi }{16}

Option D (

\frac{\pi}{16}

) is correct because the integral is solved using a trigonometric substitution and Walli's formula.

698 Easy

698 of 995

Analyzing +4 | -1

I_1=\int\limits_{0}^1\left(1-x^{50}\right)^{100} d x, I_2=\int\limits_{0}^1\left(1-x^{50}\right)^{101}

and

I_2=\alpha I_1

, then

\alpha

equals to

View answer and explanation

Correct option: C

\frac{5050}{5051}

The correct answer is C (

\frac{5050}{5051}

) because by using integration by parts,

I_2

can be expressed in terms of

I_1

, leading to the given ratio.

699 Easy

699 of 995

Application +4 | -1

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

is equal to

View answer and explanation

Correct option: D

2 / \pi

Option D (

2 / \pi

) is correct because the integral of the absolute value of sin(2πx) from 0 to 1 is calculated by splitting the integral into two parts and evaluating each part separately.

700 Medium

700 of 995

Understanding +4 | -1

\int \frac{\log \left(t+\sqrt{1+t^2}\right)}{\sqrt{1+t^2}} d t=\frac{1}{2}(g(t))^2+C

, where

\mathrm{C}

is a constant, then

g(b)

is equal to :

View answer and explanation

Correct option: D

\log (2+\sqrt{5})

Option D (

\log (2+\sqrt{5})

) is correct. By substituting

u = \log(t + \sqrt{1+t^2})

, the integral simplifies, and g(t) can be found.

701 Easy

701 of 995

Application +4 | -1

\int\limits_{0}^a \frac{1}{1+4 x^2} d x=\frac{\pi}{8}

, then

a

equals

View answer and explanation

Correct option: B

\frac{\pi}{4}

Option D is incorrect. The correct answer is 1/2. The integral is evaluated using the arctangent function, and solving for 'a' gives 1/2.

702 Easy

702 of 995

Analyzing +4 | -1

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

is equal to

View answer and explanation

Correct option: A

- \,\left( {\frac{\mathrm{\pi }}{2}} \right)\,\,\log \,\,2

Option A is correct because by using properties of definite integrals and trigonometric identities, the integral can be simplified and evaluated to

- \left( {\frac{\mathrm{\pi }}{2}} \right)\,\log \,2

703 Easy

703 of 995

Application +4 | -1

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

is equal to

View answer and explanation

Correct option: A

Option A is correct because the integral is solved using trigonometric identities and properties of definite integrals.

704 Easy

704 of 995

Application +4 | -1

The value of the integral

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

, is

View answer and explanation

Correct option: B

\frac{3}{2}

Option B (

\frac{3}{2}

) is correct because the integral of

x[x]

from 0 to 2, where

[x]

is the greatest integer function, is

\frac{3}{2}

705 Easy

705 of 995

Application +4 | -1

The integral

\int_1^2 {{e^x} \cdot {x^x}\left( {2 + {{\log }_e}x} \right)dx}

equals

View answer and explanation

Correct option: C

e\left( {4e - 1} \right)

Option C (

e(4e - 1)

) is correct because it is the result of evaluating the definite integral after applying the substitution method.

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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

706 Medium

706 of 995

Application +4 | -1

Let

F(x)

be the area bounded by the curve

f(t)=\frac{e^t}{t}

between

t=a(a > 1), t=x

and axis of abscissa then the area bounded by

g(t)=\frac{e^t}{t+t_0}\left(t_0 > 0\right)

between

t=a

t=x

and axis of abscissa is

View answer and explanation

Correct option: C

e^{-t_0}\left[F\left(x+t_0\right)-F\left(a+t_0\right)\right]

Option C (

e^{-t_0}[F(x+t_0)-F(a+t_0)]

) is correct because it represents the area calculated using a substitution and the given function F(x).

707 Easy

707 of 995

Application +4 | -1

\phi(x)=\int\limits_{{1 / x}}^{\sqrt{x}} \sin \left(t^2\right) \, d t

, then

\phi^{\prime}(1)

is equal to

View answer and explanation

Correct option: C

(3 / 2) \sin 1

Option C (

(3 / 2) \sin 1

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

708 Easy

708 of 995

Understanding +4 | -1

Statement-I :

\int e^x \sin x d x=\frac{e^x}{2}(\sin x-\cos x)+c

.

Statement-II :

\int e^x\left(f(x)+f^{\prime}(x)\right) d x=e^x f(x)+c

View answer and explanation

Correct option: A

If both Statement - I and Statement - II are true, and Statement - II is the correct explanation of Statement- I.

Option A is correct because both statements are true, and Statement II correctly explains Statement I, as Statement I is a specific case of Statement II.

709 Easy

709 of 995

Application +4 | -1

Let

g(x)=\int_0^x f(t) d t

, where

f

is such that

\frac{1}{2} \leq f(t) \leq 1

for

t \in[0,1]

and,

0 \leq f(t) \leq \frac{1}{2}

for

t \in[1,2]

Then,

g(2)

satisfies the inequality,

View answer and explanation

Correct option: B

0 \le g(2) < 2

Option B (

0 \le g(2) < 2

) is correct because it is the range that

g(2)

satisfies based on the given inequalities for

f(t)

710 Easy

710 of 995

Understanding +4 | -1

f(x)

is a periodic function, with period

T

, then

View answer and explanation

Correct option: C

\int\limits_{a}^b f(x) d x=\int\limits_{{a+T}}^{b+T} f(x) d x

Option C is correct because shifting the integration limits by the period T of a periodic function does not change the value of the definite integral.

711 Medium

711 of 995

Understanding +4 | -1

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

712 Hard

712 of 995

Application +4 | -1

\{x\}

denotes the fractional part of

x

, then

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

is equal to

View answer and explanation

Correct option: B

\frac{1}{2}\{x\}(\{x\}-1)

Option B (

\frac{1}{2}\{x\}(\{x\}-1)

) is correct because it represents the general form of the integral of the fractional part of x minus 1/2 from 0 to x.

713 Medium

713 of 995

Understanding +4 | -1

n \in \mathbf{N}

, the value of

\int\limits_{0}^n[x] \mathrm{d} x

(where

[x]

is the greatest integer function) is

View answer and explanation

Correct option: B

\frac{n(n-1)}{2}

Option B (

\frac{{n(n-1)}}{{2}}

) is correct because the integral of the greatest integer function from 0 to n can be expressed as a sum of integers from 0 to n-1.

714 Medium

714 of 995

Application +4 | -1

The volume of the solid generated by revolving the region bounded by

y = {x^2} + 1

and

y = 2x + 1

about

x -

axis is

View answer and explanation

Correct option: A

\frac{{104\mathrm{\pi }}}{{15}}\mathrm{\;cu\;units\;}

Option A (

\frac{{104\mathrm{\pi }}}{{15}}\mathrm{\;cu\;units\;}

) is correct because it represents the volume of the solid generated by revolving the region between the two curves about the x-axis, calculated using the disk method.

715 Medium

715 of 995

Application +4 | -1

Let

f\left( x \right)

be a non-negative continuous function such that the area bounded by the curve

y = f\left( x \right),\;x

-axis and the ordinates

x = \frac{\mathrm{\pi }}{4}\mathrm{\;}

and

x = \mathrm{\beta } > \frac{\mathrm{\pi }}{4}\,\mathrm{is}\,\mathrm{\pi }

\left( {\mathrm{\beta }\sin \mathrm{\beta } + \frac{\mathrm{\pi }}{4}\cos \mathrm{\beta } + \sqrt 2 \mathrm{\beta }} \right)\;\mathrm{the\;}f\left( {\frac{\mathrm{\pi }}{2}} \right)\mathrm{s}

View answer and explanation

Correct option: A

\left( {1 - \frac{\mathrm{\pi }}{4} + \sqrt 2 } \right)

Option A (

\left( {1 - \frac{\mathrm{\pi }}{4} + \sqrt 2 } \right)

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

\beta = \frac{{\pi }}{2}

716 Easy

716 of 995

Application +4 | -1

The value of

\int \nolimits_2^4 \,\,\left\{ {\left| {x - 2} \right|\,\, + \,\,\left| {x - 3} \right|} \right\}dx\,\,\mathrm{is}

View answer and explanation

Correct option: C

\mathrm{3}

Option C (3) is correct because the integral is evaluated by splitting the interval based on the absolute value functions and then integrating.

717 Easy

717 of 995

Application +4 | -1

\int \,\,\frac{1}{{1\,\, + \,\,\cos \,\,ax}}\,\,\;dx

is equal to

View answer and explanation

Correct option: B

\frac{1}{a}\,\,\tan \,\,\frac{{ax}}{2}\,\, + \,\,c

Option B,

\frac{1}{a}\,\,\tan \,\,\frac{{ax}}{2}\,\, + \,\,c

, is correct because using trigonometric identities, the integral can be simplified and solved, resulting in this expression.

718 Easy

718 of 995

Application +4 | -1

f\left( y \right) = {e^y},\,\,\;g\left( y \right) = y;\;\;\,\;y > 0

and

F\left( t \right) = \int \nolimits_0^t \,\,f\left( {t - y} \right)\,\,g\,\,\left( y \right)\;\,\,dy

, then

View answer and explanation

Correct option: B

F\,\,\left( t \right) = {e^t} - \,\,\left( {1 + t} \right)

Option B (

F\,\left( t \right) = {e^t} - \,\left( {1 + t} \right)

) is correct because it is the result of evaluating the convolution integral using integration by parts.

719 Easy

719 of 995

Application +4 | -1

\frac{d}{d x} G(x)=\frac{e^{\tan x}}{x}, x \in(0, \pi / 2)

, then

\int\limits_{{1 / 4}}^{1 / 2} \frac{2}{x} \cdot e^{\tan \left(\pi x^2\right)} d x

is equal to

View answer and explanation

Correct option: A

G(\pi / 4)-G(\pi / 16)

Option A (

G(\pi / 4)-G(\pi / 16)

) is correct because after substituting and changing variables, the integral can be expressed in terms of G(x) and evaluated at the limits.

720 Easy

720 of 995

Application +4 | -1

\int \nolimits_0^{1.5} \,\,\left[ {{x^2}} \right]\,\,dx

View answer and explanation

Correct option: C

2\,\, - \,\,\sqrt 2

Option C (

2\,\, - \,\,\sqrt 2

) is correct because the integral of the greatest integer function of x² from 0 to 1.5 is calculated by splitting the integral into intervals where the greatest integer function has a constant value.

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