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JEE • MATHEMATICS • DEFINITE INTEGRATION

Definite integration practice · Easy

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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.

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Showing 781-791 of 791 questions
781 Easy
781 of 791
Analyzing +4 | -1
If f(x)f(x) is a continuous function in [0,π][0, \pi] such that f(0)=f(π)=0f(0)=f(\pi)=0, then the value of ∫0π/2{f(2x)+f′′(2x)}sin⁡xcos⁡xdx\int_0^{\pi / 2}\left\{f(2 x)+f^{\prime \prime}(2 x)\right\} \sin x \cos x d x is equal to
View answer and explanation
Correct option: D
0
Option D (0) is correct because integration by parts simplifies the integral to zero, utilizing the given boundary conditions.
782 Easy
782 of 791
Application +4 | -1
The value of the integral ∫−π/2π/2  cos⁡  x  −  cos⁡3  x dx\int \nolimits_{ - \pi /2}^{\pi /2} \,\,\sqrt {\cos \,\,x\,\, - \,\,{{\cos }^3}\,\,x} \,dx is
View answer and explanation
Correct option: B
43\frac{4}{3}
Option B (43\frac{4}{3}) is correct because the integral simplifies to an integral of sin(x)sqrt(cos(x)) which can be solved by substitution.
783 Easy
783 of 791
Application +4 | -1
Area bounded by curve xy=c,xx y=c, x-axis between x=1x=1 and x=4x=4, is
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Correct option: A
clog⁡3c \log 3 sq. unit
Option C (2clog⁡22 c \log 2 sq. unit) is correct because the area is calculated by integrating y=cxy = \frac{{c}}{{x}} from x=1x=1 to x=4x=4, which simplifies to clog⁡4=2clog⁡2c \log 4 = 2c \log 2.
784 Easy
784 of 791
Application +4 | -1
The value of the integral ∫e−1e2∣log⁡exx∣dx\int\limits_ {e^{-1}}^{e^2}\left|\frac{\log _e x}{x}\right| d x, is
View answer and explanation
Correct option: B
52\frac{5}{2}
Option B (52\frac{5}{2}) is correct because the integral is evaluated by splitting it into two parts based on the sign of the logarithm and then integrating.
785 Easy
785 of 791
Application +4 | -1
The value of ∫−23∣1−x2∣dx\int_{-2}^3\left|1-x^2\right| d x, is
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Correct option: D
283\frac{28}{3}
Option D (28/3) is correct because the integral of the absolute value function |1-x2| is computed by splitting the integral into intervals where the function inside the absolute value is positive or negative, and then integrating each part separately.
786 Easy
786 of 791
Application +4 | -1
The value of the integral ∫0π/4sin⁡x+cos⁡x3+sin⁡2x dx\int\limits_{0}^{\pi / 4} \frac{\sin x+\cos x}{3+\sin 2 x} \mathrm{~d} x is
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Correct option: C
(1/4)log⁡3(1 / 4) \log 3
Option D ((1/8)log⁡3(1 / 8) \log 3) is correct because it is the result of evaluating the definite integral after a suitable substitution.
787 Easy
787 of 791
Application +4 | -1
The value of the integral I=∫01x(1−x)ndxI=\int\limits_{0}^1 x(1-x)^n d x is
View answer and explanation
Correct option: C
1n+1−1n+2\frac{1}{n+1}-\frac{1}{n+2}
Option C (1n+1−1n+2\frac{1}{n+1}-\frac{1}{n+2}) is correct because it is the result of integrating by parts and using the properties of definite integrals.
788 Easy
788 of 791
Analyzing +4 | -1
The absolute value of ∫1019cos⁡x1+x8dx\int\limits_{{10}}^{19} \frac{\cos x}{1+x^8} d x, is
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Correct option: A
less than 10−710^{-7}
Option A (less than 10−710^{-7}) is correct because the absolute value of the integral is bounded by the integral of the absolute value of the integrand, which is less than 10−710^{-7}.
789 Easy
789 of 791
Application +4 | -1
The value I=∫0π/211+cot⁡xdxI=\int\limits_{0}^{\pi / 2} \frac{1}{1+\cot x} d x is
View answer and explanation
Correct option: C
π/4\pi / 4
Option C (π/4\pi/4) is correct because using the property ∫0af(x)dx=∫0af(a−x)dx\int_0^a f(x) dx = \int_0^a f(a-x) dx simplifies the integral to π/4\pi/4.
790 Easy
790 of 791
Application +4 | -1
If ∫0∞ e−x2  dx=π2\int \limits_0^\infty \,{e^{ - {x^2}}}\,\,dx = \sqrt {\frac{\pi }{2}} then ∫0∞  e−ax2  dx,    a>0\int \limits_0^\infty \,\,{e^{ - a{x^2}}}\,\,dx,\,\,\;a > 0 is

View answer and explanation
Correct option: D
12  πa\frac{1}{2}\,\,\frac{{\sqrt \pi }}{a}
Option D (12 πa\frac{1}{2}\,\frac{{\sqrt {\pi } }}{a}) is correct because it follows from a u-substitution and the given integral value.
791 Easy
791 of 791
Application +4 | -1
lim⁡n→∞  {1na  +  1na+1  +  1na+2  +  …+  1nb}\lim\limits_{n \to \infty } \,\,\left\{ {\frac{1}{{na}}\,\, + \,\,\frac{1}{{na + 1}}\,\, + \,\,\frac{1}{{na + 2}}\,\, + \,\, \ldots + \,\,\frac{1}{{nb}}} \right\} is equal to
View answer and explanation
Correct option: A
log⁡  (ba)\log \,\,\left( {\frac{b}{a}} \right)
Option A (log⁡  (ba)\log \,\,\left( {\frac{b}{a}} \right)) is correct because the limit of the Riemann sum is correctly converted to a definite integral and evaluated.
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