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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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331 Easy
331 of 791
Application +4 | -1
Let f(x)={x}f(x)=\{x\}, the fractional part of xx then ∫−11f(x)dx\int\limits_{{-1}}^1 f(x) d x is equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because the integral of the fractional part function over the interval [-1, 1] can be calculated by splitting the integral and using the definition of the fractional part function.
332 Easy
332 of 791
Application +4 | -1
The area bounded by the y−y -axis, y=cos⁡xy = \cos x and y=sin⁡x,0≤x≤π/4y = \sin x,0 \le x \le \pi /4 is
View answer and explanation
Correct option: B
2−1\sqrt 2 - 1
Option B (2−1\sqrt 2 - 1) is correct because it represents the definite integral of the difference between cos⁡x\cos x and sin⁡x\sin x from 0 to π4\frac{\pi}{4}.
333 Easy
333 of 791
Application +4 | -1
If a<∫02π110+3cos⁡xdxdx<ba<\int\limits_{0}^{2 \pi} \frac{1}{10+3 \cos x} d x d x < b, then the ordered pair (a,b)(a, b) is
View answer and explanation
Correct option: B
(2π13,2π7)\left(\frac{2 \pi}{13}, \frac{2 \pi}{7}\right)
Option B ((2π13,2π7)\left(\frac{{2 \pi}}{{13}}, \frac{{2 \pi}}{{7}}\right)) is correct because it correctly bounds the integral based on the minimum and maximum values of the integrand.
334 Easy
334 of 791
Application +4 | -1
∫−22∣1−x2∣dx=\int\limits_{{-2}}^2\left|1-x^2\right| d x=
View answer and explanation
Correct option: B
4
Option B (4) is correct because the integral of the absolute value function is evaluated by splitting the integral into intervals where the function inside the absolute value is positive or negative.
335 Easy
335 of 791
Application +4 | -1
The value of the integral ∫0100sin⁡  (x  −  [x]) π dx\int \limits_0^{100} \sin \,\,\left( {x\,\, - \,\,\left[ x \right]} \right)\,\pi \,dx, is
View answer and explanation
Correct option: B
200/π200/\pi
Option B is correct because the integral can be simplified using the periodic property of the function, leading to the result 200/π\pi.
336 Easy
336 of 791
Application +4 | -1
∫−π/2π/2  cos⁡  x1+  ex  dx=\int \limits_{ - \pi /2}^{\pi /2} \,\,\frac{{\cos \,\,x}}{{1 + \,\,{e^x}}}\,\,dx =
View answer and explanation
Correct option: A
1\mathrm{1}
Option A (1) is correct because the integral can be simplified by using the property of definite integrals and substituting x = -t.
337 Easy
337 of 791
Analyzing +4 | -1
Let f(x)f(x) be an odd continuous function which is periodic with period 2. If g(x)=∫0xf(t)dtg(x)=\int_0^x f(t) d t, then
View answer and explanation
Correct option: C
g(2n)=0g(2 n)=0 for all n∈Nn \in N
Option C (g(2n)=0g(2 n)=0 for all n∈Nn \in N) is correct because g(x) is periodic with period 2 and g(2) = 0, implying g(2n) = 0 for all n in N.
338 Easy
338 of 791
Analyzing +4 | -1
If f(a+b+1−x)=f(x)f(a + b + 1 - x) = f(x), for all x, where a and b are fixed positive real numbers, then 1a+b ∫abx(f(x)+f(x+1))dx\frac{1}{{a + b}}\,\int_a^b {x(f(x) + f(x + 1))dx} is equal to
View answer and explanation
Correct option: C
∫a−1b−1f(x+1)dx\int_{a - 1}^{b - 1} {f(x + 1)dx}
Option D (∫a−1b−1f(x+1)dx\int_{a - 1}^{b - 1} {f(x + 1)dx}) is correct because by using the given property and applying integration properties, the integral simplifies to the given form.
339 Easy
339 of 791
Application +4 | -1
∫  1(a2+x2)3/2  dx\int \,\,\frac{1}{{{{\left( {{a^2} + {x^2}} \right)}^{3/2}}}}\,\,dx is equal to
View answer and explanation
Correct option: A
xa2 a2  +  x2  +  C\frac{x}{{{a^2}\,\sqrt {{a^2}\,\, + \,\,{x^2}} }}\,\, + \,\,C
Option A (xa2 a2 + x2 + C\frac{x}{{{a^2}\,\sqrt {{a^2}\, + \,{x^2}} }}\, + \,C) is correct because it results from a trigonometric substitution, specifically x=atan⁡θx = a \tan \theta.
340 Easy
340 of 791
Understanding +4 | -1
If f(t)f\left( t \right) is an odd function, then ∫0x  f(t)  dt\int \nolimits_0^x \,\,f\left( t \right)\,\,dt is
View answer and explanation
Correct option: B
An even function
Option B (An even function) is correct because the integral of an odd function is an even function.
341 Easy
341 of 791
Application +4 | -1
If ∫2x2x3(ln⁡x)f(t)dt=x2−2x+5\int_{2 x^2}^{x^3}(\ln x) f(t) d t=x^2-2 x+5, then f(8)=f(8)=
View answer and explanation
Correct option: B
12ln⁡2\frac{1}{2 \ln 2}
The correct answer is B (12ln⁡2\frac{1}{2 \ln 2}) because after differentiating both sides and substituting x=2, we can solve for f(8).
342 Easy
342 of 791
Application +4 | -1
If for every integer n∫nn+1  f(x)  dx=n2n\int \limits_n^{n + 1} \,\,f\left( x \right)\,\,dx = {n^2}, then the value of ∫−24 f(x)  dx\int \limits_{ - 2}^4 \,f\left( x \right)\,\,dx, is
View answer and explanation
Correct option: C
19\mathrm{19}
Option C (19) is correct because the integral from -2 to 4 can be broken down into a sum of integrals over intervals of length 1, each of which is given by n².
343 Easy
343 of 791
Application +4 | -1
∫01x13/2(4−x5)3/2dx\int\limits_{0}^1 \frac{x^{13 / 2}}{\left(4-x^5\right)^{3 / 2}} d x
View answer and explanation
Correct option: D
25[13−π6]\frac{2}{5}\left[\frac{1}{\sqrt{3}}-\frac{\pi}{6}\right]
Option D (25[13−π6]\frac{2}{5}\left[\frac{1}{\sqrt{3}}-\frac{\pi}{6}\right]) is correct because after substitution and integration, the definite integral evaluates to this value.
344 Easy
344 of 791
Application +4 | -1
A line tangent to the graph of the function y=f(x)y=f(x) at the point x=ax=a forms an π/3\pi / 3 with yy-axis and at x=bx=b an angle π/4\pi / 4 with xx-axis then ∫abf′′(x)dx\int\limits_{a}^b f^{\prime \prime}(x) d x is
View answer and explanation
Correct option: A
1−131-\frac{1}{\sqrt{3}}
Option A (1−131-\frac{{1}}{{\sqrt{{3}}}}) is correct because it represents the difference between the derivatives at points b and a, calculated using the tangent angles.
345 Easy
345 of 791
Application +4 | -1
If for a real number y1[y]y_1[y] is the greatest integer less than or equal to yy, then the value of the integral ∫π/23π/2[2sin⁡x]dx\int\limits_{{\pi / 2}}^{3 \pi / 2}[2 \sin x] d x, is
View answer and explanation
Correct option: C
−π/2-\pi / 2
Option C (−π/2-\pi / 2) is correct because the integral of the greatest integer function of 2sin⁡x2\sin x from π/2\pi/2 to 3π/23\pi/2 evaluates to −π/2-\pi/2.

Explore Other Chapters in MATHEMATICS

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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
346 Easy
346 of 791
Application +4 | -1
∫0∞[2ex]dx\int_0^{\infty}\left[\frac{2}{e^x}\right] d x (where [.] denotes the greatest integer function) equals
View answer and explanation
Correct option: A
log⁡e2\log _e 2
Option A (log⁡e2\log _e 2) is correct because the greatest integer function simplifies the integral to a simple logarithmic value.
347 Easy
347 of 791
Application +4 | -1
∫−12∣x∣dx\int\limits_{{-1}}^2|x| d x
View answer and explanation
Correct option: A
5/25 / 2
Option A (5/25 / 2) is correct because the integral of ∣x∣|x| from -1 to 2 is equal to 5/25/2.
348 Easy
348 of 791
Application +4 | -1
∫01000  ex−[x]  dx    is\int \nolimits_0^{1000} \,\,{e^{x - \left[ x \right]}}\,\,dx\,\,\;\mathrm{is}
View answer and explanation
Correct option: C
1000  (e  −  1)1000\,\,\left( {e\,\, - \,\,1} \right)
Option C is correct because the integral is evaluated using the periodic property of the function.
349 Easy
349 of 791
Understanding +4 | -1
The value of ∫02πcos⁡99  x dx\int \limits_0^{2\pi } {\cos ^{99}}\,\,x\,dx, is
View answer and explanation
Correct option: D
0\mathrm{0}
Option D (0) is correct because the integral of cos99(x) from 0 to 2pi evaluates to zero due to the properties of even and odd functions over symmetric intervals.
350 Easy
350 of 791
Application +4 | -1
If f′(x)=dx(1+x2)3/2  and    f(0)=0,   then   f(1)f'\left( x \right) = \frac{{dx}}{{{{\left( {1 + {x^2}} \right)}^{3/2}}}}\,\,\mathrm{and}\;\,\,f\left( 0 \right) = 0,\;\,\mathrm{then}\;\,f\left( 1 \right) is equal to

View answer and explanation
Correct option: C
12\frac{1}{{\sqrt 2 }}
Option C (12\frac{1}{{\sqrt 2 }}) is correct because integrating f′(x)f'(x) and using the initial condition f(0)=0f(0) = 0, we find f(1)=12f(1) = \frac{1}{{\sqrt 2 }}.
351 Easy
351 of 791
Application +4 | -1
The value of ∫016π/3∣sin⁡x∣dx\int\limits_{0}^{16 \pi / 3}|\sin x| d x is
View answer and explanation
Correct option: C
21/221 / 2
Option D (21 / 2) is correct because it is the result of integrating the absolute value of sin(x) over the given interval.
352 Easy
352 of 791
Application +4 | -1
If ∫0πxf(sin⁡x)dx=A∫0π/2f(sin⁡x)dx\int\limits_{0}^\pi x f(\sin x) d x=A \int\limits_{0}^{\pi / 2} f(\sin x) d x, then AA is
View answer and explanation
Correct option: B
143\frac{14}{3}
The correct answer is not among the options provided. The solution should lead to A = π\pi.
353 Easy
353 of 791
Analyzing +4 | -1
Let f(x)=2+∣x∣−∣x−1∣+∣x+1∣,x∈Rf(x)=2+|x|-|x-1|+|x+1|, x \in R. Consider
(S1):f′(−32)+f′(−12)+f′(12)+f′(32)=2(S 1): f^{\prime}\left(-\frac{3}{2}\right)+f^{\prime}\left(-\frac{1}{2}\right)+f^{\prime}\left(\frac{1}{2}\right)+f^{\prime}\left(\frac{3}{2}\right)=2

(S2):∫−22f(x)dx=12(S 2): \int\limits_{ - 2}^2 {f\left( x \right)dx = 12}

Then,
View answer and explanation
Correct option: D
Only (S2) is correct
Only (S2) is correct because the integral of the piecewise function f(x) from -2 to 2 equals 12, while the sum of the derivatives at the specified points equals 4, not 2.
354 Easy
354 of 791
Application +4 | -1
The value of ∫0[x]2x2[x]dx\int\limits_{0}^{[x]} \frac{2^x}{2^{[x]}} d x, is
View answer and explanation
Correct option: B
[x]log⁡2\frac{[x]}{\log 2}
Option B ([x]log⁡2\frac{[x]}{\log 2}) is correct because the integral simplifies to this expression using the properties of periodic functions and the greatest integer function.
355 Easy
355 of 791
Application +4 | -1
The value of ∫−π/2π/2sin⁡2xdx1+ex\int_{-\pi / 2}^{\pi / 2} \frac{\sin ^2 x d x}{1+e^x} is
View answer and explanation
Correct option: B
π4\frac{\pi}{4}
Option B is correct because using the property of definite integrals and symmetry, the integral simplifies to an integral of sin2(x) from 0 to pi/2, which evaluates to pi/4.
356 Easy
356 of 791
Application +4 | -1
If f′(x)=g(x)f^{\prime}(x)=g(x) for a≤x≤ba \leq x \leq b then ∫abf(x)g(x)dx\int\limits_{a}^b f(x) g(x) \mathrm{d} x is equal to
View answer and explanation
Correct option: C
(f(b))2−(f(a))22\frac{(f(b))^2-(f(a))^2}{2}
Option C ((f(b))2−(f(a))22\frac{(f(b))^2-(f(a))^2}{2}) is correct because it results from direct integration using the substitution g(x) = f'(x).
357 Easy
357 of 791
Application +4 | -1
Let f:R→Rf: R \rightarrow R defined by f(x)=cos⁡x+xf(x)=\cos x+x if ∫0πf−1(x)dx=π22−k\int\limits_{0}^\pi f^{-1}(x) d x=\frac{\pi^2}{2}-k, then the value of kk is ____.
View answer and explanation
Correct option: A
0
Option A (0) is correct because, after applying integration by parts and simplifying, the value of k is found to be 0.
358 Easy
358 of 791
Application +4 | -1
∫π/3π/2  cosec2  x  dx\int \nolimits_{\pi /3}^{\pi /2} \,\,\mathrm{cose}{\mathrm{c}^2}\,\,x\,\,dx is equal to
View answer and explanation
Correct option: B
1\mathrm{1}
Option B (1\mathrm{1}) is correct because it is the result of evaluating the definite integral of cosec2(x)cosec^2(x) from π/3\pi/3 to π/2\pi/2.
359 Easy
359 of 791
Application +4 | -1
The value of ∫exdx(sin⁡ex+e−xcos⁡ex)2\int \frac{e^x d x}{\left(\sin e^x+e^{-x} \cos e^x\right)^2} is equal to
View answer and explanation
Correct option: A
tan⁡(ex−tan⁡−1ex)+C\tan \left(e^x-\tan ^{-1} e^x\right)+C
Option A (tan⁡(ex−tan⁡−1ex)+C\tan \left(e^x-\tan ^{-1} e^x\right)+C) is correct because it is the result of integrating the given expression using substitution and trigonometric identities.
360 Easy
360 of 791
Application +4 | -1
∫0πsin⁡2xdx\int_0^\pi \sin ^2 x d x is equal to
View answer and explanation
Correct option: B
π2\frac{\pi}{2}
Option B (π2\frac{\pi}{2}) is correct because the integral of sin⁡2x\sin^2 x from 0 to π\pi is equal to π2\frac{\pi}{2}.
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