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JEE • MATHEMATICS • DEFINITE INTEGRATION

Definite integration practice · Mixed difficulty

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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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451 Easy
451 of 995
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.
452 Easy
452 of 995
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.
453 Easy
453 of 995
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.
454 Easy
454 of 995
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).
455 Easy
455 of 995
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.
456 Medium
456 of 995
Application +4 | -1
The value of ∑n=1100∫n−1nex−[x]dx,\sum\limits_{n = 1}^{100} {\int_{n - 1}^n {{e^{x - \left[ x \right]}}dx,} } where [x]\left[ x \right] is the greatest integer ≤x,\le x, is
View answer and explanation
Correct option: A
100 (e−1)100\,(e - 1)
The correct answer is A (100 (e−1)100\,(e - 1)) because it correctly evaluates the integral and uses the properties of the greatest integer function.
457 Easy
457 of 995
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.
458 Easy
458 of 995
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.
459 Easy
459 of 995
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.
460 Easy
460 of 995
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.
461 Easy
461 of 995
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.
462 Easy
462 of 995
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.
463 Easy
463 of 995
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 }}.
464 Medium
464 of 995
Application +4 | -1
lim⁡n→∞∑r=14n1n+r\lim \limits_{{n \rightarrow \infty}} \sum\limits_{{r=1}}^{4 n} \frac{1}{n+r} is
View answer and explanation
Correct option: A
log⁡e5\log _e 5
Option A (log⁡e5\log _e 5) is correct because the limit of the sum can be expressed as a definite integral, which evaluates to log⁡e5\log_e 5.
465 Easy
465 of 995
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.

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
466 Easy
466 of 995
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.
467 Medium
467 of 995
Application +4 | -1
∫0π/2xsin⁡xcos⁡xcos⁡4x+sin⁡4xdx=\int\limits_{0}^{\pi / 2} \frac{x \sin x \cos x}{\cos ^4 x+\sin ^4 x} d x=
View answer and explanation
Correct option: D
π216\frac{\pi^2}{16}
Option D (π216\frac{{\pi^2}}{{16}}) is correct because it is the result of evaluating the definite integral using trigonometric substitution and properties of definite integrals.
468 Easy
468 of 995
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.
469 Medium
469 of 995
Application +4 | -1
For m>0,n>0m>0, n>0, let Im,n=∫01xm(log⁡x)nI_{m, n}=\int\limits_{0}^1 x^m(\log x)^n dx\mathrm{d} x, then I5,5I_{5,5} is given by
View answer and explanation
Correct option: C
−5!66-\frac{5 !}{6^6}
Option C (−5!66-\frac{{5 !}}{{6^6}}) is correct because the integral can be solved by repeated integration by parts.
470 Easy
470 of 995
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).
471 Medium
471 of 995
Analyzing +4 | -1
∫0∞  x  log⁡  x    dx(1+x2)2\int \nolimits_0^\infty \,\,\frac{{x\,\,\log \,\,x\;\,\,dx}}{{{{\left( {1 + {x^2}} \right)}^2}}} is equal to
View answer and explanation
Correct option: A
0\mathrm{0}
Option A (0) is correct because the integral evaluates to zero due to the symmetry of the integrand after a suitable substitution.
472 Easy
472 of 995
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.
473 Medium
473 of 995
Application +4 | -1
Let α>0\alpha > 0. If ∫0αxx+α−xdx=16+20215\int\limits_{{0}}^{\alpha} \frac{x}{\sqrt{x+\alpha}-\sqrt{x}} d x=\frac{16+20 \sqrt{2}}{15}, then α\alpha is equal to:
View answer and explanation
Correct option: C
2\sqrt{2}
Option C (2\sqrt{2}) is correct because it involves solving a definite integral after rationalizing the denominator and then comparing the result with the given value to find α\alpha.
474 Medium
474 of 995
Application +4 | -1
∫04{x}\int\limits_{0}^4\{\sqrt{x}\} is equal to, where {x}\{x\} denotes the fraction part of xx.
View answer and explanation
Correct option: D
73\frac{7}{3}
Option D (73\frac{7}{3}) is correct because the integral of the fractional part of the square root of x from 0 to 4 is equal to 73\frac{7}{3}.
475 Medium
475 of 995
Application +4 | -1
∫01/2  ∣ sin⁡  π  x∣  dx\int \limits_0^{1/2} \,\,\left| {\,\sin \,\,\pi \,\,x} \right|\,\,dx is equal to
View answer and explanation
Correct option: D
1  /  π1\,\,/\,\,\pi
Option D (1/π1/\pi) is correct because the integral is evaluated using substitution and the properties of the sine function.
476 Easy
476 of 995
Application +4 | -1
The value of I=∫−π/2π/2   ∣  sin⁡  x∣  dx  isI = \int \nolimits_{ - \pi /2}^{\pi /2} \,\,\,|\,\,\sin \,\,x|\,\,dx\,\,is
View answer and explanation
Correct option: B
2\mathrm{2}
Option B (2) is correct because the integral of |sin x| from -π/2 to π/2 is equal to 2.
477 Medium
477 of 995
Application +4 | -1
∫0πxlog⁡sin⁡xdx=\int\limits_{0}^\pi x \log \sin x d x=
View answer and explanation
Correct option: B
π22log⁡12\frac{\pi^2}{2} \log \frac{1}{2}
Option B (π22log⁡12\frac{\pi^2}{2} \log \frac{1}{2}) is correct because the integral can be solved using the properties of definite integrals and trigonometric functions, specifically exploiting the symmetry of the sine function.
478 Easy
478 of 995
Application +4 | -1
If In=∫0π/4tan⁡n  θ    dθ{I_n} = \int \nolimits_0^{\pi /4} {\tan ^n}\,\,\mathrm{\theta }\;\,\,d\mathrm{\theta }, then I8+I6{I_8} + {I_6} is equal to
View answer and explanation
Correct option: A
17\frac{1}{7}
The correct answer is A (17\frac{1}{7}) because after applying the reduction formula and evaluating the definite integral, the result is 17\frac{1}{7}.
479 Easy
479 of 995
Analyzing +4 | -1
The value of [∫0sin⁡2θsin⁡−1  ϕ dϕ  +  ∫0cos⁡2θ  cos⁡−1  ϕ dϕ]\left[ {\int \nolimits_0^{{{\sin }^2}\theta } {{\sin }^{ - 1}}\,\,\sqrt {\phi \,} d\phi \,\, + \,\,\int \nolimits_0^{{{\cos }^2}\theta } \,\,{{\cos }^{ - 1}}\,\,\sqrt {\phi \,} d\phi } \right] is equal to
View answer and explanation
Correct option: D
π/4\pi /4
Option D (π/4\pi/4) is correct because the derivative of the function inside the brackets is zero, making the function constant. Evaluating at θ=π/4\theta = \pi/4 gives π/4\pi/4.
480 Easy
480 of 995
Understanding +4 | -1
The value of ∫−11  x  ∣ x ∣  dx    is\int \nolimits_{ - 1}^1 \,\,x\,\,\left| {\,x\,} \right|\,\,dx\,\,\;\mathrm{is} is
View answer and explanation
Correct option: C
0\mathrm{0}
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.
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