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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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61 Medium
61 of 995
Application +4 | -1
The area bounded by the parabola y2=4xy^2=4 x and the line 2x−3y+4=02 x-3 y+4=0, in square unit, is
View answer and explanation
Correct option: B
13\frac{1}{3}
Option B (1/3) is correct because the area bounded by the parabola and the line is calculated by integrating the difference between the two functions.
62 Medium
62 of 995
Application +4 | -1
∫0π x  sin⁡  x  cos⁡4x dx=\int \limits_0^\pi \,x\,\,\sin \,\,x\,\,{\cos ^4}x\,dx =
View answer and explanation
Correct option: B
π5\frac{\pi }{{5}}
Option B (π5\frac{\pi }{{5}}) is correct. The integral is solved by using the property ∫0af(x)dx=∫0af(a−x)dx\int \limits_0^a f(x) dx = \int \limits_0^a f(a-x) dx and then integrating by substitution.
63 Medium
63 of 995
Application +4 | -1
∫0πxsin⁡x1+cos⁡2xdx=\int\limits_{0}^\pi \frac{x \sin x}{1+\cos ^2 x} d x=
View answer and explanation
Correct option: A
π24\frac{\pi^2}{4}
Option A (π24\frac{\pi^2}{4}) is correct because the integral is solved using the property of definite integrals and trigonometric substitution.
64 Easy
64 of 995
Application +4 | -1
∫0π/4sin⁡  (x  −  [x])  dx\int \limits_0^{\pi /4} \sin \,\,\left( {x\,\, - \,\,\left[ x \right]} \right)\,\,dx is equal to
View answer and explanation
Correct option: B
1−121 - \frac{1}{{\sqrt 2 }}
Option B (1 - 1/√2) is correct because the integral simplifies to the integral of sin(x) from 0 to π/4, which evaluates to 1 - 1/√2.
65 Medium
65 of 995
Analyzing +4 | -1
The value of ∫01  tan⁡−1  (2x−11+x−x2)  dx\int \nolimits_0^1 \,\,{\tan ^{ - 1}}\,\,\left( {\frac{{2x - 1}}{{1 + x - {x^2}}}} \right)\,\,dx is
View answer and explanation
Correct option: B
0\mathrm{0}
Option B (0) is correct because the integral simplifies to zero after applying properties of inverse trigonometric functions and definite integrals.
66 Easy
66 of 995
Application +4 | -1
∫0π    xdxa2  cos⁡2  x  +  b2  sin⁡2  x  is equal to\int \nolimits_0^\pi \,\,\,\,\frac{{x\mathrm{ }dx}}{{{a^2}\,\,{{\cos }^2}\,\,x\,\, + \,\,{b^2}\,\,{{\sin }^2}\,\,x}}\,\,\mathrm{is }\,\mathrm{equal }\,\mathrm{to}
View answer and explanation
Correct option: C
π22 ab\frac{{{\pi ^2}}}{{2\,ab}}
Option C (π22 ab\frac{{{\pi ^2}}}{{2\,ab}}) is correct because the integral is solved using properties of definite integrals and trigonometric substitutions.
67 Medium
67 of 995
Application +4 | -1
The value of 2∫sin⁡xdxsin⁡(x−π4)\sqrt{2} \int \frac{\sin x d x}{\sin \left(x-\frac{\pi}{4}\right)} is
View answer and explanation
Correct option: C
x+log⁡∣sin⁡(x−π4)∣+cx+\log \left|\sin \left(x-\frac{\pi}{4}\right)\right|+c
Option C (x+log⁡∣sin⁡(x−π4)∣+cx+\log |\sin (x-\frac{\pi}{4})|+c) is correct because after applying a substitution and simplifying, the integral evaluates to this expression.
68 Medium
68 of 995
Application +4 | -1
If 0<  a  <10 < \,\,a\,\, < 1, then ∫−11  11−2 ax+a2  dx\int \limits_{ - 1}^1 \,\,\frac{1}{{\sqrt {1 - 2\,ax + {a^2}} }}\,\,dx is equal to
View answer and explanation
Correct option: D
None of these
The correct answer is D (None of these) because the integral evaluates to 2, which is not among the provided options.
69 Medium
69 of 995
Application +4 | -1
lim⁡n→∞(1n+1+1n+2+⋯+12n)\lim \limits_{{n \rightarrow \infty}}\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n}\right) equals:
View answer and explanation
Correct option: A
ln⁡2\ln 2
Option A (ln⁡2\ln 2) is correct because the limit of the given sum can be expressed as a definite integral, which evaluates to ln⁡2\ln 2.
70 Easy
70 of 995
Application +4 | -1
lim⁡n→∞[11+n+12+n+13+n+…+12n]\lim \limits_{{n \rightarrow \infty}}\left[\frac{1}{1+n}+\frac{1}{2+n}+\frac{1}{3+n}+\ldots+\frac{1}{2 n}\right] is equal to
View answer and explanation
Correct option: C
log⁡e2\log _{e} 2
Option C (log⁡e2\log _{e} 2) is correct because the limit can be expressed as a definite integral which evaluates to log⁡e2\log_e 2.
71 Easy
71 of 995
Application +4 | -1
lim⁡n→∞{1n+1+1n+2+1n+3+…+1n+n},\lim\limits_{{n \rightarrow \infty}} \left\{\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\ldots+\frac{1}{n+n}\right\}, is equal to
View answer and explanation
Correct option: C
log⁡e2\log _e 2
Option C (log⁡e2\log _e 2) is correct because the limit of the sum can be expressed as a definite integral.
72 Medium
72 of 995
Application +4 | -1
If ∫  x  tan⁡−1x1+x2  dx=  1+x2  f(x)  +  A  log⁡  (x+x2+1)+C\int \,\,\frac{{x\,\,{{\tan }^{ - 1}}x}}{{\sqrt {1 + {x^2}} }}\,\,dx = \,\,\sqrt {1 + {x^2}} \,\,f\left( x \right)\,\, + \,\,A\,\,\log \,\,\left( {x + \sqrt {{x^2} + 1} } \right) + C, then
View answer and explanation
Correct option: C
f  (x)=2  tan⁡−1  x,  A=−1f\,\,\left( x \right) = 2\,\,{\tan ^{ - 1}}\,\,x,\,\,A = - 1
Option A is correct because after integrating by parts and simplifying, the resulting function f(x) and constant A match the values given in option A.
73 Easy
73 of 995
Application +4 | -1
The value of ∫032π/31+cos⁡2x dx\int\limits_{0}^{32 \pi / 3} \sqrt{1+\cos 2 x} \, d x, is
View answer and explanation
Correct option: B
44−32\frac{44-\sqrt{3}}{\sqrt{2}}
Option B (44−32\frac{44-\sqrt{3}}{\sqrt{2}}) is correct because the integral is evaluated by breaking it into intervals where cos(x) is either positive or negative.
74 Medium
74 of 995
Application +4 | -1
∫−3π/2−π/2[(x+π)3+cos⁡2(x+3π)]dx\int\limits_{{-3 \pi / 2}}^{-\pi / 2}\left[(x+\pi)^3+\cos ^2(x+3 \pi)\right] {dx} 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 simplifies using substitution and properties of even and odd functions.
75 Easy
75 of 995
Understanding +4 | -1
∫−1010  log⁡  (a  +  xa  −  x)  dx\int \nolimits_{ - 10}^{10} \,\,\log \,\,\left( {\frac{{a\,\, + \,\,x}}{{a\,\, - \,\,x}}} \right)\,\,dx is equal to
View answer and explanation
Correct option: A
0\mathrm{0}
Option A (0) is correct because the integrand is an odd function and the limits of integration are symmetric about the origin.
76 Medium
76 of 995
Analyzing +4 | -1
Let I1=∫aπ−a  xf  (sin⁡  x)    dx,    I2=∫aπ−a  f  (sin⁡  x)    dx{I_1} = \int \nolimits_a^{\pi - a} \,\,xf\,\,(\sin \,\,x)\;\,\,dx,\;\,\,{I_2} = \int \nolimits_a^{\pi - a} \,\,f\;(\sin \,\,x)\;\,\,dx, then I2{I_2} is equal to

View answer and explanation
Correct option: C
2π  I1\frac{2}{\mathrm{\pi }}\,\,{I_1}
Option C (2π\I1\frac{2}{\mathrm{\pi }}\I_1) is correct because it is derived by using the property of definite integrals and solving for I2 in terms of I1.
77 Medium
77 of 995
Analyzing +4 | -1
The value of ∫0π  (∑r=03  ar cos⁡3−r  x  sin⁡r  x)dx\int \nolimits_0^\pi \,\,\left( {\sum \nolimits_{r = 0}^3 \,\,{a_r}\,{{\cos }^{3 - r}}\,\,x\,\,{{\sin }^r}\,\,x} \right)dx depends upon
View answer and explanation
Correct option: D
a1{a_1} and a3{a_3}
Option D (a1{a_1} and a3{a_3}) is correct because only the integrals involving cos⁡2xsin⁡x\cos^2 x \sin x and sin⁡3x\sin^3 x are non-zero over the interval [0,π][0, \pi].
78 Medium
78 of 995
Understanding +4 | -1
Consider I=∫01dx1+x2I=\int\limits_{0}^1 \frac{d x}{1+x^2} Then II satisfies
View answer and explanation
Correct option: C
I<1{I} < 1
Option C (I<1{I} < 1) is correct because the integrand 11+x2\frac{{1}}{{1+x^2}} is less than 1 over the interval [0, 1], thus the integral must be less than 1.
79 Medium
79 of 995
Application +4 | -1
Let f:R→Rf: R \rightarrow R defined as f(x)=x+sin⁡xf(x)=x+\sin x then value of ∫02πf−1(x)dx=\int\limits_{0}^{2 \pi} f^{-1}(x) d x=
View answer and explanation
Correct option: A
2π22 \pi^2
Option D ((x-4)²+(y-5)²=25) is incorrect. Option D is incorrect because the incentre is (1,1) which does not lie on the circle (x-4)²+(y-5)²=25. The incentre of the triangle formed by the lines xy=0 and x/3+y/4=1 is (1,1).
80 Easy
80 of 995
Application +4 | -1
If ∫0π/2cos⁡xdx6−5sin⁡x+sin⁡2x=log⁡K\int\limits_{0}^{\pi / 2} \frac{\cos x d x}{6-5 \sin x+\sin ^2 x}=\log K then KK is equal to
View answer and explanation
Correct option: B
43\frac{4}{3}
Option B (4/3) is correct because the integral evaluates to log(4/3), thus K = 4/3.
81 Easy
81 of 995
Analyzing +4 | -1
The value of ∫e2x (2  sin⁡  3x  +  3  cos⁡  3x) dx\int {e^{2x}}\,(2\,\,\sin \,\,3x\,\, + \,\,3\,\,\cos \,\,3x)\,dx is
View answer and explanation
Correct option: A
e2x  sin⁡  3x  +  c{e^{2x}}\,\,\sin \,\,3x\,\, + \,\,c
Option A (e2xsin⁡,3x,+,c{e^{2x}}\sin \\,3x\\, + \\,c) is correct because the integral can be solved by recognizing a pattern where the cosine term cancels out after integration by parts.
82 Medium
82 of 995
Application +4 | -1
If ∫0∞  e−ax  dx=1a\int \limits_0^\infty \,\,{e^{ - ax}}\,\,dx = \frac{1}{a}, then ∫0∞ xne−ax  dx\int \limits_0^\infty \,{x^n}{e^{ - ax}}\,\,dx, is
View answer and explanation
Correct option: C
n!an+1\frac{{n!}}{{{a^{n + 1}}}}
Option C (n!an+1\frac{{n!}}{{{a^{n + 1}}}}) is correct because it is obtained by differentiating the given integral n times with respect to 'a'.
83 Easy
83 of 995
Application +4 | -1
The value of ∫01x3(1−x)11 dx\int\limits_{0}^1 x^3(1-x)^{11} \mathrm{~d} x is equal to
View answer and explanation
Correct option: D
111−312+314−114\frac{1}{11}-\frac{3}{12}+\frac{3}{14}-\frac{1}{14}
Option D (111−312+313−114\frac{1}{11}-\frac{3}{12}+\frac{3}{13}-\frac{1}{14}) is correct because using the property of definite integrals and expanding the polynomial simplifies the integral to the given expression.
84 Medium
84 of 995
Application +4 | -1
Let f(x)=x−[x]f(x)=x-[x], where [x][x] is the integral part of xx. Then∫−11f(x)dx=\int\limits_{{-1}}^1 f(x) d x=
View answer and explanation
Correct option: A
1
Option A (1) is correct because the integral of the fractional part function from -1 to 1 equals 1.
85 Medium
85 of 995
Application +4 | -1
The set of values of α  (α>0)\alpha \,\,(\alpha > 0) for which the inequality ∫−αα  ex dx  >  32\int \nolimits_{ - \mathrm{\alpha }}^\mathrm{\alpha } \,\,{e^x}\,dx\,\, > \,\,\frac{3}{2} holds true, is
View answer and explanation
Correct option: C
(log⁡  2,  ∞)\left( {\log \,\,2,\,\,\infty } \right)
Option C ((log⁡ 2, ∞)\left( {\log \,2,\,\infty } \right)) is correct because the inequality leads to eα>2{e^\alpha } > 2, which implies α>log⁡2\alpha > \log 2.
86 Easy
86 of 995
Application +4 | -1
If f(x)={ecos⁡xsin⁡x for ∣x∣≤22 otherwise, f(x)=\left\{\begin{array}{cl}e^{\cos x} \sin x & \text { for }|x| \leq 2 \\ 2 & \text { otherwise, }\end{array}\right. then
∫−23f(x)dx=\int\limits_{{-2}}^3 f(x) d x=
View answer and explanation
Correct option: C
2
Option C is correct because the integral is split into two parts, and the first part is zero due to the integrand being an odd function, leaving only the second part to be evaluated.
87 Easy
87 of 995
Application +4 | -1
Suppose a continuous function f:[0,∞)→Rf:[0, \infty) \rightarrow {R} satisfies f(x)=2∫0xtf(t)dt+1 for all x≥0.f(x)=2 \int\limits_{0}^x t f(t) d t+1 \text { for all } x \geq 0. Then f(1)f(1) equals
View answer and explanation
Correct option: A
ee
Option A (e) is correct because the function f(x) is found to be ex2e^{x^2}, and thus f(1) = e.
88 Easy
88 of 995
Application +4 | -1
The value of the integral ∫0π/4sin⁡x+cos⁡x3+sin⁡2xdx\int\limits_{0}^{\pi / 4} \frac{\sin x+\cos x}{3+\sin 2 x} d x, is
View answer and explanation
Correct option: C
14log⁡3\frac{1}{4} \log 3
Option D is correct. The integral is solved by using substitution and simplifying the expression, resulting in (1/8)log 3.
89 Easy
89 of 995
Analyzing +4 | -1
∫011−x1+xdx=\int\limits_{0}^1 \sqrt{\frac{1-x}{1+x}} d x=
View answer and explanation
Correct option: B
π2−1\frac{\pi}{2}-1
Option B (π2−1\frac{{\pi}}{{2}}-1) is correct because by rationalizing the denominator and using basic integration formulas, the integral evaluates to this value.
90 Easy
90 of 995
Application +4 | -1
∫015[x2]dx\int\limits_{0}^{15}\left[x^2\right] d x is equal to
View answer and explanation
Correct option: B
2−22-\sqrt{2}
Option B (2−22-\sqrt{2}) is correct because the integral of the greatest integer function of x squared from 0 to 1.5 is equal to 2−22-\sqrt{2}.
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