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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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211 Easy
211 of 791
Application +4 | -1
The value of the integral ∫01  1(x2+1)3/2  dx\int \limits_0^1 \,\,\frac{1}{{{{\left( {{x^2} + 1} \right)}^{3/2}}}}\,\,dx, is
View answer and explanation
Correct option: B
2/2\sqrt 2 /2
Option B (2/2\sqrt 2 /2) is correct because the integral is solved using trigonometric substitution and simplification.
212 Easy
212 of 791
Application +4 | -1
If In=∫0∞  e−x xn−1  dx{I_n} = \int \limits_0^\infty \,\,{e^{ - x}}\,{x^{n - 1}}\,\,dx, then ∫0∞  e−λx  xn−1  dx\int \limits_0^\infty \,\,{e^{ - \lambda x}}\,\,{x^{n - 1}}\,\,dx is equal to
View answer and explanation
Correct option: C
Inλn\frac{{{I_n}}}{{{\lambda ^n}}}
Option C (Inλn\frac{{{I_n}}}{{{\lambda ^n}}}) is correct because it is the result of a correct u-substitution and simplification.
213 Easy
213 of 791
Application +4 | -1
The value of ∫−ππcos⁡2x1+axdx,a>0\int_{-\pi}^\pi \frac{\cos ^2 x}{1+a^x} d x, a > 0 is
View answer and explanation
Correct option: C
π2\frac{\pi}{2}
Option C (π2\frac{{\pi}}{{2}}) is correct because, after using the property of definite integrals and simplifying, the value of the integral is π2\frac{{\pi}}{{2}}.
214 Easy
214 of 791
Application +4 | -1
∫0π/2max⁡(sin⁡x,cos⁡x)dx\int\limits_{0}^{\pi / 2} \max (\sin x, \cos x) d x is equal to
View answer and explanation
Correct option: D
2\sqrt{2}
Option D (2\sqrt{{2}}) is correct because the integral ∫0π/2max⁡(sin⁡x,cos⁡x)dx\int_0^{\pi/2} \max(\sin x, \cos x) dx can be split into two integrals based on where sin⁡x>cos⁡x\sin x > \cos x and vice versa, and the result is 2\sqrt{{2}}.
215 Easy
215 of 791
Application +4 | -1
If ∫12dx(x2−2x+4)32=kk+5\int\limits_{1}^2 \frac{d x}{\left(x^2-2 x+4\right)^{\frac{3}{2}}}=\frac{k}{k+5} then kk is equal to :
View answer and explanation
Correct option: A
1
Option A (1) is correct because it is the value of k that satisfies the given integral equation after evaluating the integral and solving for k.
216 Easy
216 of 791
Application +4 | -1
∫0π/2cos⁡  x−sin⁡  x1+  cos⁡  x  sin⁡  x  dx\int \nolimits_0^{\pi /2} \frac{{\cos \,\,x - \sin \,\,x}}{{1 + \,\,\cos \,\,x\,\,\sin \,\,x}}\,\,dx is equal to
View answer and explanation
Correct option: A
0\mathrm{0}
Option A (0\mathrm{0}) is correct because using the property of definite integrals ∫0af(x)dx=∫0af(a−x)dx\int_0^a f(x) dx = \int_0^a f(a-x) dx leads to the integral being equal to its negative.
217 Easy
217 of 791
Application +4 | -1
If f(θ)=43(1−cos⁡6θ−sin⁡6θ)f(\theta)=\frac{4}{3}\left(1-\cos ^6 \theta-\sin ^6 \theta\right), then lim⁡n→∞1n[f(1n)+f(2n)+f(3n)+…+f(nn)]=\lim \limits_{{n \rightarrow \infty}} \frac{1}{n}\left[\sqrt{f\left(\frac{1}{n}\right)}+\sqrt{f\left(\frac{2}{n}\right)}+\sqrt{f\left(\frac{3}{n}\right)}+\ldots+\sqrt{f\left(\frac{n}{n}\right)}\right]=
View answer and explanation
Correct option: D
1−cos⁡22\frac{1-\cos 2}{2}
Option D (1−cos⁡22\frac{{1-\cos 2}}{{2}}) is correct because the limit can be expressed as a definite integral, which simplifies to ∫01sin⁡2xdx\int_0^1 \sin 2x dx, and evaluating this integral yields 1−cos⁡22\frac{{1-\cos 2}}{{2}}.
218 Easy
218 of 791
Application +4 | -1
The area of the loop between the curve y=asin⁡xy = a\sin x and x−x - axis is
View answer and explanation
Correct option: B
2a2a
Option B (2a2a) is correct because it represents the definite integral of asin⁡xa\sin x from 0 to π\pi.
219 Easy
219 of 791
Application +4 | -1
If ∫ab  x3    dx=0\int \nolimits_a^b \,\,{x^3}\,\,\;dx = 0 and if ∫ab  x2   dx=23\int \nolimits_a^b \,\,{x^2}\;\,dx = \frac{2}{3}, then the values of a    and    ba\;\,\,\mathrm{and}\,\,\;b are respectively
View answer and explanation
Correct option: D
−1,  1- 1,\,\,1
Option D (−1,,,1- 1,\\,\\,1) is correct because the integral of an odd function from -a to a is 0, and the integral of x² from -1 to 1 is 2/3.
220 Easy
220 of 791
Application +4 | -1
The equation of the tangent to the curve
y=∫xx2log⁡t dt at x=2 is y=\int\limits_{x}^{x^2} \log t \mathrm{~d} t \quad \text { at } x=2 \text { is }
View answer and explanation
Correct option: D
y+8log⁡2+2=(7log⁡2)xy+8 \log 2+2=(7 \log 2) x
The correct answer is D (y+8log⁡2+2=(7log⁡2)xy+8 \log 2+2=(7 \log 2) x) because it represents the equation of the tangent line to the given curve at x=2, found by evaluating the integral and its derivative.
221 Easy
221 of 791
Application +4 | -1
∫2k  (2x+1)  dx=6\int \nolimits_2^k \,\,\left( {2x + 1} \right)\,\,dx = 6, then kk is equal to
View answer and explanation
Correct option: D
3\mathrm{3}
Option A (12log⁡tan⁡(x2+π12)+c\frac{1}{2}\log \tan \left( {\frac{x}{2}\\ + \\ \frac{\pi }{{12}}} \right) + c) is correct because the integral can be solved by rewriting the denominator and using trigonometric identities.
222 Easy
222 of 791
Application +4 | -1
If a<∫02π110+3cos⁡xdx  dx<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 the integral's value lies between these bounds, derived from the minimum and maximum values of the cosine function.
223 Easy
223 of 791
Application +4 | -1
∫−10  dxx2+2x+2\int \nolimits_{ - 1}^0 \,\,\frac{{dx}}{{{x^2} + 2x + 2}} is equal to
View answer and explanation
Correct option: B
π/4\mathrm{\pi }/4
Option B (π4\frac{\mathrm{\pi }}{4}) is correct because the definite integral of dxx2+2x+2\frac{{dx}}{{{x^2} + 2x + 2}} from -1 to 0 is π4\frac{\pi}{4}.
224 Easy
224 of 791
Analyzing +4 | -1
The value of the integral ∫0π  sin⁡  k   xsin⁡  x  dx\int \limits_0^\pi \,\,\frac{{\sin \,\,k\,\;x}}{{\sin \,\,x}}\,\,dx is an even integer) is equal to
View answer and explanation
Correct option: D
0\mathrm{0}
Option D (0\mathrm{0}) is correct because the integral evaluates to 0 due to the symmetry of the sine function.
225 Easy
225 of 791
Application +4 | -1
The value of ∫01  dxx  +  1  −  x2\int \nolimits_0^1 \,\,\frac{{dx}}{{x\,\, + \,\,\sqrt {1\,\, - \,\,{x^2}} }} is
View answer and explanation
Correct option: D
π4\frac{\mathrm{\pi }}{4}
Option D (π4\frac{\mathrm{\pi }}{4}) is correct because the integral is solved using a trigonometric substitution, and the result is π4\frac{\pi}{4}.

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
226 Easy
226 of 791
Application +4 | -1
The value of ∫−13  ( ∣x−2∣+ [x] )  dx\int \limits_{ - 1}^3 \,\,\left( {\,\left| {x - 2} \right| + \,\left[ x \right]\,} \right)\,\,dx is ( [x]\left[ x \right] stands for greatest integer less than or equal to xx)
View answer and explanation
Correct option: A
7\mathrm{7}
Option A (7) is correct because it is the result of evaluating the definite integral by splitting the integral at the points where the absolute value and greatest integer functions change their behavior.
227 Easy
227 of 791
Application +4 | -1
If f(x)=∫−1x  ∣ t ∣  dtf\left( x \right) = \int \nolimits_{ - 1}^x \,\,\left| {\,t\,} \right|\,\,dt, then for any x≥0,  f(x)x \ge 0,\,\,f\left( x \right) is equal to
View answer and explanation
Correct option: B
12  (1  +  x2)\frac{1}{2}\,\,\left( {1\,\, + \,\,{x^2}} \right)
Option B is correct because the integral of the absolute value function is split into two parts and then evaluated.
228 Easy
228 of 791
Application +4 | -1
The value of ∫0π/2  sin⁡8  x dx\int \nolimits_0^{\pi /2} \,\,{\sin ^8}\,\,x\,dx is
View answer and explanation
Correct option: A
105  π32  (4  !)\frac{{105\,\,\pi }}{{32\,\,\left( {4\,\,!} \right)}}
Option A is correct because the integral is evaluated using Walli's formula, resulting in the given expression.
229 Easy
229 of 791
Application +4 | -1
∫02π(sin⁡x+∣sin⁡x∣)dx=\int\limits_{0}^{2 \pi}(\sin x+|\sin x|) d x=
View answer and explanation
Correct option: B
4
Option B (4) is correct because the integral of sin(x) + |sin(x)| from 0 to 2π is calculated by splitting the integral at π and evaluating.
230 Easy
230 of 791
Application +4 | -1
The value of ∫01  log⁡  [x]  dx\int \nolimits_0^1 \,\,\log \,\,\left[ x \right]\,\,dx is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the logarithm of 0 is undefined, making the integral undefined.
231 Easy
231 of 791
Analyzing +4 | -1
If f(x)=2−x cos⁡ x2+ x cos⁡ xf(x) = \frac{{2 - x\,\cos \,x}}{{2 + \,x\,\cos \,x}} and g(x)=log⁡e x,  (x > 0)g(x) = {\log _e}\,x,\,\,(x\, > \,0) then the value of the integral ∫−π/4π/4 g(f(x))dx\int_{ - \pi /4}^{\pi /4} \, g(f(x))dx is
View answer and explanation
Correct option: D
log⁡e1{\log _e}1
Option D (log⁡e1{\log _e}1) is correct because the integral evaluates to 0, which is the natural logarithm of 1.
232 Easy
232 of 791
Application +4 | -1
Let f:(0,∞)→Rf:(0, \infty) \rightarrow \mathbf{R} and let
F(x)=∫0xf(t)dtF(x)=\int\limits_{0}^x f(t) d t

If F(x2)=x2(1+x)F\left(x^2\right)=x^2(1+x), then f(4)f(4) equals
View answer and explanation
Correct option: C
4
Option C (4) is correct because by using the fundamental theorem of calculus and differentiating the given equation, we can find f(4).
233 Easy
233 of 791
Application +4 | -1
If g(x)=∫0xcos⁡4t dtg(x)=\int\limits_{0}^x \cos ^4 t \mathrm{~d} t then g(x+π)g(x+\pi) equals
View answer and explanation
Correct option: A
g(x)+g(π)g(x)+g(\pi)
Option A (g(x)+g(π)) is correct because the integral from 0 to x+π can be split into integrals from 0 to π and π to x+π, and a substitution can be used to show the latter equals g(x).
234 Easy
234 of 791
Application +4 | -1
If n∈Nn \in N, then ∫−nn(−1)[x]dx\int\limits_{{-n}}^n(-1)^{[x]} d x equals
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the integrand is an odd function, and the integral of an odd function over a symmetric interval is zero.
235 Easy
235 of 791
Application +4 | -1
The value of ∫−13{∣x−2∣+[x]}dx\int\limits_{{-1}}^3\{|x-2|+[x]\} d x, where [x][x] denotes the greatest integer less than or equal to xx is
View answer and explanation
Correct option: B
7
Option B (7) is correct because the integral is evaluated by splitting it into intervals based on the absolute value and greatest integer functions.
236 Easy
236 of 791
Application +4 | -1
The value of ∫−π/2π/2sin⁡2x1+2xdx\int\limits_{{-\pi / 2}}^{\pi / 2} \frac{\sin ^2 x}{1+2^x} d x is :
View answer and explanation
Correct option: C
π4\frac{\pi}{4}
Option C (π4\frac{\pi}{4}) is correct because the integral is solved using the property of definite integrals and trigonometric identities.
237 Easy
237 of 791
Understanding +4 | -1
∫−100100  f  (x)  dx\int \nolimits_{ - 100}^{100} \,\,f\,\,\left( x \right)\,\,dx is equal to
View answer and explanation
Correct option: D
∫−100100  f(−x)  dx\int \nolimits_{ - 100}^{100} \,\,f\left( { - x} \right)\,\,dx
Option D (∫−100100 f(−x) dx\int \nolimits_{ - 100}^{100} \,f\left( { - x} \right)\,dx) is correct because substituting x=−tx = -t results in the same integral.
238 Easy
238 of 791
Application +4 | -1
∫0[x]/38x2[3x]dx\int_0^{[x] / 3} \frac{8^x}{2^{[3 x]}} d x where [⋅][\cdot] denotes the greatest integer function, is equal to
View answer and explanation
Correct option: D
[x]ln⁡8\frac{[x]}{\ln 8}
The correct answer is D ([x]ln⁡8\frac{[x]}{\ln 8}) because after simplifying the integrand and using the properties of the greatest integer function, the integral evaluates to [x]ln⁡8\frac{[x]}{\ln 8}.
239 Easy
239 of 791
Analyzing +4 | -1
Let J=∫01cos⁡xxdxJ=\int\limits_{0}^1 \frac{\cos x}{\sqrt{x}} d x and K=∫01sin⁡xxdxK=\int\limits_{0}^1 \frac{\sin x}{\sqrt{x}} \mathrm{dx}. Then which one of the following is true?
View answer and explanation
Correct option: A
K<2/3\mathrm{K}<2 / 3 and J<2\mathrm{J}<2
Option A (K < 2/3 and J < 2) is correct because by comparing the integrands with simpler functions, we can establish the inequalities for the integrals.
240 Easy
240 of 791
Analyzing +4 | -1
If I=∫0111+xπ/2dx,I=\int\limits_{0}^1 \frac{1}{1+x^{\pi / 2}} d x, then
View answer and explanation
Correct option: A
ln⁡2<I<π4\ln 2 < I < \frac{\pi}{4}
Option A (ln⁡2<I<π4\ln 2 < I < \frac{\pi}{4}) is correct because it correctly bounds the integral by comparing it to integrals of simpler functions.
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