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JEE • MATHEMATICS • DEFINITE INTEGRATION

Definite integration practice · Mixed difficulty

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481 Easy
481 of 995
Application +4 | -1
The value of ∫−π/2π/211+esin⁡xdx\int\limits_{{-\pi / 2}}^{\pi / 2} \frac{1}{1+e^{\sin x}} d x is
View answer and explanation
Correct option: B
π2\frac{\pi}{2}
Option B (π2\frac{\pi}{2}) is correct because applying the property ∫−aaf(x)dx=∫0a[f(x)+f(−x)]dx\int_{-a}^a f(x) dx = \int_0^a [f(x) + f(-x)] dx simplifies the integral to ∫0π/21dx\int_0^{\pi/2} 1 dx, which evaluates to π/2\pi/2.
482 Medium
482 of 995
Application +4 | -1
∫0π  cos⁡  2x  +  12    dx\int \nolimits_0^\pi \,\,\sqrt {\frac{{\cos \,\,2x\,\, + \,\,1}}{2}} \;\,\,dx is equal to
View answer and explanation
Correct option: B
2\mathrm{2}
Option B (2) is correct. The integral simplifies to ∫0π∣cos⁡x∣dx\int \nolimits_0^{\pi } |\cos x| dx, which evaluates to 2.
483 Medium
483 of 995
Application +4 | -1
The integral ∫sin⁡2xcos⁡2x(sin⁡3x+cos⁡3x)2dx\int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^3 x+\cos ^3 x\right)^2} d x is equal to:
View answer and explanation
Correct option: B
−13(1+tan⁡3x)+c-\frac{1}{3\left(1+\tan ^3 x\right)}+c
Option B (−13(1+tan⁡3x)+c-\frac{1}{3(1+\tan ^3 x)}+c) is correct because the integral simplifies to this expression after substitution and integration.
484 Easy
484 of 995
Application +4 | -1
∫cos⁡2x(cos⁡x+sin⁡x)2dx=\int \frac{\cos 2 x}{(\cos x+\sin x)^2} d x= ?
View answer and explanation
Correct option: C
log⁡(cos⁡x+sin⁡x)+c\log (\cos x+\sin x)+c
Option C (log⁡(cos⁡x+sin⁡x)+c\log (\cos x+\sin x)+c) is correct because it is the result of integrating the simplified expression after applying trigonometric identities and substitution.
485 Easy
485 of 995
Application +4 | -1
If f  (x)=∫x2x4  sin⁡  t dtf\,\,\left( x \right) = \int \limits_{{x^2}}^{{x^4}} \,\,\sin \,\,\sqrt {t\,} dt, then f′  (x)f'\,\,\left( x \right) equals
View answer and explanation
Correct option: B
4 x3  sin⁡  x2  −  2 x  sin⁡  x4\,{x^3}\,\,\sin \,\,{x^2}\,\, - \,\,2\,x\,\,\sin \,\,x
Option B (4 x3  sin⁡  x2  −  2 x  sin⁡  x4\,{x^3}\,\,\sin \,\,{x^2}\,\, - \,\,2\,x\,\,\sin \,\,x) is correct because it is the result of applying the Leibniz rule for differentiating under the integral sign.
486 Medium
486 of 995
Application +4 | -1
∫0π/8  cos⁡3  4  θ  dθ\int \nolimits_0^{\pi /8} \,\,{\cos ^3}\,\,4\,\,\mathrm{\theta }\,\,d\mathrm{\theta } is equal to
View answer and explanation
Correct option: D
1 / 6\mathrm{1}\,\mathrm{/}\,\mathrm{6}
Option D (1 / 6\mathrm{1}\,\mathrm{/}\,\mathrm{6}) is correct because it is the result of evaluating the definite integral of cos⁡3 4 θ{\cos ^3}\,4\,\theta from 0 to π/8\pi/8.
487 Easy
487 of 995
Application +4 | -1
Solution of the equation ∫log⁡2xdxex−1=π6\int\limits_{{\log 2}}^x \frac{d x}{\sqrt{e^x-1}}=\frac{\pi}{6} are
View answer and explanation
Correct option: B
x=2log⁡2x=2 \log 2
The correct answer is B (x=2log⁡2x=2 \log 2) because it satisfies the given integral equation after solving the integral and applying the given condition.
488 Medium
488 of 995
Application +4 | -1
The value ∫1a[x]f′(x)dx,a>1\int\limits_{1}^a[x] f^{\prime}(x) d x, a > 1, where [x][x] denotes the greatest integer not exceeding xx, is
View answer and explanation
Correct option: B
[a]f(a)−{f(1)+f(2)+…+f([a])}[a] f(a)-\{f(1)+f(2)+\ldots+f([a])\}
The correct answer is B ([a]f(a)−{f(1)+f(2)+…+f([a])}[a] f(a)-\{f(1)+f(2)+\ldots+f([a])\}) because the integral is evaluated by splitting the range and using integration by parts.
489 Medium
489 of 995
Application +4 | -1
∫0π/2log⁡sin⁡xdx=\int\limits_{0}^{\pi / 2} \log \sin x d x=
View answer and explanation
Correct option: A
−(π2)log⁡2-\left(\frac{\pi}{2}\right) \log 2
Option A (−(π2)log⁡2-\left(\frac{\pi}{2}\right) \log 2) is the correct answer because it is the result of evaluating the definite integral using properties of logarithms and trigonometric functions.
490 Easy
490 of 995
Application +4 | -1
If g(x)=∫0xcos⁡4t dtg(x)=\int\limits_{0}^x \cos ^4 t \, 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)g(x)+g(\pi)) is correct because the integral from 0 to x+π\pi can be split into integrals from 0 to π\pi and π\pi to x+π\pi, and the second integral can be transformed to an integral from 0 to x.
491 Medium
491 of 995
Analyzing +4 | -1
Let f(x)f\left( x \right) be an integrable function defined on [a,  b],  b   >   a   >  0\left[ {a,\,\,b} \right],\,\,b\,\,\, > \,\,\,a\,\,\, > \,\,0 If I1=∫π/6π/3  f  (tan⁡  θ  +  cot⁡  θ)sec⁡2  θ  dθ{I_1} = \int \limits_{\pi /6}^{\pi /3} \,\,f\,\,\left( {\tan \,\,\theta \,\, + \,\,\cot \,\,\theta } \right){\sec ^2}\,\,\theta \,\,d\theta and, I2=∫π/6π/3  f(tan⁡  θ  +  cot⁡  θ)  cosec2  θ  dθ{I_2} = \int \limits_{\pi /6}^{\pi /3} \,\,f\left( {\tan \,\,\theta \,\, + \,\,\cot \,\,\theta } \right)\,\,\mathrm{cose}{\mathrm{c}^2}\,\,\theta \,\,d\theta, then I1I2=\frac{{{I_1}}}{{{I_2}}} =
View answer and explanation
Correct option: A
A positive integer
Option A (A positive integer) is correct because the integral I1 equals I2, making the ratio equal to 1, which is a positive integer.
492 Easy
492 of 995
Analyzing +4 | -1
Let In=∫0π/4  tan⁡nx  dx,  (n>1  and  n∈N){I_n} = \int \limits_0^{\pi /4} \,\,{\tan ^n}x\,\,dx,\,\,\left( {n > 1\,\,\mathrm{and}\,\,n \in N} \right), then

View answer and explanation
Correct option: B
In  +  In−2=  1n−1{I_n}\,\, + \,\,{I_{n - 2}} = \,\,\frac{1}{{n - 1}}
Option B is correct because the recurrence relation between In and In-2 is found to be In  + In−2=  1n−1{I_n}\,\,+ \,{I_{n - 2}} = \,\,\frac{1}{{n - 1}}.
493 Medium
493 of 995
Application +4 | -1
∫03  ∣x3  +  x2  +  3x∣dx\int \nolimits_0^3 \,\,\left| {{x^3}\,\, + \,\,{x^2}\,\, + \,\,3x} \right|dx is equal to
View answer and explanation
Correct option: B
1714\frac{{171}}{4}
Option B (1714\frac{{171}}{4}) is correct because the integrand is always non-negative in the interval [0, 3], so the absolute value can be dropped, and the resulting polynomial can be integrated directly.
494 Easy
494 of 995
Application +4 | -1
The value of the integral I=∫−π/2π/2cos⁡2x1+3xdxI=\int\limits_{{-\pi / 2}}^{\pi / 2} \frac{\cos ^2 x}{1+3^x} d x is
View answer and explanation
Correct option: A
π4\frac{\pi}{4}
Option A (π4\frac{\pi}{4}) is correct because by applying the property ∫−aaf(x)dx=∫0a[f(x)+f(−x)]dx\int_{-a}^a f(x) dx = \int_0^a [f(x) + f(-x)] dx, the integral simplifies to ∫0π/2cos⁡2xdx\int_0^{\pi/2} \cos^2 x dx, which evaluates to π/4\pi/4.
495 Easy
495 of 995
Understanding +4 | -1
∫1e  1x  dx\int \nolimits_1^e \,\,\frac{1}{x}\,\,dx is equal to
View answer and explanation
Correct option: C
1\mathrm{1}
Option C (1) is correct because the integral of 1/x from 1 to e is equal to ln(x) evaluated from 1 to e, which is ln(e) - ln(1) = 1 - 0 = 1.

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
496 Medium
496 of 995
Application +4 | -1
∫−11  [x  sin⁡  πx]  dx\int \nolimits_{ - 1}^1 \,\,[x\,\,\sin \,\,\mathrm{\pi }x]\,\,dx is equal to
View answer and explanation
Correct option: D
0\mathrm{0}
Option D (0) is correct because the integrand is an even function and the value of the function inside the integral is zero within the integration limits.
497 Easy
497 of 995
Application +4 | -1
∫021[x]4dx\int\limits_{0}^{21}[x]^4 d x is equal
View answer and explanation
Correct option: A
44100
Option A (44100) is correct because the sum of the fourth powers of integers from 0 to 20 equals 44100.
498 Easy
498 of 995
Application +4 | -1
If the value of the integral ∫01/2x2(1−x2)3/2dx\int\limits_{0}^{1 / 2} \frac{x^2}{\left(1-x^2\right)^{3 / 2}} d x is k6\frac{k}{6}, then kk is equal to
View answer and explanation
Correct option: B
23−π2 \sqrt{3}-\pi
Option B is correct. The integral is solved using trigonometric substitution, resulting in k = 23\sqrt{3} - π\pi.
499 Easy
499 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 from 0 to x results in an even function.
500 Medium
500 of 995
Analyzing +4 | -1
Let In=∫0π/2  sin⁡n  x  dx,  n∈N{I_n} = \int \limits_0^{\pi /2} \,\,{\sin ^n}\,\,x\,\,dx,\,\,n \in N. Then
View answer and explanation
Correct option: C
n (In−2−In)=In−2n\,\left( {{I_{n - 2}} - {I_n}} \right) = {I_{n - 2}}
Option C (n (In−2−In)=In−2n\,\left( {{I_{n - 2}} - {I_n}} \right) = {I_{n - 2}}) is correct because it represents the correct reduction formula derived from integration by parts.
501 Easy
501 of 995
Analyzing +4 | -1
Which of the following statements is correct for the function g(α)g(\alpha ) for α∈R,\alpha \in R, such that
g(α)=∫π/6π/3sin⁡αxcos⁡αx+sin⁡αx dxg(\alpha ) = \int\limits_{\pi /6}^{\pi /3} {\frac{{{{\sin }^\alpha }x}}{{{{\cos }^\alpha }x + {{\sin }^\alpha }x}}\,} dx
View answer and explanation
Correct option: D
g(α)g(\alpha ) is an even function
Option D (g(α)g(\alpha) is an even function) is correct because the function g(α)g(\alpha) simplifies to a constant, which is an even function.
502 Easy
502 of 995
Analyzing +4 | -1
If P=∫03π  f  (cos⁡2  x)  dx  and  Q=∫0π  f(cos⁡2  x)  dxP = \int \nolimits_0^{3\pi } \,\,f\,\,({\cos ^2}\,\,x)\,\,dx\,\,\mathrm{and}\,\,{\cal Q} = \int \nolimits_0^\pi \,\,f({\cos ^2}\,\,x)\,\,dx , then
View answer and explanation
Correct option: C
P  −  3 Q=0P\,\, - \,\,3\,{\cal Q} = 0
Option C (P  −  3 Q=0P\,\, - \,\,3\,{\cal Q} = 0) is correct because the integral from 0 to 3π\pi of f(cos²x) is three times the integral from 0 to π\pi of f(cos²x).
503 Easy
503 of 995
Application +4 | -1
lim⁡n→∞  ∑r=1n  ∣ r3r4  +  n4 ∣\lim\limits_{n \to \infty } \,\,\sum \limits_{r = 1}^n \,\,\left| {\,\frac{{{r^3}}}{{{r^4}\,\, + \,\,{n^4}}}\,} \right| equals
View answer and explanation
Correct option: D
14  log⁡  2\frac{1}{4}\,\,\log \,\,2
Option D (14 log⁡ 2\frac{1}{4}\,\log \,2) is correct because the limit of the Riemann sum can be expressed as a definite integral, which evaluates to 14 log⁡ 2\frac{1}{4}\,\log \,2.
504 Medium
504 of 995
Application +4 | -1
∫αβ  x  −  αβ  −  x  dx\int \nolimits_\alpha ^\beta \,\,\sqrt {\frac{{x\,\, - \,\,\alpha }}{{\beta \,\, - \,\,x}}} \,\,dx is equal to
View answer and explanation
Correct option: B
π2  (β−α)\frac{\pi }{2}\,\,\left( {\beta - \alpha } \right)
Option D (π(β−α)\pi \left( {\beta - \alpha } \right)) is incorrect. Option B (π2 (β−α)\frac{\pi }{2}\,\left( {\beta - \alpha } \right)) is correct because the integral evaluates to π2(β−α)\frac{\pi}{2}(\beta - \alpha) after substitution and simplification.
505 Medium
505 of 995
Application +4 | -1
∫π/6π/3  11  +  tan⁡3  x  dx   is\int \nolimits_{\pi /6}^{\pi /3} \,\,\frac{1}{{1\,\, + \,\,{{\tan }^3}\,\,x}}\,\,dx\;\,\mathrm{is}
View answer and explanation
Correct option: A
π12\frac{\pi }{{12}}
Option A (π12\frac{\pi}{{12}}) is correct because using the property of definite integrals and adding the two forms of the integral leads to a simple result.
506 Easy
506 of 995
Application +4 | -1
The value of lim⁡n→∞1n(sin⁡πn+sin⁡2πn+⋯+sin⁡nπn)\lim \limits_{{n \rightarrow \infty}} \frac{1}{n}\left(\sin \frac{\pi}{n}+\sin \frac{2 \pi}{n}+\cdots+\sin \frac{n \pi}{n}\right) is
View answer and explanation
Correct option: D
2π\frac{2}{\pi}
Option D (2π\frac{2}{\pi}) is correct. The limit is evaluated by recognizing it as a Riemann sum, converting it to a definite integral, and then evaluating the integral.
507 Medium
507 of 995
Application +4 | -1
If In=∫1e(log⁡ex)ndxI_n=\int\limits_{1}^e\left(\log _e x\right)^n d x ( nn is a positive integer), then I2012+(2012)I2011=I_{2012}+(2012) I_{2011}=
View answer and explanation
Correct option: B
I2013+(2013)I2012I_{2013}+(2013) I_{2012}
Option B (I2013+(2013)I2012I_{2013}+(2013) I_{2012}) is correct because it is derived from the recurrence relation obtained by integration by parts.
508 Medium
508 of 995
Analyzing +4 | -1
Let f(x)f(x) be a quadratic function satisfying the equation ∫0xf(t)dt=xf(x)−4x33+ax2 ∀ x∈R\int\limits_{0}^x f(t) d t=x f(x)-\frac{4 x^3}{3}+a x^2 \, \forall \, x \in R and aa ∈\in R. Also f(0)=a2−a−6f(0)=a^2-a-6, then number of integral values of aa such that f(x)=0f(x)=0 has real roots is/are
View answer and explanation
Correct option: C
7
The correct answer is 7 because solving the integral equation gives the quadratic function, and finding the range of 'a' for real roots leads to 7 integral values.
509 Easy
509 of 995
Application +4 | -1
If ∫0π/3tan⁡θ2ksec⁡θdθ=1−12,k>0\int_0^{\pi / 3} \frac{\tan \theta}{\sqrt{2 k \sec \theta}} d \theta=1-\frac{1}{\sqrt{2}}, k>0. Then the value of kk is
View answer and explanation
Correct option: A
2
The correct answer is A (2) because after evaluating the integral and solving for k, the value of k is found to be 2.
510 Easy
510 of 995
Application +4 | -1
The value of ∫01x2α−1log⁡x dx\int\limits_{0}^1 \frac{x^{2 \alpha}-1}{\log x} \mathrm{~d} x, if α=2n−12\alpha=\frac{2 n-1}{2} is
View answer and explanation
Correct option: C
log⁡2n\log 2 n
Option C is correct because the integral can be solved using Leibniz rule for differentiation under the integral sign.
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