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JEE • MATHEMATICS • DEFINITE INTEGRATION

Definite integration practice · Easy

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361 Easy
361 of 791
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.
362 Easy
362 of 791
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}.
363 Easy
363 of 791
Analyzing +4 | -1
If ∫0xf(t)dt=x+∫x1tf(t) dt\int\limits_{0}^x f(t) d t=x+\int\limits_{x}^1 tf(t) \, d t, then the value of f(1)f(1), is
View answer and explanation
Correct option: A
1/21 / 2
Option A (1/2) is correct because differentiating both sides of the equation using Leibniz's rule allows us to solve for f(x) and then find f(1).
364 Easy
364 of 791
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.
365 Easy
365 of 791
Analyzing +4 | -1
If In=∫0π/4tan⁡nxdxI_n=\int_0^{\pi / 4} \tan ^n x d x, then 1I2+I4,1I3+I5,1I4+I6,…\frac{1}{I_2+I_4}, \frac{1}{I_3+I_5}, \frac{1}{I_4+I_6}, \ldots. Form
View answer and explanation
Correct option: A
an A.P.
Option A (an A.P.) is correct because the reciprocals of the given expression form an arithmetic progression with a common difference of 1.
366 Easy
366 of 791
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.
367 Easy
367 of 791
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.
368 Easy
368 of 791
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.
369 Easy
369 of 791
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.
370 Easy
370 of 791
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.
371 Easy
371 of 791
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.
372 Easy
372 of 791
Analyzing +4 | -1
The value of definite integral ∫−ππx21+sin⁡x+1+sin⁡2xdx\int\limits_{{-\pi}}^\pi \frac{x^2}{1+\sin x+\sqrt{1+\sin ^2 x}} d x is
View answer and explanation
Correct option: B
π33\frac{\pi^3}{3}
Option B (π33\frac{\pi^3}{3}) is correct because by applying the property of definite integrals and simplifying, the integral evaluates to π33\frac{\pi^3}{3}.
373 Easy
373 of 791
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}}.
374 Easy
374 of 791
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.
375 Easy
375 of 791
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
376 Easy
376 of 791
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.
377 Easy
377 of 791
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.
378 Easy
378 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 from 0 to x results in an even function.
379 Easy
379 of 791
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.
380 Easy
380 of 791
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).
381 Easy
381 of 791
Analyzing +4 | -1
f(x)=∫dxsin⁡6xf(x)=\int \frac{d x}{\sin ^6 x} is a polynomial of degree
View answer and explanation
Correct option: A
5 in cot⁡x\cot x
Option A (5 in cot⁡x\cot x) is correct because after applying the reduction formula and simplifying, the resulting expression is a polynomial of degree 5 in cot⁡x\cot x.
382 Easy
382 of 791
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.
383 Easy
383 of 791
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.
384 Easy
384 of 791
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.
385 Easy
385 of 791
Analyzing +4 | -1
lim⁡n→∞{(n+1)(n+2)(n+3)……3nn2n}1n\lim \limits_{{n \rightarrow \infty}}\left\{\frac{(n+1)(n+2)(n+3) \ldots \ldots 3 n}{n^{2 n}}\right\}^{\frac{1}{n}} is equal to
View answer and explanation
Correct option: C
9e2\frac{9}{e^2}
Option C (9e2\frac{9}{e^2}) is correct because the limit can be evaluated by taking the logarithm and converting the resulting sum to a definite integral.
386 Easy
386 of 791
Application +4 | -1
Area bounded by y=5x2 & y−9=2x2y=5 x^2 \, \& \, y-9=2 x^2 is
View answer and explanation
Correct option: D
12312 \sqrt{3}
Option D (12312 \sqrt{{3}}) is correct because it represents the area bounded by the two parabolas after finding their intersection points and integrating the difference of the functions.
387 Easy
387 of 791
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.
388 Easy
388 of 791
Application +4 | -1
The value of ∫−ππ  sin⁡3  x  cos⁡2  x    dx\int \nolimits_{ - \pi }^\pi \,\,{\sin ^3}\,\,x\,\,{\cos ^2}\,\,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 odd function, and the integral is evaluated over a symmetric interval [-π, π].
389 Easy
389 of 791
Application +4 | -1
For x>0x > 0, if f(x)=∫1xlog⁡et(1+t)dtf(x) = \int\limits_1^x {\frac{{{{\log }_e}t}}{{\left( {1 + t} \right)}}dt} then f(e)+f(1e) f(e) + f\left( {\frac{1}{e}} \right)\, is equal to
View answer and explanation
Correct option: C
12\frac{1}{2}
Option C (12\frac{1}{2}) is correct because after evaluating the integrals and using substitution, the expression simplifies to 1/2.
390 Easy
390 of 791
Analyzing +4 | -1
The integral of x2−xx3−x2+x−1\frac{x^2-x}{x^3-x^2+x-1} w.r.t. xx is
View answer and explanation
Correct option: A
12log⁡(x2+1)+C\frac{1}{2} \log \left(x^2+1\right)+C
Option A (12log⁡(x2+1)+C\frac{1}{2} \log \left(x^2+1\right)+C) is correct because factoring the denominator and simplifying the integrand leads to a simple integral.
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