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JEE • MATHEMATICS • DEFINITE INTEGRATION

Definite integration practice · Medium

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391 Medium
391 of 596
Application +4 | -1
The value of ∫−π/2π/2cos⁡tsin⁡(2t−π/4)dt\int\limits_{{-\pi / 2}}^{\pi / 2} \cos t \sin (2 t-\pi / 4) \mathrm{d} t is
View answer and explanation
Correct option: D
−2/3-\sqrt{2} / 3
Option D (-2\sqrt{2} / 3) is correct because it is the result of evaluating the definite integral after applying trigonometric identities and simplifying.
392 Medium
392 of 596
Analyzing +4 | -1
The value of ∫0π/2  x  +  sin⁡  x1  +  cos⁡  x  dx\int \limits_0^{\pi /2} \,\,\frac{{x\,\, + \,\,\sin \,\,x}}{{1\,\, + \,\,\cos \,\,x}}\,\,dx, is
View answer and explanation
Correct option: C
π/2\pi /2
The correct answer is C (π/2\pi /2) because the integral can be solved by splitting the integrand and using integration by parts.
393 Medium
393 of 596
Application +4 | -1
Evaluate ∫0π/4sin⁡x+cos⁡x9+16sin⁡2xdx\int\limits_{0}^{\pi / 4} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x.
View answer and explanation
Correct option: A
120ln⁡3\frac{1}{20} \ln 3
Option A (120ln⁡3\frac{{1}}{{20}} \ln 3) is correct because the integral is solved by using a substitution to simplify the trigonometric expression.
394 Medium
394 of 596
Analyzing +4 | -1
The value of the integral
∫410[x2]dx[x2−28x+196]+[x2], where [x]\int\limits_{4}^{10} \frac{\left[x^2\right] d x}{\left[x^2-28 x+196\right]+\left[x^2\right]} \text {, where }[x]
denotes the greatest integer less than or equal to x\mathrm{x}, is :
View answer and explanation
Correct option: D
3
Option D (3) is correct because the integral simplifies to 3 after using the property ∫abf(x)dx=∫abf(a+b−x)dx\int_a^b f(x) dx = \int_a^b f(a+b-x) dx.
395 Medium
395 of 596
Application +4 | -1
∫0πxsin⁡3xdx=\int\limits_{0}^\pi x \sin ^3 x d x=
View answer and explanation
Correct option: B
2π3\frac{2 \pi}{3}
Option B (2π3\frac{2 \pi}{3}) is correct because using the property ∫0af(x)dx=∫0af(a−x)dx\int_0^a f(x) dx = \int_0^a f(a-x) dx simplifies the integral, allowing for direct calculation.
396 Medium
396 of 596
Application +4 | -1
lim⁡n→∞∑k=1nnn2+k2x2,x>0\lim \limits_{{n \rightarrow \infty}} \sum\limits_{{k=1}}^n \frac{n}{n^2+k^2 x^2}, x>0 is equal to
View answer and explanation
Correct option: C
tan⁡−1(x)x\frac{\tan ^{-1}(x)}{x}
Option C is correct because converting the limit to an integral and evaluating it results in tan⁡−1(x)x\frac{\tan^{-1}(x)}{x}.
397 Medium
397 of 596
Application +4 | -1
If a curve y=ax+bxy = a\sqrt x + bx passes through the point (1, 2) and the area bounded by the curves, line x=4x = 4 and xx-axis is 8 sq unit, then
View answer and explanation
Correct option: A
a=3,  b=−1a = 3,\;b = - 1
Option A, a=3,  b=−1a = 3,\;b = - 1, is the correct solution because these values satisfy both the point (1, 2) lying on the curve and the area condition.
398 Medium
398 of 596
Analyzing +4 | -1
If f(x)=ex1+ex,I1=∫f(−a)f(a)xg{x(1−x)}dxf(x)=\frac{e^x}{1+e^x}, I_1=\int\limits_{{f(-a)}}^{f(a)} x g\{x(1-x)\} d x and I2=∫f(−a)f(a)g{x(1−x)}dxI_2=\int\limits_{{f(-a)}}^{f(a)} g\{x(1-x)\} d x then the value of I2I1\frac{I_2}{I_1} is
View answer and explanation
Correct option: D
2
Option D (2) is correct. By using the property ∫abf(x)dx=∫abf(a+b−x)dx\int_a^b f(x) dx = \int_a^b f(a+b-x) dx and the given function's property f(x)+f(−x)=1f(x) + f(-x) = 1, we find that 2I1=I22I_1 = I_2, hence I2/I1=2I_2/I_1 = 2.
399 Medium
399 of 596
Application +4 | -1
\lim \limits_\limits {n \rightarrow \infty}\left(\frac{(n+1)^{1 / 3}}{n^{4 / 3}}+\frac{(n+2)^{1 / 3}}{n^{4 / 3}}+\ldots .+\frac{(2 n)^{1 / 3}}{n^{4 / 3}}\right)\lim \limits_\limits {n \rightarrow \infty}\left(\frac{(n+1)^{1 / 3}}{n^{4 / 3}}+\frac{(n+2)^{1 / 3}}{n^{4 / 3}}+\ldots .+\frac{(2 n)^{1 / 3}}{n^{4 / 3}}\right) is equal to :
View answer and explanation
Correct option: A
34(2)4/3−34\frac{3}{4}(2)^{4 / 3}-\frac{3}{4}
Option A (1p+1\frac{1}{p+1}) is correct because the limit of the sum is converted to a definite integral and evaluated.
400 Medium
400 of 596
Application +4 | -1
If f(x)=∫0g(x)dt1+t3,g(x)=∫0cos⁡x(1+sin⁡t)2dtf(x)=\int\limits_{0}^{g(x)} \frac{d t}{\sqrt{1+t^3}}, g(x)=\int\limits_{0}^{\cos x}(1+\sin t)^2 d t, then the value of f′(π2)f^{\prime}\left(\frac{\pi}{2}\right) is equal to:
View answer and explanation
Correct option: B
−1-1
Option B (-1) is correct because it is found by applying the chain rule and the Fundamental Theorem of Calculus to find the derivative of f(x) and then evaluating it at x = π2\frac{\pi}{2}.
401 Medium
401 of 596
Application +4 | -1
The value of ∫−ππcos⁡2x1+axdx,a>0\int\limits_{{-\pi}}^\pi \frac{\cos ^2 x}{1+a^x} d x, a > 0, is
View answer and explanation
Correct option: A
π/2\pi / 2
Option A (π/2\pi / 2) is correct because the integral can be solved by using the property of definite integrals and combining the integral with a transformed version of itself.
402 Medium
402 of 596
Application +4 | -1
The value of integral ∫−11log⁡(x+x2+1)dx\int\limits_{{-1}}^1 \log \left(x+\sqrt{x^2+1}\right) d x is
View answer and explanation
Correct option: A
0
Option A (0) is correct because the integrand is an odd function, and the integral is evaluated over a symmetric interval [-1, 1].
403 Medium
403 of 596
Application +4 | -1
The area bounded by the ellipse 3x2+2y2=63 x^2+2 y^2=6 with the co-ordinate axes in sq. units is
View answer and explanation
Correct option: A
6π\sqrt{6} \pi
Option A (6π\sqrt{6} \pi) is correct because the area of the ellipse is calculated using the formula πab\pi ab where a and b are the semi-major and semi-minor axes.
404 Medium
404 of 596
Analyzing +4 | -1
If ∫01  et dtt+1=a,\int \nolimits_0^1 \,\,\frac{{{e^t}\,dt}}{{t + 1}} = a,, then ∫b−1b  e−t dtt−b−1\int \nolimits_{b - 1}^b \,\,\frac{{{e^{ - t}}\,dt}}{{t - b - 1}} is equal to
View answer and explanation
Correct option: B
− ae−b- \,a{e^{ - b}}
Option B (−ae−b-ae^{-b}) is correct because after substitution and manipulation of the integral, it simplifies to −ae−b-ae^{-b}.
405 Medium
405 of 596
Analyzing +4 | -1
The area bounded by ∣x−1∣≤2  and  x2−y2=1\left| {x - 1} \right| \le 2\;\mathrm{and}\;{x^2} - {y^2} = 1, is
View answer and explanation
Correct option: C
62−log⁡∣3+22∣6\sqrt 2 - \log |3 + 2\sqrt 2 |
Option D (None of these) is incorrect; the correct answer is 62−log⁡∣3+22∣6\sqrt 2 - \log |3 + 2\sqrt 2 | which is obtained by evaluating the definite integral of the function within the given bounds.

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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
406 Medium
406 of 596
Application +4 | -1
∫0ax4a2−x2dx=\int\limits_{0}^a x^4 \sqrt{a^2-x^2} d x=
View answer and explanation
Correct option: B
π32a6\frac{\pi}{32} a^6
Option B (π/32 a6) is correct because the integral evaluates to π/32 a6 after using trigonometric substitution and the Gamma function.
407 Medium
407 of 596
Application +4 | -1
∫010π  ∣sin⁡  x∣  dx    is\int \nolimits_0^{10\pi } \,\,|\sin \,\,x|\,\,dx\;\,\,\mathrm{is}
View answer and explanation
Correct option: A
20
Option A (20) is correct because the integral of |sin x| over one period (π) is 2, and there are 10 such periods in the interval [0, 10π].
408 Medium
408 of 596
Application +4 | -1
The value of ∫02π∣sin⁡3θ∣dθ\int\limits_{0}^{2 \pi}\left|\sin ^3 \theta\right| d \theta is
View answer and explanation
Correct option: C
8/38 / 3
Option C (8/38 / 3) is correct because the integral of the absolute value of sin⁡3θ\sin^3 \theta from 0 to 2π2\pi is 8/38/3.
409 Medium
409 of 596
Application +4 | -1
If ∫−14  f (x)  dx=4\int \nolimits_{ - 1}^4 \,\,f\,\left( x \right)\,\,dx = 4 and ∫24  {3  −  f(x)}  dx=7\int \nolimits_2^4 \,\,\left\{ {3\,\, - \,\,f\left( x \right)} \right\}\,\,dx = 7, then the value of
∫−12  f(x)  dx\int \nolimits_{ - 1}^2 \,\,f\left( x \right)\,\,dx
is
View answer and explanation
Correct option: D
5\mathrm{5}
Option D (5) is correct because the integral from -1 to 2 is found by subtracting the integral from 2 to 4 from the integral from -1 to 4.
410 Medium
410 of 596
Analyzing +4 | -1
The value of the integral ∫−π4π4x+π42−cos⁡2xdx\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{2-\cos 2 x} d x is:
View answer and explanation
Correct option: B
π263\frac{\pi^{2}}{6 \sqrt{3}}
The correct answer is B (π263\frac{\pi^{2}}{6 \sqrt{3}}) because it results from evaluating the definite integral using properties of definite integrals and trigonometric substitutions.
411 Medium
411 of 596
Application +4 | -1
∫0∞[2ex]dx\int\limits_{0}^{\infty}\left[\frac{2}{e^x}\right] d x is [where [] represents the greatest integer function]
View answer and explanation
Correct option: D
log⁡e2\log _e 2
Option D (log_e 2) is correct because the integral of the greatest integer function [2/e^x] from 0 to infinity equals log_e 2.
412 Medium
412 of 596
Application +4 | -1
If 2 f(x)  −  3 f(1/x)=x,    then  ∫12  f(x)dx2\,f\left( x \right)\,\, - \,\,3\,f(1/x) = x,\;\,\,\mathrm{then}\,\,\int \nolimits_1^2 \,\,f\left( x \right)dx is equal to

View answer and explanation
Correct option: B
(− 3  /  5)  (1+log⁡  2)\left( { - \,3\,\,/\,\,5} \right)\,\,(1 + \log \,\,2)
Option B ((-3/5)(1 + log 2)) is correct because it is the result of solving the given functional equation and then evaluating the definite integral.
413 Medium
413 of 596
Application +4 | -1
∫−π/2π/2  ∣ x ∣8  cos⁡2  2x  +  1 dx\int \limits_{ - \pi /2}^{\pi /2} \,\,\frac{{\left| {\,x\,} \right|}}{{8\,\,{{\cos }^2}\,\,2x\,\, + \,\,1}}\,dx has the value
View answer and explanation
Correct option: B
π212\frac{{{\pi ^2}}}{12}
Option B is correct because the integral is evaluated using properties of even functions and trigonometric substitutions.
414 Medium
414 of 596
Application +4 | -1
∫815  dx(x−3)  x+1\int \nolimits_8^{15} \,\,\frac{{dx}}{{\left( {x - 3} \right)\,\,\sqrt {x + 1} }} is equal to
View answer and explanation
Correct option: A
12  log⁡  53\frac{1}{2}\,\,\log \,\,\frac{5}{3}
Option A (12log⁡53\frac{1}{2}\log \frac{5}{3}) is correct because the integral evaluates to this value after a suitable trigonometric substitution and simplification.
415 Medium
415 of 596
Application +4 | -1
If f(x)=∫−1x  ∣t∣  dtf\left( x \right) = \int \limits_{ - 1}^x \,\,\left| t \right|\,\,dt, then for any x≥0,    f(x)x \ge 0,\,\,\;f\left( x \right) equals

View answer and explanation
Correct option: C
12  (1+x2)\frac{1}{2}\,\,\left( {1 + {x^2}} \right)
Option C (12 (1+x2)\frac{1}{2}\,\left( {1 + {x^2}} \right)) is correct because the integral of the absolute value function from -1 to x, for x≥0x \ge 0, is 12(1+x2)\frac{1}{2}(1 + x^2).
416 Medium
416 of 596
Application +4 | -1
If [t][t] stands for the integral part of tt, then ∫05π/12[tan⁡x]dx=\int\limits_{0}^{5 \pi / 12}[\tan x] d x=
View answer and explanation
Correct option: C
π4\frac{\pi}{4}
Option C (π4\frac{\pi}{4}) is correct because after using the properties of the greatest integer function and trigonometric functions, the integral evaluates to π4\frac{\pi}{4}.
417 Medium
417 of 596
Analyzing +4 | -1
The integral ∫(1+x−1x)x+1xdx\int\left(1+x-\frac{1}{x}\right)^{x+\frac{1}{x}} d x is equal to
View answer and explanation
Correct option: D
xex+1x+cx e^{x+\frac{1}{x}}+c
Option D (xex+1x+cx e^{x+\frac{1}{x}}+c) is correct because integrating by parts simplifies the expression to the given result.
418 Medium
418 of 596
Application +4 | -1
Let y=f(x)y=f(x) be a thrice differentiable function in (−5,5).(-5,5). Let the tangents to the curve y=f(x)y=f(x) at (1,f(1))(1, f(1)) and (3,f(3))(3, f(3)) make angles π6\frac{\pi}{6} and π4,\frac{\pi}{4}, respectively with positive xx-axis. If 27∫13((f′(t))2+1)f′′(t)dt=α+β3,27 \int\limits_{1}^3\left(\left(f^{\prime}(t)\right)^2+1\right) f^{\prime \prime}(t) d t=\alpha+\beta \sqrt{3}, where α,β\alpha, \beta are integers, then the value of α+β\alpha+\beta equal
View answer and explanation
Correct option: D
2626
The correct answer is D (26) because by using substitution and integration, we can find the values of alpha and beta and then calculate their sum.
419 Medium
419 of 596
Application +4 | -1
∫01  tan⁡−1  (1x2−x+1)  dx\int \nolimits_0^1 \,\,{\tan ^{ - 1}}\,\,\left( {\frac{1}{{{x^2} - x + 1}}} \right)\,\,dx is equal to
View answer and explanation
Correct option: D
π2  −  log⁡  2\frac{\mathrm{\pi }}{2}\,\, - \,\,\log \,\,2
The correct answer is D (π2−log⁡,2\frac{\mathrm{\pi }}{2}\\ - \log \\,2) because the integral can be simplified using trigonometric identities and integration by parts.
420 Medium
420 of 596
Application +4 | -1
The value of the integral ∫−π2π2sin⁡4x(1+log⁡(2+sin⁡x2−sin⁡x))dx\int\limits_ {-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^4 x\left(1+\log \left(\frac{2+\sin x}{2-\sin x}\right)\right) d x is
View answer and explanation
Correct option: C
38π\frac{3}{8} \pi
Option C (38π\frac{3}{8} \pi) is correct because it is the result obtained after using the property of definite integrals to eliminate the logarithmic term and then evaluating the remaining integral.
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