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JEE • MATHEMATICS • DEFINITE INTEGRATION

Definite integration practice · Medium

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61 Medium
61 of 596
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.
62 Medium
62 of 596
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.
63 Medium
63 of 596
Application +4 | -1
Let f(x)f(x) be a function satisfying f′(x)=f(x)f^{\prime}(x)=f(x) with f(0)=1f(0)=1 and gg be the function satisfying f(x)+g(x)=x2f(x)+g(x)=x^2.

The value of the integral ∫01f(x)g(x)dx\int_0^1 f(x) g(x) d x is
View answer and explanation
Correct option: D
e−12e2−32e-\frac{1}{2} e^2-\frac{3}{2}
Option D is correct because by finding f(x) and g(x), we can evaluate the integral ∫01f(x)g(x)dx\int_0^1 f(x) g(x) d x which results in e−12e2−32e-\frac{1}{2} e^2-\frac{3}{2}.
64 Medium
64 of 596
Application +4 | -1
∫0π/4log⁡(1+tan⁡θ)dθ=\int\limits_{0}^{\pi / 4} \log (1+\tan \theta) d \theta=
View answer and explanation
Correct option: C
π8log⁡2\frac{\pi}{8} \log 2
Option C (π8log⁡2\frac{\pi}{8} \log 2) is correct because it is the result of evaluating the definite integral using properties of logarithms and trigonometric functions after applying a substitution.
65 Medium
65 of 596
Application +4 | -1
The value of the definite integral ∫−(π/2)π/2cos⁡2x1+5x\int\limits_{{-(\pi / 2)}}^{\pi / 2} \frac{\cos ^2 x}{1+5^x} equals to:
View answer and explanation
Correct option: D
π4\frac{\pi}{4}
Option D (π4\frac{\pi}{4}) is correct because 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, the integral simplifies to ∫0π/2cos⁡2xdx=π4\int_0^{\pi/2} \cos^2 x dx = \frac{\pi}{4}.
66 Medium
66 of 596
Application +4 | -1
The value of lim⁡n→∞[1n+1n+1+1n+2+…+13n]\lim _{n \rightarrow \infty}\left[\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{3 n}\right] is
View answer and explanation
Correct option: C
log⁡e3\log _e 3
Option C (log⁡e3\log _e 3) is correct because the limit represents the definite integral of 11+x\frac{1}{1+x} from 0 to 2, which evaluates to log⁡e3\log_e 3.
67 Medium
67 of 596
Application +4 | -1
∫2π/2  cos⁡  x(1  +  sin⁡  x)  (2  +  sin⁡  x)    dx\int \nolimits_2^{\pi /2} \,\,\frac{{\cos \,\,x}}{{(1\,\, + \,\,\sin \,\,x)\,\,(2\,\, + \,\,\sin \,\,x)}}\;\,\,dx
is equal to
View answer and explanation
Correct option: A
log⁡  43\log \,\,\frac{4}{3}
Option A (log⁡ 43\log \,\frac{4}{3}) is correct because after substitution and partial fraction decomposition, the definite integral evaluates to this value.
68 Medium
68 of 596
Application +4 | -1
∫03(12ddx(tan⁡−12x1−x2))dx\int\limits_{0}^{\sqrt{3}}\left(\frac{1}{2} \frac{d}{d x}\left(\tan ^{-1} \frac{2 x}{1-x^2}\right)\right) d x equals to:
View answer and explanation
Correct option: A
π3\frac{\pi}{3}
Option A (π3\frac{\pi}{3}) is correct because the integral evaluates to 12[2tan⁡−1x]03=π3\frac{1}{2} [2 \tan^{-1} x]_0^{\sqrt{3}} = \frac{\pi}{3}.
69 Medium
69 of 596
Application +4 | -1
Let f(x)f(x) be a continuous function in RR such that f(x)+f(y)=f(x+y)f(x)+f(y)=f(x+y). If ∫03f(x)dx=k\int_0^3 f(x) d x=k, then ∫−33f(x)dx\int\limits_{{-3}}^3 f(x) d x is equal to -
View answer and explanation
Correct option: B
0
Option B (0) is correct because the given functional equation implies that f(x) is of the form kx, which is an odd function, and the integral of an odd function over a symmetric interval is zero.
70 Medium
70 of 596
Application +4 | -1
If ∫0100f(x)dx=a\int\limits_{0}^{100} f(x) d x=a, then ∑r=1100(∫01f(r−1+x)dx)\sum\limits_{{r=1}}^{100}\left(\int\limits_{0}^1 f(r-1+x) d x\right) is equal to
View answer and explanation
Correct option: B
aa
Option B (aa) is correct because the sum of the integrals can be combined into a single integral from 0 to 100, which is given as aa.
71 Medium
71 of 596
Application +4 | -1
∫01x(1−x)ndx=\int\limits_{0}^1 x(1-x)^n d x=
View answer and explanation
Correct option: D
1n+1−1n+2\frac{1}{n+1}-\frac{1}{n+2}
Option D is correct because by using the property ∫0af(x)dx=∫0af(a−x)dx\int_0^a f(x) dx = \int_0^a f(a-x) dx and simplifying the integral, we get the result 1n+1−1n+2\frac{1}{n+1} - \frac{1}{n+2}.
72 Medium
72 of 596
Application +4 | -1
The value of the integral ∫αβ  1(x−α)  (β−x)  dx\int \limits_\alpha ^\beta \,\,\frac{1}{{\sqrt {\left( {x - \alpha } \right)\,\,\left( {\beta - x} \right)} }}\,\,dx, is
View answer and explanation
Correct option: C
π\pi
Option C π\pi is correct because the integral evaluates to π\pi after a suitable trigonometric substitution.
73 Medium
73 of 596
Application +4 | -1
The value of the integral ∫1nan−1nxa−x+xdx\int\limits_ {\frac{1}{n}}^{\frac{an-1}{n}} \frac{\sqrt{x}}{\sqrt{a-x}+\sqrt{x}} d x is
View answer and explanation
Correct option: C
na−22n\frac{n a-2}{2 n}
Option C (na−22n\frac{{n a-2}}{{2 n}}) is correct because the integral can be solved 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 adding the two integrals.
74 Medium
74 of 596
Application +4 | -1
The value of ∫−ππ2y(1+sin⁡y)1+cos⁡2ydy\int\limits_{{-\pi}}^\pi \frac{2 y(1+\sin y)}{1+\cos ^2 y} d y is:
View answer and explanation
Correct option: A
π2\pi^2
The correct answer is A (π2\pi^2). The integral is solved by using properties of definite integrals and trigonometric substitutions.
75 Medium
75 of 596
Application +4 | -1
∫−ππ2x(1+sin⁡x)1+cos⁡2xdx=\int\limits_{{-\pi}}^\pi \frac{2 x(1+\sin x)}{1+\cos ^2 x} d x=
View answer and explanation
Correct option: B
π2\pi^2
Option B (π2\pi^2) is correct because after simplifying the integral using properties of even and odd functions, the final result is π2\pi^2.

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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
76 Medium
76 of 596
Analyzing +4 | -1
The value of ∫01 tan⁡−1  (2x  −  11+  x−  x2)  dx\int \limits_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 can be simplified using the property tan⁡−1(a)−tan⁡−1(b)=tan⁡−1(a−b1+ab)\tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}(\frac{a-b}{1+ab}) and the substitution x→1−xx \rightarrow 1-x.
77 Medium
77 of 596
Application +4 | -1
If nn is a positive integer and [x][x] is the greatest integer not exceeding xx, then ∫0n{x−[x]}dx\int\limits_{0}^n\{x-[x]\} d x equals
View answer and explanation
Correct option: C
n/2n / 2
Option C (n/2n / 2) is correct because the integral of the fractional part of x from 0 to n is equal to n/2 due to the periodic nature of the function.
78 Medium
78 of 596
Application +4 | -1
∫−π/3−π/6dx1+tan⁡4x=\int\limits_{{-\pi / 3}}^{-\pi /6} \frac{d x}{1+\tan ^4 x}=
View answer and explanation
Correct option: C
π12\frac{\pi}{12}
Option C (π12\frac{\pi}{12}) is correct because after applying the property of definite integrals and simplifying, the integral evaluates to π12\frac{\pi}{12}.
79 Medium
79 of 596
Application +4 | -1
For which of following values of mm area bounded between y=x−x2&y=mxy=x-x^{2} \& y=m x equals 92\frac{9}{2}
View answer and explanation
Correct option: B
-2
Option B (-2) is correct because when m = -2, the area between the curves y = x - x2 and y = mx equals 9/2.
80 Medium
80 of 596
Application +4 | -1
If nn is an odd natural number, then
∫−π/6π/6  π  +  4xn1−sin⁡  (pq  +  π6)  dx=\int \limits_{ - \pi /6}^{\pi /6} \,\,\frac{{\pi \,\, + \,\,4{x^n}}}{{1 - \sin \,\,\left( {pq\,\, + \,\,\frac{\pi }{6}} \right)}}\,\,dx =
View answer and explanation
Correct option: A
4  π4\,\,\pi
Option A (4π4\pi) is correct. The integral simplifies due to the odd function and trigonometric identities, resulting in the value 4π4\pi.
81 Medium
81 of 596
Application +4 | -1
If [x][x] denotes the greatest integer less than or equal to x\mathrm{x}, then the value of the integral ∫−π/2π/2[[x]−sin⁡x]dx\int\limits_{-\pi / 2}^{\pi / 2}[[x]-\sin x] d x is equal to:
View answer and explanation
Correct option: D
−π-\pi
Option D (−π- \pi) is correct because the integral simplifies to integrating -1 over the given interval due to the properties of the greatest integer function and sine function.
82 Medium
82 of 596
Analyzing +4 | -1
The value of lim⁡n→∞  [nn2+12  +  nn2+22  +  ...  +  nn2+n2]\lim\limits_{n \to \infty } \,\,\left[ {\frac{n}{{{n^2} + {1^2}}}\,\, + \,\,\frac{n}{{{n^2} + {2^2}}}\,\, + \,\,...\,\, + \,\,\frac{n}{{{n^2} + {n^2}}}} \right]
View answer and explanation
Correct option: A
π4\frac{\pi }{4}
Option A (π4\frac{\pi}{4}) is correct because the limit of the sum can be expressed as a definite integral, which evaluates to π4\frac{\pi}{4}.
83 Medium
83 of 596
Application +4 | -1
If ∫f(x)  dx=F(x),    then  ∫ x3 f(x2)  dx\int f\left( x \right)\,\,dx = F\left( x \right),\;\,\,\mathrm{then}\,\,\int \,{x^3}\,f\left( {{x^2}} \right)\,\,dx is equal to

View answer and explanation
Correct option: A
12  { x2 F  (x2)  −  ∫F  (x2)  dx2}  \frac{1}{2}\,\,\{ \,{x^2}\,F\,\,\left( {{x^2}} \right)\,\, - \,\,\int F\,\,\left( {{x^2}} \right)\,\,d{x^2}\} \;
Option A (\frac{1}{2}\\ {\x^2}\\ F\\,\left( {\x^2} \right)\\ - \\ \int F\\,\left( {\x^2} \right)\\ d{x^2}\\}\;\frac{1}{2}\\ {\x^2}\\ F\\,\left( {\x^2} \right)\\ - \\ \int F\\,\left( {\x^2} \right)\\ d{x^2}\\}\;) is correct because it is derived using integration by parts with appropriate substitution.
84 Medium
84 of 596
Application +4 | -1
The value of the definite integral, ∫θ1θ2dθ1+tan⁡θ=501πK\int\limits_{{\theta_1}}^{\theta_2} \frac{{d} \theta}{1+\tan \theta}=\frac{501 \pi}{{K}} where θ2=1003π2008\theta_2=\frac{1003 \pi}{2008} and θ1=π2008\theta_1=\frac{\pi}{2008}. The value of KK equals
View answer and explanation
Correct option: D
2008
Option D (2008) is correct because the integral is evaluated 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 then comparing with the given form.
85 Medium
85 of 596
Analyzing +4 | -1
If f(x)=max⁡{sin⁡x,cos⁡x,12}f(x)=\max \left\{\sin x, \cos x, \frac{1}{2}\right\}, then the area of the region bounded by the curves y=f(x),xy=f(x), x-axis, yy-axis and x=5π3x=\frac{5 \pi}{3}, is
View answer and explanation
Correct option: B
2+32+5π12\sqrt{2}+\frac{\sqrt{3}}{2}+\frac{5 \pi}{12}
Option D (4/e4 / e) is correct because the limit can be evaluated by converting the product to a sum using logarithms and then expressing the sum as a definite integral.
86 Medium
86 of 596
Application +4 | -1
∫02  [x2]  dx\int \nolimits_0^2 \,\,\left[ {{x^2}} \right]\,\,dx is
View answer and explanation
Correct option: D
− 2−3+5- \,\sqrt 2 - \sqrt 3 + 5
Option D ([−  2−  3+5]\left[ { - \,\,\sqrt 2 - \,\,\sqrt 3 + 5} \right]) is correct because the integral of the greatest integer function of x squared over the interval [0, 2] evaluates to this expression.
87 Medium
87 of 596
Analyzing +4 | -1
The value of ∫0π/2  cos⁡  3x  +  12  cos⁡  x  −  1  dx\int \limits_0^{\pi /2} \,\,\frac{{\cos \,\,3x\,\, + \,\,1}}{{2\,\,\cos \,\,x\,\, - \,\,1}}\,\,dx, is
View answer and explanation
Correct option: B
1\mathrm{1}
Option B (1) is correct because the integral simplifies to 1 after using trigonometric identities and evaluating the definite integral.
88 Medium
88 of 596
Application +4 | -1
If ∫−14f(x)dx=4\int\limits_{{-1}}^4 f(x) d x=4 and ∫24(3−f(x))dx=7\int\limits_{2}^4(3-f(x)) d x=7, the value of ∫2−1f(x)dx\int\limits_{2}^{-1} f(x) d x is
View answer and explanation
Correct option: C
-5
Option C (-5) is correct because it is the result of using the properties of definite integrals to find the value of the required integral.
89 Medium
89 of 596
Application +4 | -1
The area of the region bounded by the curve 9x2+4y2−36=09{x^2} + 4{y^2} - 36 = 0 is
View answer and explanation
Correct option: D
6  π  sq  unit6\;\mathrm{\pi \;sq\;unit}
Option D, 6  π  sq  unit6\;\mathrm{\pi \;sq\;unit}, is the correct area of the ellipse because the area is calculated using the formula πab\pi ab, where a and b are the semi-major and semi-minor axes.
90 Medium
90 of 596
Application +4 | -1
The value of ∫02  ∣cos⁡  (πx2) ∣\int \limits_0^2 \,\,\left| {\cos \,\,\left( {\frac{{\pi x}}{2}} \right)\,} \right|, is
View answer and explanation
Correct option: D
4/π4/\pi
Option D (4/π) is correct because the integral of |cos(πx/2)| from 0 to 2 is calculated by splitting the integral at x=1 where cos(πx/2) changes sign.
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