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JEE • MATHEMATICS • DEFINITE INTEGRATION

Definite integration practice · Medium

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91 Medium
91 of 596
Analyzing +4 | -1
Let In=∫0π4tan⁡nx dx,n∈NI_n=\int\limits_{0}^{\frac{\pi}{4}} \tan ^n x \, d x, n \in N. Then
View answer and explanation
Correct option: A
I1=I3+2I5{I}_1={I}_3+2 {I}_5
The correct answer is A (I1=I3+2I5{I}_1={I}_3+2 {I}_5) because it represents the correct recurrence relation between the given integrals, derived through integration by parts or reduction formula.
92 Medium
92 of 596
Application +4 | -1
The smallest interval [a,b][a, b] such that ∫01dx1+x4∈[a,b]\int\limits_{0}^1 \frac{d x}{\sqrt{1+x^4}} \in[a, b] is given by
View answer and explanation
Correct option: A
[12,1]\left[\frac{1}{\sqrt{2}}, 1\right]
Option A ([12,1]\left[\frac{1}{\sqrt{2}}, 1\right]) is correct because it provides the smallest interval that contains the value of the definite integral, which is bounded by the minimum and maximum values of the integrand.
93 Medium
93 of 596
Application +4 | -1
∫−3π2π2[(x+π)3+cos⁡2(x+3π)]dx\int\limits_ {-\frac{3 \pi}{2}}^{\frac{\pi}{2}}\left[(x+\pi)^3+\cos ^2(x+3 \pi)\right] d x is equal to
View answer and explanation
Correct option: C
π2\frac{\pi}{2}
Option C (π2\frac{\pi}{2}) is correct because the integral simplifies to an integral of an even function after substitution, and the odd function integrates to zero.
94 Medium
94 of 596
Analyzing +4 | -1
For y=f(x)=∫0x 2 ∣ t ∣ dty = f\left( x \right) = \int \limits_0^x \,2\,\left| {\,t\,} \right|\,dt, the tangent lines parallel to the bisector of the first quadrant angle are
View answer and explanation
Correct option: A
y=x  ±  14y = x\,\, \pm \,\,\frac{1}{4}
Option A (y=x  ±  14y = x\,\, \pm \,\,\frac{1}{4}) is correct because the tangent lines parallel to the bisector of the first quadrant are found by setting the derivative equal to 1 and solving for x.
95 Medium
95 of 596
Application +4 | -1
Value of ∫0x[sin⁡t]dt,(2n+1)π<x<(2n+2)π,n∈N\int\limits_{0}^x[\sin t] d t,(2 n+1) \pi < x < (2 n+2) \pi, n \in N is equal to
View answer and explanation
Correct option: D
(n+1)π−x(n+1) \pi-x
Option D ((n+1)π−x(n+1) \pi - x) is correct because the integral of [sin t] over the given interval simplifies to this expression after considering the intervals where sin t is negative.
96 Medium
96 of 596
Application +4 | -1
If [.] denotes the greatest integer function, then the value of the integral ∫−π/2π/2([xπ]+0.5)dx\int_{-\pi / 2}^{\pi / 2}\left(\left[\frac{x}{\pi}\right]+0.5\right) d x, is
View answer and explanation
Correct option: C
0
Option C (0) is correct because the integral evaluates to zero due to the symmetry and properties of the greatest integer function.
97 Medium
97 of 596
Application +4 | -1
If [x][x] is the greatest integer ≤x\leq x, then the value of the integral ∫−0.90.9([x2]+log⁡(2−x2+x))dx\int\limits_{{-0.9}}^{0.9}\left(\left[x^2\right]+\log \left(\frac{2-x}{2+x}\right)\right) d x is
View answer and explanation
Correct option: D
0
Option D (0) is correct because the integral simplifies to the sum of an even function and an odd function integrated over a symmetric interval, where the integral of the odd function is zero.
98 Medium
98 of 596
Analyzing +4 | -1
Limn→∞[1n2sec⁡21n2+2n2sec⁡24n2……+.+1nsec⁡21]  equals\mathop {{\text{Lim}}}\limits_{n \to \infty } \left[ {\frac{1}{{{n^2}}}{{\sec }^2}\frac{1}{{{n^2}}} + \frac{2}{{{n^2}}}{{\sec }^2}\frac{4}{{{n^2}}} \ldots \ldots + . + \frac{1}{n}{{\sec }^2}1} \right]\,\,{\text{equals}}
View answer and explanation
Correct option: D
12tan⁡1\frac{1}{2} \tan 1
Option D (12tan⁡1\frac{1}{2} \tan 1) is correct because the limit can be expressed as a definite integral, which evaluates to 12tan⁡1\frac{1}{2} \tan 1.
99 Medium
99 of 596
Application +4 | -1
Let f(x)=∫−1xet2dtf(x)=\int_{-1}^x e^{t^2} d t and h(x)=f(1+g(x))h(x)=f(1+g(x)) where g(x)g(x) is defined for all x,g′(x)x, g^{\prime}(x) exists for all xx, and g(x)<0g(x)<0 for x>0x>0. If h′(1)=eh^{\prime}(1)=e and g′(1)=1g^{\prime}(1)=1, then the possible values which g(1)g(1) can take
View answer and explanation
Correct option: C
-2
Option C is correct because by differentiating h(x) and using the given conditions, we can solve for g(1), which is -2.
100 Medium
100 of 596
Analyzing +4 | -1
If I=∫01  dx1+xπ/2I = \int \nolimits_0^1 \,\,\frac{{dx}}{{1 + {x^{\pi /2}}}}, then
View answer and explanation
Correct option: A
log⁡e  2<I<π/4{\log _e}\,\,2 < I < \pi /4
Option A (log⁡e,2<I<π/4{\log _e}\\,2 < I < \pi /4) is correct because it correctly bounds the value of the integral using inequalities.
101 Medium
101 of 596
Application +4 | -1
If f  (x)=ae2x + bex  +  cxf\,\,\left( x \right) = a{e^{2x}}\, + \,b{e^x}\,\, + \,\,c\mathrm{ }x satisfies the conditions f(0)=−1,  f′(log⁡  2)=31,  ∫0log⁡4  ( f  (x)  −  cx)  dx=392f\left( 0 \right) = - 1,\,\,f'\left( {\log \,\,2} \right) = 31,\,\,\int \limits_0^{\log 4} \,\,\left( {\,f\,\,\left( x \right)\,\, - \,\,c\mathrm{ }x} \right)\,\,dx = \frac{{39}}{2}, then

View answer and explanation
Correct option: A
a=5,  b=−6,  c=− 7a = 5,\,\,b = - 6,\,\,c = - \,7
Option A (a=5,  b=−6,  c=− 7a = 5,\,\,b = - 6,\,\,c = - \,7) is correct because it satisfies all three given conditions when substituted into the function and its derivative.
102 Medium
102 of 596
Application +4 | -1
Let ∫αlog⁡e4dxex−1=π6.\int\limits_ \alpha^{\log _e 4} \frac{{dx}}{\sqrt{{e}^{{x}}-1}}=\frac{\pi}{6}. Then eα{e}^\alpha and e−α{e}^{-\alpha} are the roots of the equation:
View answer and explanation
Correct option: A
2x2−5x+2=02 x^2-5 x+2=0
Option A (2x2−5x+2=02 x^2-5 x+2=0) is correct because eα{e}^\alpha and e−α{e}^{-\alpha} are the roots of this equation, derived from the given integral.
103 Medium
103 of 596
Analyzing +4 | -1
The integral ∫π/6π/3tan⁡3x⋅sin⁡23x(2sec⁡2x⋅sin⁡23x+3tan⁡x⋅sin⁡6x)dx\int\limits_{{\pi / 6}}^{\pi / 3} \tan ^3 x \cdot \sin ^2 3 x\left(2 \sec ^2 x \cdot \sin ^2 3 x+3 \tan x \cdot \sin 6 x\right) d x is equal to
View answer and explanation
Correct option: D
−118-\frac{1}{18}
Option D (-118\frac{{1}}{{18}}) is correct because the integral simplifies to the derivative of a product, which can be easily evaluated at the limits of integration.
104 Medium
104 of 596
Application +4 | -1
Value of ∫π23π2[2sin⁡x]dx\int_{\frac{\pi}{2}}^{\frac{3 \pi}{2}}[2 \sin x] d x is [.] →\rightarrow g.t.
View answer and explanation
Correct option: A
−π2-\frac{\pi}{2}
Option A (−π2\frac{{-\pi}}{2}) is correct because the integral of the greatest integer function of 2sin⁡(x)2\sin(x) from π2\frac{{\pi}}{2} to 3π2\frac{{3\pi}}{2} evaluates to −π2-\frac{{\pi}}{2}.
105 Medium
105 of 596
Analyzing +4 | -1
lim⁡t→0(11sin⁡2t+21sin⁡2t+…+n1sin⁡2t)sin⁡2t\lim \limits_{{t \rightarrow 0}}\left(1^{\frac{1}{\sin ^{2} t}}+2^{\frac{1}{\sin ^{2} t}}+\ldots+n^{\frac{1}{\sin ^{2} t}}\right)^{\sin ^{2} t} is equal to
View answer and explanation
Correct option: D
nn
Option D (n) is correct because the limit evaluates to n by considering the dominant term as t approaches 0.

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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
106 Medium
106 of 596
Application +4 | -1
∫01  x3   dx(x2  +  1)3/2\int \nolimits_0^1 \,\,\frac{{{x^3}\;\,dx}}{{{{\left( {{x^2}\,\, + \,\,1} \right)}^{3/2}}}} is equal to
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the calculated value of the integral ∫01  x3 dx(x2  + 1)3/2\int \nolimits_0^1 \,\,\frac{{{x^3}\,dx}}{{{{\left( {{x^2}\,\,+\,1} \right)}^{3/2}}}} does not match any of the given options.
107 Medium
107 of 596
Analyzing +4 | -1
The value of ∫06(x+12x−36+x−12x−36)dx\int_0^6(\sqrt{x+\sqrt{12 x-36}}+\sqrt{x-\sqrt{12 x-36}}) d x is equal to
View answer and explanation
Correct option: A
636 \sqrt{3}
Option C (ax2+bx+c=0a x^2+b x+c=0 has atleast one root in (−2,−1)(-2,-1)) is the correct answer because the given conditions imply that there must be a root in the interval (-2, -1).
108 Medium
108 of 596
Application +4 | -1
If A=∫0505π∣cos⁡x∣dxA=\int\limits_{0}^{505 \pi}|\cos x| d x and B=∫505π1007π∣sin⁡x∣dxB=\int\limits_{{505 \pi}}^{1007 \pi}|\sin x| d x, then A+BA+B is equal to
View answer and explanation
Correct option: B
2014
Option B (2014) is correct because evaluating the integrals A and B separately and then summing them gives 2014.
109 Medium
109 of 596
Application +4 | -1
The area bounded by y=xe∣x∣y = x{e^{\left| x \right|}} and lines ∣x∣=1,  y=0\left| x \right| = 1,\;y = 0 is
View answer and explanation
Correct option: D
2 sq unit
Option D (2 sq unit) is correct because it represents the sum of the absolute values of the definite integrals of xe∣x∣xe^{|x|} from -1 to 0 and 0 to 1.
110 Medium
110 of 596
Application +4 | -1
Area of the region bounded by the curve y=ex,x=0  &  y=ey=e^x, x=0 \, \, \& \, \, y=e is
View answer and explanation
Correct option: D
∫1eeydy\int\limits_{1}^e e^y d y
Option D (∫1eeydy\int\limits_{1}^e e^y d y) is correct because the area can be found by integrating with respect to y, with the limits of integration being the y-values of the region.
111 Medium
111 of 596
Analyzing +4 | -1
∫1/etan⁡xt1+t2dt+∫1/ecot⁡xdtt(1+t2)=\int\limits_{{1 / e}}^{\tan x} \frac{t}{1+t^2} d t+\int\limits_{{1 / e}}^{\cot x} \frac{d t}{t\left(1+t^2\right)}=
View answer and explanation
Correct option: C
1
Option C (1) is correct because the derivative of the given expression is zero, implying it's a constant, and evaluating it at x = pi/4 gives 1.
112 Medium
112 of 596
Application +4 | -1
The value of lim⁡n→∞1n∑j=1n(2j−1)+8n(2j−1)+4n\lim \limits_{{n \rightarrow \infty}} \frac{1}{n} \sum\limits_{{j=1}}^n \frac{(2 j-1)+8 n}{(2 j-1)+4 n} is equal to:
View answer and explanation
Correct option: D
1+2log⁡e(32)1+2 \log _e\left(\frac{3}{2}\right)
Option D (1+2log⁡e(32)1+2 \log _e(\frac{3}{2})) is correct because it is the result of evaluating the limit by converting it to a definite integral and solving the integral.
113 Medium
113 of 596
Analyzing +4 | -1
Area of region ∣cos⁡x−sin⁡x∣≤y≤sin⁡x|\cos x-\sin x| \leq y \leq \sin x for x∈(0,π/2)x \in(0, \pi / 2) is
View answer and explanation
Correct option: C
(5+1−22)(\sqrt{5}+1-2 \sqrt{2}) sq. unit
The correct answer is C (5+1−22\sqrt{5}+1-2 \sqrt{2}) sq. unit because it represents the area of the region calculated by integrating the difference between the functions.
114 Medium
114 of 596
Application +4 | -1
If pp and qq are different roots of the equation tan⁡x=x\tan x=x then ∫012sin⁡(pt)sin⁡(qt)dt\int\limits_{0}^1 2 \sin (p t) \sin (q t) d t is equal to
View answer and explanation
Correct option: A
0
Option A (0) is correct because the integral evaluates to 0 due to the roots satisfying tan x = x.
115 Medium
115 of 596
Application +4 | -1
If ∫dxx+x7=p(x)\int \frac{d x}{x+x^7}=p(x) then, ∫x6x+x7dx\int \frac{x^6}{x+x^7} d x is equal to:
View answer and explanation
Correct option: A
ln⁡∣x∣−p(x)+c\ln |x|-p(x)+c
Option A (ln⁡∣x∣−p(x)+c\ln |x|-p(x)+c) is correct because the integral can be separated and simplified using the given definition of p(x)p(x).
116 Medium
116 of 596
Application +4 | -1
The value of α∈(− π,   0)\alpha \in \left( { - \,\pi ,\;\,0} \right) satisfying sin⁡  α+∫α2αcos⁡  2  x dx=0\sin \,\,\alpha + \int \limits_\alpha ^{2\alpha } \cos \,\,2\,\,x\,dx = 0, is

View answer and explanation
Correct option: C
− π/3- \,\pi /3
Option C ( − π/3\ - \,\pi /3) is correct because it satisfies the given equation after evaluating the integral and solving the trigonometric equation.
117 Medium
117 of 596
Analyzing +4 | -1
The maximum value of ∫a−1a+1e−(x−1)2dx\int\limits_{{a-1}}^{a+1} e^{-(x-1)^2} d x is attained ( aa is real) at
View answer and explanation
Correct option: B
a=1a=1
Option B (a=1) is correct because the maximum value is attained when the derivative of the integral with respect to 'a' is zero, which occurs at a=1.
118 Medium
118 of 596
Application +4 | -1
∫0π/4  (cos⁡  x  −  sin⁡  x)  dx  +  ∫π/45π/4  (sin⁡  x  −  cos⁡  x)  dx  +  ∫2ππ/4  (cos⁡  x  −  sin⁡  x)  dx\int \nolimits_0^{\pi /4} \,\,(\cos \,\,x\,\, - \,\,\sin \,\,x)\,\,dx\,\, + \,\,\int \nolimits_{\pi /4}^{5\pi /4} \,\,(\sin \,\,x\,\, - \,\,\cos \,\,x)\,\,dx\,\, + \,\,\int \nolimits_{2\pi }^{\pi /4} \,\,(\cos \,\,x\,\, - \,\,\sin \,\,x)\,\,dx is equal to
View answer and explanation
Correct option: D
4  2−24\,\,\sqrt 2 - 2
Option D (4√2 - 2) is correct because the integral is evaluated by computing the definite integrals and summing the results, leading to the correct value.
119 Medium
119 of 596
Analyzing +4 | -1
The value of {∫−1/21/2cos⁡2x⋅log⁡(1+x1−x)dx}{∫01/2cos⁡2x⋅log⁡(1+x1−x)dx}\frac{\left\{\int\limits_{{-1 / 2}}^{1 / 2} \cos 2 x \cdot \log \left(\frac{1+x}{1-x}\right) d x\right\}}{\left\{\int\limits_{0}^{1 / 2} \cos 2 x \cdot \log \left(\frac{1+x}{1-x}\right) d x\right\}} is
View answer and explanation
Correct option: A
0
Option A (0) is correct because the numerator integral evaluates to zero since the integrand is an odd function integrated over a symmetric interval.
120 Medium
120 of 596
Application +4 | -1
If g(x)=∫0x  cos⁡4  t dt\mathrm{g}\left( x \right) = \int \nolimits_0^x \,\,{\cos ^4}\,\,t\,dt, then g(x  +  π)\mathrm{g}\left( {x\,\, + \,\,\pi } \right) equals
View answer and explanation
Correct option: A
g(x)  +  g(π)\mathrm{g}\left( x \right)\,\, + \,\,\mathrm{g}\left( \pi \right)
Option A (g(x) + g(π)) is correct because substituting t = π + y allows us to separate the integral into g(x) + g(π).
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