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JEE • MATHEMATICS • DEFINITE INTEGRATION

Definite integration practice · Mixed difficulty

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121 Easy
121 of 995
Analyzing +4 | -1
If f (x)f\,\left( x \right) is continuous for all real values of xx, then ∑r=110  ∫01(r−1+x)  dx\sum \limits_{r = 1}^{10} \mathop {\,\,\int }\nolimits_0^1 \left( {r - 1 + x} \right)\,\,dx is equal to
View answer and explanation
Correct option: A
∫010  f(x)  dx\int \nolimits_0^{10} \,\,f\left( x \right)\,\,dx
Option A (∫010  f(x)  dx\int\nolimits_0^{10} \,\,f\left( x \right)\,\,dx) is correct because the sum of the integrals simplifies to a single integral from 0 to 10 due to the continuity of f(x).
122 Medium
122 of 995
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.
123 Medium
123 of 995
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.
124 Hard
124 of 995
Application +4 | -1
The value of ∫−11ddx(tan⁡−11x)dx\int\limits_{{-1}}^1 \frac{d}{d x}\left(\tan ^{-1} \frac{1}{x}\right) d x, is
View answer and explanation
Correct option: C
−π/2-\pi / 2
Option C (-π2\frac{\pi}{2}) is correct because the derivative of tan⁡−1(1x)\tan^{-1}(\frac{1}{x}) is -11+x2\frac{1}{1+x^2}, and integrating this from -1 to 1 gives -π2\frac{\pi}{2}.
125 Medium
125 of 995
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.
126 Medium
126 of 995
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.
127 Easy
127 of 995
Application +4 | -1
If In=∫01xne−x dxI_n=\int\limits_{0}^1 x^n e^{-x} \, d x for n∈Nn \in N, then In−nIn−1=I_n-n I_{n-1}=
View answer and explanation
Correct option: C
−1/e-1 / e
Option C (-1 / e) is correct because it is obtained by applying integration by parts to the integral and simplifying the resulting expression.
128 Easy
128 of 995
Application +4 | -1
The value of ∫0π[2sin⁡x]dx\int\limits_{0}^\pi[2 \sin x] \mathrm{d} x, where [,] represents the greatest integer function is
View answer and explanation
Correct option: D
2π/32 \pi / 3
Option D (2π/32 \pi / 3) is correct. The integral is solved by breaking it into intervals where the greatest integer function has constant values and then evaluating the resulting integrals.
129 Medium
129 of 995
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.
130 Medium
130 of 995
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.
131 Medium
131 of 995
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.
132 Medium
132 of 995
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.
133 Medium
133 of 995
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.
134 Easy
134 of 995
Application +4 | -1
∫020π(∣sin⁡x∣+∣cos⁡x∣)2dx\int_0^{20 \pi}(|\sin x|+|\cos x|)^2 d x is equal to
View answer and explanation
Correct option: D
20(π+2)20(\pi+2)
The correct answer is D (20(π+2)20(\pi+2)) because it is the result of evaluating the definite integral ∫020π(∣sin⁡x∣+∣cos⁡x∣)2dx\int_0^{20 \pi}(|\sin x|+|\cos x|)^2 dx as shown in the step-by-step solution.
135 Medium
135 of 995
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.

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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
136 Medium
136 of 995
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).
137 Easy
137 of 995
Application +4 | -1
∫0π  x  dx1  +  sin⁡  x  is equal to\int \nolimits_0^\pi \,\,\frac{{x\,\,dx}}{{1\,\, + \,\,\sin \,\,x}}\,\,\mathrm{is }\,\mathrm{equal}\,\mathrm{ to}
View answer and explanation
Correct option: C
π\pi
Option C (π\pi) is correct because after applying the property of definite integrals and simplifying, the integral evaluates to π\pi.
138 Easy
138 of 995
Application +4 | -1
The value of ∫1/π2/πsin⁡1/xx2 dx\int\limits_{{1 / \pi}}^{2 / \pi} \frac{\sin 1 / x}{x^2} \mathrm{~d} x is equal
View answer and explanation
Correct option: A
1
Option A (1) is correct because substituting t=1xt = \frac{{1}}{{x}} transforms the integral to ∫π/2πsin⁡(t)dt\int_{\pi/2}^{\pi} \sin(t) dt, which evaluates to 1.
139 Easy
139 of 995
Application +4 | -1
If ∫01exex+e−x dx=log⁡e+A1+2\int\limits_{0}^1 \frac{\sqrt{e^x}}{\sqrt{e^x+e^{-x}}} \mathrm{~d} x=\log \frac{e+A}{1+\sqrt{2}} then AA is equal to
View answer and explanation
Correct option: B
e2+1\sqrt{e^2+1}
Option B (e2+1\sqrt{{e^2+1}}) is correct because after substituting ex=te^x = t, the integral evaluates to log⁡e+e2+11+2\log \frac{{e+\sqrt{{e^2+1}}}}{{1+\sqrt{{2}}}}, thus A=e2+1A = \sqrt{{e^2+1}}.
140 Medium
140 of 995
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.
141 Easy
141 of 995
Analyzing +4 | -1
∫−ππ∣π−∣x∣∣dx\int\limits_{ - \pi }^\pi {|\pi - |x||dx} is equal to
View answer and explanation
Correct option: C
π2{\pi ^2}
Option C π2{\pi ^2} is correct because the definite integral evaluates to this value after considering the absolute value function.
142 Easy
142 of 995
Application +4 | -1
The area bounded by y=−x2+2x+3  and  y=0  isy = - {x^2} + 2x + 3\;\mathrm{and}\;y = 0\;\mathrm{is}
View answer and explanation
Correct option: B
323\frac{{32}}{3} sq units
Option B (32/3 sq units) is correct because it is the result of integrating the quadratic function between its roots.
143 Medium
143 of 995
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.
144 Easy
144 of 995
Application +4 | -1
For k≥0,lim⁡n→∞1k+3k+….+(2n−1)knk+1=k \geq 0, \lim \limits_{{n \rightarrow \infty}} \frac{1^k+3^k+\ldots .+(2 n-1)^k}{n^{k+1}}=
View answer and explanation
Correct option: B
2kk+1\frac{2^k}{k+1}
Option B (2kk+1\frac{2^k}{k+1}) is correct because the limit of the sum can be expressed as a definite integral, which evaluates to 2kk+1\frac{2^k}{k+1}.
145 Easy
145 of 995
Application +4 | -1
A polynomial PP is positive for x>0x > 0, and the area of the region bounded by P(x)P(x), the xx-axis, and the vertical lines x=0x=0 and x=Kx=K is K2(K+3)/3K^2(K+3) / 3. The polynomial P(x)P(x) is
View answer and explanation
Correct option: A
x2+2xx^2+2 x
Option A (x2+2xx^2+2x) is correct because differentiating the given area function with respect to K yields the polynomial P(K), which simplifies to x2+2xx^2+2x.
146 Easy
146 of 995
Application +4 | -1
∫0π/2  cot⁡  xcot⁡  x  +  tan⁡  x  dx\int \nolimits_0^{\pi /2} \,\,\frac{{\sqrt {\cot \,\,x} }}{{\sqrt {\cot \,\,x\,\,} + \,\,\sqrt {\tan \,\,x} }}\,\,dx is equal to
View answer and explanation
Correct option: D
π4\frac{\mathrm{\pi }}{4}
Option D (π4\frac{\mathrm{\pi }}{4}) is correct because the integral can be solved using the property of definite integrals where ∫0af(x)dx=∫0af(a−x)dx\int \nolimits_0^a f(x) dx = \int \nolimits_0^a f(a-x) dx.
147 Medium
147 of 995
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.
148 Medium
148 of 995
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).
149 Easy
149 of 995
Analyzing +4 | -1
The value of ∫0π/2  sin⁡2  x  −  cos⁡2  xsin⁡3  x  +  cos⁡3  x  dx\int \nolimits_0^{\pi /2} \,\,\frac{{{{\sin }^2}\,\,x\,\, - \,\,{{\cos }^2}\,\,x}}{{{{\sin }^3}\,\,x\,\, + \,\,{{\cos }^3}\,\,x}}\,\,dx is
View answer and explanation
Correct option: A
0\mathrm{0}
Option A is correct because the integral evaluates to 0 due to the symmetry and properties of definite integrals.
150 Medium
150 of 995
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.
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