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JEE • MATHEMATICS • DEFINITE INTEGRATION

Definite integration practice · Medium

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481 Medium
481 of 596
Application +4 | -1
If for non-zero x,af(x)+bf(1x)=1x−5x, a f(x)+b f\left(\frac{1}{x}\right)=\frac{1}{x}-5, where a≠ba \neq b, then ∫12f(x)dx=\int\limits_{1}^2 f(x) d x=____.
View answer and explanation
Correct option: A
1a2−b2[aln⁡2+7b2−5a]\frac{1}{a^2-b^2}\left[a \ln 2+\frac{7 b}{2}-5 a\right]
Option A (1a2−b2[aln⁡2+7b2−5a]\frac{{1}}{{a^2-b^2}}[a \ln 2+\frac{{7b}}{{2}}-5a]) is correct because it is the result of solving the system of equations for f(x) and then integrating from 1 to 2.
482 Medium
482 of 596
Analyzing +4 | -1
If f(x+f(y))=f(x)+yf(x+f(y))=f(x)+y for all x,y∈Rx, y \in R and f(0)=1f(0)=1, then ∫010f(10−x)dx\int_0^{10} f(10-x) d x is equal to
View answer and explanation
Correct option: D
10∫01f(x)dx10 \int_0^1 f(x) d x
The correct answer is D (10∫01f(x)dx10 \int_0^1 f(x) d x) because the function is periodic with period 1, allowing simplification of the integral.
483 Medium
483 of 596
Application +4 | -1
The value of ∫0π∣cos⁡x∣3 dx\int\limits_{0}^\pi|\cos x|^3 \mathrm{~d} x is:
View answer and explanation
Correct option: B
43\frac{4}{3}
Option B (43\frac{4}{3}) is correct because it is the value obtained after evaluating the definite integral of the absolute value of cosine cubed over the given interval.
484 Medium
484 of 596
Application +4 | -1
The area between y=xy=x and y=x2,0≤x≤2y=x^2, 0 \leq x \leq 2 is
View answer and explanation
Correct option: A
1
Option C (1/3) is incorrect, the correct answer is 1 because it is the sum of two integrals, one from 0 to 1 and another from 1 to 2, representing the area between the curves.
485 Medium
485 of 596
Analyzing +4 | -1
If ∫sin⁡x1t2f(t)dt=1−sin⁡x\int\limits_{{\sin x}}^1 t^2 f(t) d t=1-\sin x for all x∈[0,π/2]x \in[0, \pi / 2], then f(13)=f\left(\frac{1}{\sqrt{3}}\right)=
View answer and explanation
Correct option: A
3
Option C (3) is correct because differentiating both sides using Leibniz's rule and solving for f(x)f(x) gives f(13)=3f(\frac{{1}}{{\sqrt{{3}}}})=3.
486 Medium
486 of 596
Application +4 | -1
∫01  ddx  [sin⁡−1  (2x1+x2)]    dx\int \nolimits_0^1 \,\,\frac{d}{{dx}}\,\,\left[ {{{\sin }^{ - 1}}\,\,\left( {\frac{{2x}}{{1 + {x^2}}}} \right)} \right]\,\,\;dx is equal to
View answer and explanation
Correct option: C
π2\frac{\mathrm{\pi }}{2}
Option C (π2\frac{\mathrm{\pi }}{2}) is correct because the integral of a derivative is the original function evaluated at the limits of integration.
487 Medium
487 of 596
Analyzing +4 | -1
If f(x)=∣ sin⁡  x  +  sin⁡  2x  +  sin⁡  3xsin⁡  2xsin⁡  3x3  +  4sin⁡  x34  sin⁡  x1  +  sin⁡  xsin⁡  x1 ∣f\left( x \right) = \left| {\,\begin{array}{cccccccccccccccccccc} {\sin \,\,x\,\, + \,\,\sin \,\,2x\,\, + \,\,\sin \,\,3x}&{\sin \,\,2x}&{\sin \,\,3x}\\ {3\,\, + \,\,4\sin \,\,x}&3&{4\,\,\sin \,\,x}\\ {1\,\, + \,\,\sin \,\,x}&{\sin \,\,x}&1 \end{array}\,} \right|, then the value of ∫0π/2 f(x)dx\int \nolimits_0^{\pi /2} \,f\left( x \right)dx is
View answer and explanation
Correct option: C
13\frac{1}{3}
Option C (13\frac{1}{3}) is correct because after simplifying the determinant, the resulting integral evaluates to 13\frac{1}{3}.
488 Medium
488 of 596
Application +4 | -1
The area bounded by curves y=cos⁡xy=\cos x and y=sin⁡xy=\sin x and ordinates x=0x=0 and x=π4x=\frac{\pi}{4} is
View answer and explanation
Correct option: C
2−1\sqrt{2}-1
Option C (2−1\sqrt{2}-1) is correct because the area bounded by the curves y=cos⁡xy=\cos x and y=sin⁡xy=\sin x and ordinates x=0x=0 and x=π4x=\frac{\pi}{4} is found by integrating the difference of the functions from 0 to π4\frac{\pi}{4}, which results in 2−1\sqrt{2}-1.
489 Medium
489 of 596
Understanding +4 | -1
If I=1π∫02πecos⁡θcos⁡(sin⁡θ)dθI=\frac{1}{\pi} \int\limits_{0}^{2 \pi} e^{\cos \theta} \cos (\sin \theta) d \theta, then find the value of II.
View answer and explanation
Correct option: A
2
Option A (2) is correct because the integral represents the real part of the power series expansion of eze^z integrated over a circle, which simplifies to 2.
490 Medium
490 of 596
Application +4 | -1
∫0πecos⁡xecos⁡x+e−cos⁡xdx\int\limits_{0}^\pi \frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} d x is equal to
View answer and explanation
Correct option: B
π2\frac{\pi}{2}
Option B (π/2) is correct because by using the property of definite integrals and adding the original integral to a transformed version, the integral simplifies to π/2.
491 Medium
491 of 596
Application +4 | -1
The value of ∫−π/2π/2    (x3  +  x  cos⁡  x  +  tan⁡5  x  +  1)  dx\int \limits_{ - \pi /2}^{\pi /2} \,\,\,\,\left( {{x^3}\,\, + \,\,x\,\,\cos \,\,x\,\, + \,\,{{\tan }^5}\,\,x\,\, + \,\,1} \right)\,\,dx is equal to
View answer and explanation
Correct option: C
π\pi
Option C (π\pi) is correct because the integral of the odd functions over the symmetric interval is zero, leaving only the integral of 1 from -π\pi/2 to π\pi/2.
492 Medium
492 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 it is the value of m for which the area between the curves equals 9/2.
493 Medium
493 of 596
Application +4 | -1
A function f(x)f(x) satisfying ∫01f(tx)dt=nf(x)\int\limits_{0}^1 f(t x) d t=n f(x), where x>0x>0, is
View answer and explanation
Correct option: A
f(x)=C⋅x1−nnf(x)=C \cdot x^{\frac{1-n}{n}}
Option A is correct because the function satisfies the given integral equation, which leads to a differential equation that can be solved.
494 Medium
494 of 596
Analyzing +4 | -1
If the value of the integral ∫05X+[X]ex−[X]dx=αe−1×β\int_0^5 {\frac{{X + \left[ X \right]}}{{{e^{x - \left[ X \right]}}}}} dx = \alpha {e^{ - 1}} \times \beta, where α,  β∈R,  5α+6β=0\alpha ,\,\,\beta \in R,\,\,5\alpha + 6\beta = 0 and [X][X] denotes the greatest integer less than or equal to xx, then the value of (α+β)2{\left( {\alpha + \beta } \right)^2} is equal to:
View answer and explanation
Correct option: B
25
Option B (25) is correct because after evaluating the integral and using the given condition 5α+6β=05\alpha + 6\beta = 0, we find that (α+β)2=25(\alpha + \beta)^2 = 25.
495 Medium
495 of 596
Analyzing +4 | -1
The minimum value of the function f(x)=∫02e∣x−t∣dtf(x) = \int\limits_{{0}}^{2} e^{|x-t|} d t is
View answer and explanation
Correct option: C
2(e−1)2(e-1)
Option C (2(e−1)2(e-1)) is correct because it represents the minimum value of the integral function, achieved within the interval [0, 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
496 Medium
496 of 596
Analyzing +4 | -1
lim⁡x→∞(x3∫−1/x1/xln⁡(1+t2)1+etdt)\lim \limits_{{x \rightarrow \infty}}\left(x^3 \int_{-1 / x}^{1 / x} \frac{\ln \left(1+t^2\right)}{1+e^t} d t\right) is equal to
View answer and explanation
Correct option: A
13\frac{1}{3}
Option A (1/3) is correct because by using King's property and L'Hopital's rule, the limit evaluates to 1/3.
497 Medium
497 of 596
Analyzing +4 | -1
The value of ∫−π3πlog⁡(sec⁡θ−tan⁡θ)dθ\int\limits_{{-\pi}}^{3 \pi} \log (\sec \theta-\tan \theta) d \theta is
View answer and explanation
Correct option: B
0
Option B (0) is correct because the integral of log⁡(sec⁡θ−tan⁡θ)\log(\sec \theta - \tan \theta) over the given interval is zero due to symmetry and properties of logarithms.
498 Medium
498 of 596
Application +4 | -1
∫0113+x+1+xdx=a+b2+c3\int\limits_{0}^1 \frac{1}{\sqrt{3+x}+\sqrt{1+x}} d x=a+b \sqrt{2}+c \sqrt{3}, then 2a−3b−4c2 a-3 b-4 c is equal to
View answer and explanation
Correct option: C
12
Option C (12) is correct. After rationalizing the integral and evaluating, the values of a, b, and c are found, and the expression 2a-3b-4c evaluates to 12.
499 Medium
499 of 596
Application +4 | -1
∫−22∣x(x−1)∣dx=\int\limits_{{-2}}^2|x(x-1)| d x=
View answer and explanation
Correct option: A
173\frac{17}{3}
Option A (173\frac{17}{3}) is correct because after splitting the integral based on the absolute value and evaluating, the result is 173\frac{17}{3}.
500 Medium
500 of 596
Application +4 | -1
Let f(x)f(x) be a continuous function such that the area bounded by the curve y=f(x)y=f(x), the xx-axis, and the lines x=0x=0 and x=ax=a is 1+a22sin⁡a.1+\frac{a^2}{2} \sin a. Then,
View answer and explanation
Correct option: C
f(a)=asin⁡a+12a2cos⁡af(a)=a \sin a+\frac{1}{2} a^2 \cos a
Option C (f(a)=asin⁡a+12a2cos⁡af(a)=a \sin a+\frac{1}{2} a^2 \cos a) is correct because it is obtained by differentiating the given integral equation using the Fundamental Theorem of Calculus and the product rule.
501 Medium
501 of 596
Application +4 | -1
If f(a+b−x)=f(x)f(a+b-x)=f(x), then ∫abxf(x)dx=\int\limits_{a}^b x f(x) d x=
View answer and explanation
Correct option: B
a+b2∫abf(x)dx\frac{a+b}{2} \int\limits_{a}^b f(x) d x
Option B (a+b2∫abf(x)dx\frac{a+b}{2} \int_a^b f(x) dx) is correct because, by using the given property f(a+b−x)=f(x)f(a+b-x)=f(x) and applying the definite integral property, the integral simplifies to a+b2∫abf(x)dx\frac{a+b}{2} \int_a^b f(x) dx.
502 Medium
502 of 596
Application +4 | -1
A function f(x)f(x) safisfying f(x)=∫0xetf(t)dt+exf(x)=\int\limits_{0}^x e^t f(t) d t+e^x is
View answer and explanation
Correct option: A
2eex−1−12 e^{e^x-1}-1
Option A (2eex−1−12 e^{e^x-1}-1) is correct because it satisfies the given integral equation after differentiation and integration.
503 Medium
503 of 596
Understanding +4 | -1
If An=∫0π/2sin⁡(2n−1)xsin⁡xdxA_n=\int\limits_{0}^{\pi / 2} \frac{\sin (2 n-1) x}{\sin x} d x then value of A1000A10\frac{A_{1000}}{A_{10}} is -
View answer and explanation
Correct option: A
1
Option D (Never less than nn) is correct because, by the AM-GM inequality, the sum of positive numbers whose product is 1 is always greater than or equal to the number of terms.
504 Medium
504 of 596
Application +4 | -1
If I=∫0100π(1−cos⁡2x)dxI=\int\limits_{0}^{100 \pi} \sqrt{(1-\cos 2 x)} d x, then the value of II is
View answer and explanation
Correct option: B
2002200 \sqrt{2}
Option B (2002200\sqrt{2}) is correct because the integral evaluates to 2002200\sqrt{2} after using the periodicity of the integrand.
505 Medium
505 of 596
Analyzing +4 | -1
Let [t][t] denote the greatest integer less than or equal to t. Then the value of the integral ∫−3101([sin⁡(πx)]+e[cos⁡(2πx)])dx\int_{-3}^{101}\left([\sin (\pi x)]+e^{[\cos (2 \pi x)]}\right) d x is equal to
View answer and explanation
Correct option: B
52e\frac{52}{e}
Option B is correct. The integral is evaluated by using the periodicity of the functions and simplifying the integral.
506 Medium
506 of 596
Application +4 | -1
If a function f:[0,27]→Rf:[0,27] \rightarrow \mathbf{R} is differentiable then for some 0<α<β<3,∫027f(x)dx0<\alpha<\beta<3, \int\limits_{0}^{27} f(x) \mathrm{d} x is equal to
View answer and explanation
Correct option: C
3[α2f(α3)+12β2f(β3)]3\left[\alpha^2 f\left(\alpha^3\right)+\frac{1}{2} \beta^2 f\left(\beta^3\right)\right]
Option C is correct because it correctly applies the Mean Value Theorem for integrals and derivatives to the given function and interval.
507 Medium
507 of 596
Analyzing +4 | -1
Let m be any integer. Then, the integral ∫0π  sin⁡2  m xsin⁡  x  dx\int \limits_0^\pi \,\,\frac{{\sin 2\,\,m\mathrm{ }\,x}}{{\sin \,\,x}}\,\,dx equals
View answer and explanation
Correct option: A
0\mathrm{0}
Option A (0) is correct because the integral evaluates to zero due to the properties of trigonometric functions and the given limits.
508 Medium
508 of 596
Application +4 | -1
The value of ∫0π/2sin⁡2/3xsin⁡2/3x+cos⁡2/3xdx\int\limits_{0}^{\pi / 2} \frac{\sin ^{2 / 3} x}{\sin ^{2 / 3} x+\cos ^{2 / 3} x} d x is
View answer and explanation
Correct option: A
π/4\pi / 4
Option A (π/4\pi / 4) is correct because the integral can be solved 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 adding the two integrals.
509 Medium
509 of 596
Application +4 | -1
Let ddxF(x)=(esin⁡xx);x>0\frac{d}{d x} F(x)=\left(\frac{e^{\sin x}}{x}\right) ; x > 0. If ∫143xesin⁡x3dx=F(k)−F(1)\int\limits_{1}^4 \frac{3}{x} e^{\sin x^3} d x=F(k)-F(1), then one of the possible value of kk is
View answer and explanation
Correct option: D
64
Option D (64) is correct because by substituting x3=tx^3 = t, the integral transforms into a form that allows direct comparison with the given function F(x).
510 Medium
510 of 596
Application +4 | -1
A minimum value of ∫0xtet2dt\int \nolimits_0^x t{e^{{t^2}}}dt is
View answer and explanation
Correct option: A
0
Option A (0) is correct because the minimum value of the integral is found by finding the critical points and checking the second derivative.
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