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JEE • MATHEMATICS • DEFINITE INTEGRATION

Definite integration practice · Mixed difficulty

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601 Easy
601 of 995
Application +4 | -1
The area between the curve y=x  exy = x\;{e^x} and y=x  e−xy = x\;{e^{ - x}} and the line x=1x = 1 in square unit, is
View answer and explanation
Correct option: D
2e\frac{2}{e} sq unit
Option D (2e\frac{2}{e} sq unit) is correct because the area between the curves is found by integrating the difference of the functions from 0 to 1.
602 Easy
602 of 995
Understanding +4 | -1
If f(x)=∫axt3etdtf(x)=\int\limits_{a}^x t^3 e^t d t, then ddxf(x)=\frac{d}{d x} f(x)=
View answer and explanation
Correct option: B
x3exx^3 e^x
Option B (x3exx^3 e^x) is correct because, by the Fundamental Theorem of Calculus, the derivative of the integral of a function is the function itself.
603 Medium
603 of 995
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.
604 Easy
604 of 995
Application +4 | -1
The value of the integral ∫−11ddx(tan⁡−11x)dx\int\limits_{{-1}}^1 \frac{\mathrm{d}}{\mathrm{d} x}\left(\tan ^{-1} \frac{1}{x}\right) \mathrm{d} x is
View answer and explanation
Correct option: C
−π/2-\pi / 2
Option C (−π/2-\pi / 2) is correct because it involves correctly evaluating the derivative and the definite integral, while accounting for the discontinuity of the arctangent function.
605 Medium
605 of 995
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}.
606 Medium
606 of 995
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.
607 Medium
607 of 995
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.
608 Easy
608 of 995
Application +4 | -1
∫−11∣x∣dx\int\limits_{{-1}}^1|x| d x is equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because the integral of |x| from -1 to 1 is the area under the curve, which is 1.
609 Easy
609 of 995
Application +4 | -1
If ∫0πxf(sin⁡x)dx=A∫0π/2f(sin⁡x)dx\int\limits_{0}^\pi x f(\sin x) d x=A \int\limits_{0}^{\pi / 2} f(\sin x) d x, then AA is
View answer and explanation
Correct option: B
π\pi
Option B (π\pi) 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, the integral transforms to A∫0π/2f(sin⁡x)dxA \int_0^{\pi/2} f(\sin x) dx, where A is found to be π\pi.
610 Easy
610 of 995
Application +4 | -1
If I=∫0π/2sin⁡θsin⁡θ+cos⁡θdθI=\int_0^{\pi / 2} \frac{\sin \theta}{\sin \theta+\cos \theta} d \theta, then I=I=
View answer and explanation
Correct option: C
π/4\pi / 4
Option C (π/4\pi / 4) is correct because the integral can be solved by using the property of definite integrals and adding the original integral to a modified version.
611 Easy
611 of 995
Application +4 | -1
∫01  ({x}  [x+1])dx\int \nolimits_0^1 \,\,\left( {\left\{ x \right\}\,\,\left[ {x + 1} \right]} \right)dx (where { ⋅  }\left\{ {\, \cdot \,\,} \right\} denotes fractional part of xx and [ ⋅ ] \left[ {\, \cdot \,} \right]\, denotes greatest integer of x)x) is equal to
View answer and explanation
Correct option: C
12\frac{1}{2}
Option C (12\frac{1}{2}) is correct because the integral simplifies to ∫01xdx\int_0^1 x dx due to the properties of fractional and greatest integer functions.
612 Medium
612 of 995
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.
613 Easy
613 of 995
Application +4 | -1
The area bounded by y=[x]y = \left[ x \right] and the two ordinates x=1 and x=1.7x = 1\,and\,x = 1.7 is
View answer and explanation
Correct option: D
710\frac{7}{{10}}
Option D, 710\frac{7}{{10}}, is the correct area because the greatest integer function [x] equals 1 between x = 1 and x = 1.7, so the area is simply the integral of 1 from 1 to 1.7.
614 Easy
614 of 995
Application +4 | -1
∫0π/2  sin⁡2  xsin⁡2x  +  cos⁡2x  dx\int \nolimits_0^{\mathrm{\pi }/2} \,\,\frac{{{{\sin }^2}\,\,x}}{{{{\sin }^2}x\,\, + \,\,{{\cos }^2}x}}\,\,dx is equal to
View answer and explanation
Correct option: A
π4\frac{\mathrm{\pi }}{4}
Option A (π4\frac{\mathrm{\pi }}{4}) is correct because the integral is solved using the property of definite integrals and symmetry.
615 Medium
615 of 995
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.

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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
616 Medium
616 of 995
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}.
617 Easy
617 of 995
Application +4 | -1
If f(x)={ecos⁡xsin⁡x for ∣x∣≤22 otherwise f(x)=\left\{\begin{array}{cc}e^{\cos x} \sin x & \text { for }|x| \leq 2 \\ 2 & \text { otherwise }\end{array}\right. then ∫−23f(x)dx=\int\limits_{{-2}}^3 f(x) d x=
View answer and explanation
Correct option: C
2
Option C is correct because the integral can be solved by splitting the integral and recognizing the symmetry of the first part.
618 Medium
618 of 995
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].
619 Easy
619 of 995
Application +4 | -1
Let 5f(x)+4f(1x)=1x+3,x>05 f(x)+4 f\left(\frac{1}{x}\right)=\frac{1}{x}+3, x > 0. Then 18∫12f(x)dx18 \int\limits_{{1}}^{2} f(x) d x is equal to
View answer and explanation
Correct option: C
10log⁡e2−610 \log _{e} 2-6
Option C (10log⁡e2−610 \log _{e} 2-6) is correct because it is the value of the definite integral after solving the functional equation.
620 Medium
620 of 995
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.
621 Easy
621 of 995
Application +4 | -1
The value of ∫011−x1+xdx\int\limits_{0}^1 \sqrt{\frac{1-x}{1+x}} d x, is
View answer and explanation
Correct option: B
π2−1\frac{\pi}{2}-1
Option B is correct. The integral is evaluated by rationalizing the denominator and then splitting the integral into two simpler integrals.
622 Easy
622 of 995
Application +4 | -1
∫−22  ∣[x]∣    dx\int \nolimits_{ - 2}^2 \,\,\left| {\left[ x \right]} \right|\;\,\,dx is equal to
View answer and explanation
Correct option: D
4\mathrm{4}
Option D (4) is correct because the integral of the absolute value of the greatest integer function over the given interval evaluates to 4.
623 Easy
623 of 995
Analyzing +4 | -1
The value of lim⁡x→1∫0(x−1)2tcos⁡t2dt(x−1)sin⁡(x−1)\lim \limits_{{x \rightarrow 1}} \frac{\int\limits_{0}^{(x-1)^2} t \cos t^2 d t}{(x-1) \sin (x-1)}
View answer and explanation
Correct option: A
is equal to 0
Option A (is equal to 0) is correct because applying L'Hopital's rule twice results in a limit of 0.
624 Easy
624 of 995
Application +4 | -1
If In=∫−nn({x+1}{x2+2}+{x2+3}{x3+4})dxI_n=\int_{-n}^n\left(\{x+1\}\left\{x^2+2\right\}+\left\{x^2+3\right\}\left\{x^3+4\right\}\right) d x, (where, ff denotes the fractional part), then I1I_1 is equal to
View answer and explanation
Correct option: B
−23-\frac{2}{3}
Option D (None of these) is correct because the integral evaluates to -2/3, which is not among the provided options. The integral is simplified using properties of fractional parts and symmetry.
625 Medium
625 of 995
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.
626 Medium
626 of 995
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.
627 Medium
627 of 995
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.
628 Easy
628 of 995
Application +4 | -1
If 1a∫1a(32x+1−1x)dx<4\frac{1}{\sqrt{a}} \int\limits_{1}^a\left(\frac{3}{2} \sqrt{x}+1-\frac{1}{\sqrt{x}}\right) dx < 4, then 'aa' may take values:
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the inequality simplifies to a+a−6<0a + \sqrt{a} - 6 < 0, which leads to a∈[0,4)a \in [0, 4), and none of the given options satisfy this condition while also being defined.
629 Easy
629 of 995
Application +4 | -1
∫−44∣x+2∣dx=\int\limits_{{-4}}^4|x+2| d x=
View answer and explanation
Correct option: C
20
Option C (20) is correct because it is the result of evaluating the definite integral of the absolute value function.
630 Easy
630 of 995
Application +4 | -1
The value of the integral ∫−π/2π/2  cos⁡  x  −  cos⁡3  x  dx\int \limits_{ - \pi /2}^{\pi /2} \,\,\sqrt {\cos \,\,x\,\, - \,\,{{\cos }^3}\,\,x\,\,} dx is
View answer and explanation
Correct option: C
4 / 3\mathrm{4}\,\mathrm{/}\,\mathrm{3}
Option C (43\frac{\mathrm{4}}{\mathrm{3}}) is correct because the integral is solved by simplifying the expression under the square root and using substitution.
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