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JEE • MATHEMATICS • DEFINITE INTEGRATION

Definite integration practice · Easy

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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481 Easy
481 of 791
Analyzing +4 | -1
Given that f(a+b+1−x)=f(x)f(a+b+1-x)=f(x) for all x∈Rx \in R. The value 1a+b∫abx{f(x)+f(x+1)}dx\frac{1}{a+b} \int_a^b x\{f(x)+f(x+1)\} d x is equal to
View answer and explanation
Correct option: D
∫a−1b−1f(x+1)dx\int_{a-1}^{b-1} f(x+1) d x
The correct answer is C (∫a−1b−1f(x+1)dx\int_{a-1}^{b-1} f(x+1) d x) because using the given property and substitution, the integral can be simplified to this form.
482 Easy
482 of 791
Application +4 | -1
The value of ∫−ππcos⁡2  x1  +  ax  dx,  a  > 0  is\int \nolimits_{ - \pi }^\pi \frac{{{{\cos }^2}\,\,x}}{{1\,\, + \,\,{a^x}}}\,\,dx,\,\,a\,\, > \,0\,\,\mathrm{is} is

View answer and explanation
Correct option: C
π2\frac{\pi }{2}
Option C (π2\frac{\pi}{2}) is correct because the integral can be solved using the property of definite integrals and trigonometric identities.
483 Easy
483 of 791
Application +4 | -1
Consider the integral
l = ∫010[x]e[x]ex−1dxl\, = \,\int\limits_0^{10} {\frac{{\left[ x \right]{e^{\left[ x \right]}}}}{{{e^{x - 1}}}}} \mathrm{ }dx, Where [x][x] denotes the greatest integer less than or equal to xx. Then, the value of ll is equal to
View answer and explanation
Correct option: C
45(e−1)45\left( {e-1} \right)
Option C (45(e−1)45(e-1)) is correct because the integral can be simplified by breaking it into intervals and using the properties of the greatest integer function.
484 Easy
484 of 791
Application +4 | -1
If x=∫0ydt1+4t2x=\int_0^y \frac{d t}{\sqrt{1+4 t^2}} and d2ydx2=ky\frac{d^2 y}{d x^2}=k y, then find kk.
View answer and explanation
Correct option: D
4
Option D (4) is correct because after differentiating the given equation twice and simplifying, we find that d2ydx2=4y\frac{{d^2 y}}{{dx^2}} = 4y.
485 Easy
485 of 791
Application +4 | -1
∫510  1(x−1)  (x−2)    dx\int \nolimits_5^{10} \,\,\frac{1}{{\left( {x - 1} \right)\,\,\left( {x - 2} \right)}}\;\,\,dx is equal to
View answer and explanation
Correct option: B
log⁡  3227\log \,\,\frac{{32}}{{27}}
Option B log⁡ 3227\log \,\frac{{32}}{{27}} is correct because it is the result of evaluating the definite integral after partial fraction decomposition.
486 Easy
486 of 791
Application +4 | -1
Suppose that the graph of y=f(x)y=f(x), contains the points (0,4)(0,4) and (2,7)(2,7). If f′f^{\prime} is continuous then ∫02f′(x)dx\int\limits_{0}^2 f^{\prime}(x) d x is equal to
View answer and explanation
Correct option: C
3
Option C (3) is correct because the integral of f′(x)f'(x) from 0 to 2 is simply f(2)−f(0)f(2) - f(0), which is given as 7 - 4 = 3.
487 Easy
487 of 791
Analyzing +4 | -1
The integral ∫π/6π/4dxsin⁡2x(tan⁡5x+cot⁡5x)\int\limits_{\pi /6}^{\pi /4} {\frac{{dx}}{{\sin 2x({{\tan }^5}x + {{\cot }^5}x)}}} equals
View answer and explanation
Correct option: C
110(π4−tan⁡−1(193))\frac{1}{{10}}\left( {\frac{\pi }{4} - {{\tan }^{ - 1}}\left( {\frac{1}{{9\sqrt 3 }}} \right)} \right)
Option C (110(π4−tan⁡−1(193))\frac{1}{{10}}\left( {\frac{\pi }{4} - {{\tan }^{ - 1}}\left( {\frac{1}{{9\sqrt 3 }}} \right)} \right)) is correct because it is the result of correctly integrating the given expression after appropriate substitution.
488 Easy
488 of 791
Understanding +4 | -1
The value of ∫−11  x∣ x ∣  dx\int \limits_{ - 1}^1 \,\,x\left| {\,x\,} \right|\,\,dx, is
View answer and explanation
Correct option: C
0\mathrm{0}
Option C (0) is correct because the function f(x)=x∣x∣f(x) = x|x| is an odd function, and the integral of an odd function over a symmetric interval [−1,1][-1, 1] is always zero.
489 Easy
489 of 791
Application +4 | -1
The area between the curves y=x2y=x^2 and y=x1/3y=x^{1 / 3}, −1≤x≤1-1 \leq x \leq 1 is
View answer and explanation
Correct option: D
32\frac{3}{2}
Option D (3/2) is correct because it is the area calculated by integrating the absolute difference between the two functions over the given interval.
490 Easy
490 of 791
Application +4 | -1
The value of ∫−11x∣x∣dx\int\limits_{{-1}}^1 x|x| d x is
View answer and explanation
Correct option: C
0
Option B (2(2π3−log⁡(2+3))2(\frac{{2 \pi}}{{3}}-\log (2+\sqrt{{3}}))) is correct because the integrand is an even function, and the integral is evaluated using integration by parts.
491 Easy
491 of 791
Application +4 | -1
lim⁡n→∞  1n2  ∑r=1n  r  er/n=\lim\limits_{n \to \infty } \,\,\frac{1}{{{n^2}}}\,\,\sum \limits_{r = 1}^n \,\,r\;{e^{r/n}} =
View answer and explanation
Correct option: B
1\mathrm{1}
The correct answer is B (1) because the limit of the Riemann sum can be evaluated as a definite integral, which equals 1 after integration by parts.
492 Easy
492 of 791
Application +4 | -1
If AA is the area of the region bounded by the curve y=3x+4,  xy = \sqrt {3x + 4} ,\;x-axis and the lines x=−1x = - 1 and x=4x = 4 and BB is that area bounded by curve y2=3x+4,  x{y^2} = 3x + 4,\;x-axis and the liens x=−1x = - 1 and x=4x = 4, then A:BA:B is equal to
View answer and explanation
Correct option: C
1:21:2
Option C (1:21:2) is correct because the area B is twice the area A since it includes both the positive and negative square root of the function.
493 Easy
493 of 791
Application +4 | -1
Area bounded by the curve y=tan⁡−11Xy=\tan ^{-1} 1_X and y=cot⁡−11Xy=\cot ^{-1} 1_X between x=0x=0 and x=1x=1 is
View answer and explanation
Correct option: A
ln⁡2\ln 2 sq. units
Option A (ln⁡2\ln 2 sq. units) is correct because the area bounded by the curves is found by integrating the difference between the two inverse trigonometric functions from 0 to 1.
494 Easy
494 of 791
Application +4 | -1
The value of ∫01x dx(x2+1)2\int\limits_{0}^1 \frac{x \mathrm{~d} x}{\left(x^2+1\right)^2} is
View answer and explanation
Correct option: D
14\frac{1}{4}
Option D (14\frac{{1}}{{4}}) is correct because by substituting x2+1=tx^2+1=t, the integral ∫01xdx(x2+1)2\int_0^1 \frac{{x dx}}{{(x^2+1)^2}} can be easily evaluated to 14\frac{{1}}{{4}}.
495 Easy
495 of 791
Understanding +4 | -1
∫−11sin⁡3xcos⁡2xdx=\int\limits_{{-1}}^1 \sin ^3 x \cos ^2 x d x=
View answer and explanation
Correct option: A
0
Option A (0) is correct because the integrand is an odd function, and the integral of an odd function over a symmetric interval is zero.

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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 Easy
496 of 791
Application +4 | -1
lim⁡n→∞  1n  ∑r=1n  rn2+  r2\lim\limits_{n \to \infty } \,\,\frac{1}{n}\,\,\sum \limits_{r = 1}^n \,\,\frac{r}{{\sqrt {{n^2} + \,\,{r^2}} }} is equal to
View answer and explanation
Correct option: B
2−1\sqrt 2 - 1
Option B (2−1\sqrt 2 - 1) is correct because it is the result of converting the limit of the sum to a definite integral and evaluating it.
497 Easy
497 of 791
Application +4 | -1
∫02π (sin⁡  x  +  ∣ sin⁡  x ∣)dx    is    equal    to\int \nolimits_0^{2\pi } \,\left( {\sin \,\,x\,\, + \,\,\left| {\,\sin \,\,x\,} \right|} \right)dx\,\,\;\mathrm{is\;}\,\,\mathrm{equal\;}\,\,\mathrm{to}
View answer and explanation
Correct option: A
4\mathrm{4}
Option A (4) is correct because the integral of the absolute value of sin(x) from 0 to 2π\pi is 4, and the integral of sin(x) from 0 to 2π\pi is 0.
498 Easy
498 of 791
Application +4 | -1
The area lying between parabola y2=4ax{y^2} = 4ax and it’s latusrectum is
View answer and explanation
Correct option: C
83a2\frac{8}{3}{a^2} sq unit
Option C (8/3 a2 sq unit) is correct because it is the area calculated by integrating the function representing the parabola from 0 to a, and multiplying by 2 to account for symmetry.
499 Easy
499 of 791
Application +4 | -1
The difference between the greatest and least values of the function
F(x)=∫0x(t+1)dtF(x)=\int\limits_{0}^x(t+1) \mathrm{d} t on [1,3][1,3] is
View answer and explanation
Correct option: C
6
Option C (6) is correct because it represents the difference between the maximum and minimum values of the integral function on the given interval, calculated by integrating and evaluating at the bounds.
500 Easy
500 of 791
Analyzing +4 | -1
If the area bounded by the xx-axis, the curve y=f(x)y = f\left( x \right) and lines x=ax = a and x=bx = b is independent of b,  ∀  b>ab,\;\forall \;b > a (a is a constant), then ff is
View answer and explanation
Correct option: A
The zero function
Option A (The zero function) is correct because if the area is independent of the upper limit of integration, the function must be zero.
501 Easy
501 of 791
Application +4 | -1
The area of the figure bounded by the curves y=ex,y=e−xy = {e^x},y = {e^{ - x}} and the straight line x=1x = 1 is
View answer and explanation
Correct option: A
e+1ee + \frac{1}{e}
Option B (e−1ee - \frac{1}{e}) is incorrect. The correct answer is C (e+1e−2e + \frac{1}{e} - 2). The provided solution is missing the subtraction of 2.
502 Easy
502 of 791
Application +4 | -1
∫02(∣2x2−3x∣+[x−12])dx,\int\limits_0^2 {\left( {\left| {2{x^2} - 3x} \right| + \left[ {x - \frac{1}{2}} \right]} \right)dx} , where [t][t] is the greatest integer function, is equal to
View answer and explanation
Correct option: B
1912\frac{19}{12}
Option B (1912\frac{19}{12}) is correct because the integral is correctly evaluated by splitting the intervals and applying the definitions of absolute value and greatest integer function.
503 Easy
503 of 791
Analyzing +4 | -1
The value of ∫23  x  +  1x2  (x−1)    dx\int \nolimits_2^3 \,\,\frac{{x\,\, + \,\,1}}{{{x^2}\,\,\left( {x - 1} \right)}}\;\,\,dx is
View answer and explanation
Correct option: B
log⁡  169  −  16\log \,\,\frac{{16}}{9}\,\, - \,\,\frac{1}{6}
Option B (log⁡  169  −  16\log \,\,\frac{{16}}{9}\,\, - \,\,\frac{1}{6}) is correct because the integral is solved by partial fraction decomposition.
504 Easy
504 of 791
Analyzing +4 | -1
Let f(x)=∫0xetf(t)dt+exf(x) = \int_0^x {{e^t}f(t)dt + {e^x}} be a differentiable function for all x∈Rx \in R. Then, f(x)f(x) equals
View answer and explanation
Correct option: A
2e(ex−1)−12{e^{({e^x} - 1)}} - 1
Option A (2e(ex−1)−12{e^{({e^x} - 1)}} - 1) is correct because it is the solution to the given integral equation after differentiation and integration.
505 Easy
505 of 791
Analyzing +4 | -1
Statement I The value of the integral ∫π/6π/3dx1+tan⁡x\int_{\pi /6}^{\pi /3} {\frac{{dx}}{{1 + \sqrt {\tan x} }}} is equal to π/6\pi /6
Statement II ∫abf(x)dx=∫abf(a+b−x)dx\int_a^b {f(x)dx = \int_a^b {f(a + b - x)dx} }
View answer and explanation
Correct option: D
Statement I is false, Statement II is true
Option D (Statement I is false, Statement II is true) is correct because the integral's value is calculated to be π12\frac{\pi }{{12}}, not π6\frac{\pi }{6}, while the integral property stated in Statement II is a standard result.
506 Easy
506 of 791
Analyzing +4 | -1
The value of the integral ∫1/n(an−1)/n  xa−x  +  x    dx\int \nolimits_{1/n}^{\left( {an - 1} \right)/n} \,\,\frac{{\sqrt x }}{{\sqrt {a - x} \,\, + \,\,\sqrt x }}\;\,\,dx〗 is
View answer and explanation
Correct option: C
na  −  22n\frac{{na\,\, - \,\,2}}{{2n}}
Option C (na − 22n\frac{{na\, - \,2}}{{2n}}) is correct because after applying the property of definite integrals and adding the two forms of the integral, the result simplifies to the given expression.
507 Easy
507 of 791
Application +4 | -1
The area of the region bounced by y2=x  and  y=∣x∣{y^2} = x\;\mathrm{and\;}y = \left| x \right| is
View answer and explanation
Correct option: B
16  sq  unit\frac{1}{6}\;\mathrm{sq\;unit}
Option B (16 squnit\frac{1}{6}\,\mathrm{sq unit}) is correct because it represents the definite integral of the difference between x\sqrt{x} and xx from 0 to 1.
508 Easy
508 of 791
Application +4 | -1
If F(x)=∫0xln⁡(1−t1+t)dtF(x)=\int\limits_{0}^x \ln \left(\frac{1-t}{1+t}\right) d t, then F′(−x)+F′(x)F^{\prime}(-x)+F^{\prime}(x) equals
View answer and explanation
Correct option: A
0
Option A (0) is correct because F′(x)F'(x) is the derivative of the integral, and F′(−x)+F′(x)F'(-x) + F'(x) simplifies to 0 due to the properties of logarithms.
509 Easy
509 of 791
Understanding +4 | -1
∫02πsin⁡6  x cos⁡5  x dx\int \nolimits_0^{2\pi } {\sin ^6}\,\,x\,{\cos ^5}\,\,x\,dx is equal to
View answer and explanation
Correct option: C
0\mathrm{0}
Option C (0) is correct because the integral of an odd function over a symmetric interval is zero.
510 Easy
510 of 791
Analyzing +4 | -1
If f(x)f(x) is an odd function of xx, then ∫−π2π2f(cos⁡x)dx\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(\cos x) d x is equal to
View answer and explanation
Correct option: C
2∫0π2f(sin⁡x)dx2 \int\limits_{0}^{\frac{\pi}{2}} f(\sin x) d x
Option C (2∫0π2f(sin⁡x)dx2 \int_0^{\frac{\pi}{2}} f(\sin x) d x) is correct because f(cos⁡x)f(\cos x) is an even function, and cos⁡x\cos x can be replaced with sin⁡x\sin x due to symmetry.
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