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JEE • MATHEMATICS • DEFINITE INTEGRATION

Definite integration practice · Mixed difficulty

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631 Medium
631 of 995
Application +4 | -1
If f(x)=∫−1x  ∣t∣  dtf\left( x \right) = \int \limits_{ - 1}^x \,\,\left| t \right|\,\,dt, then for any x≥0,    f(x)x \ge 0,\,\,\;f\left( x \right) equals

View answer and explanation
Correct option: C
12  (1+x2)\frac{1}{2}\,\,\left( {1 + {x^2}} \right)
Option C (12 (1+x2)\frac{1}{2}\,\left( {1 + {x^2}} \right)) is correct because the integral of the absolute value function from -1 to x, for x≥0x \ge 0, is 12(1+x2)\frac{1}{2}(1 + x^2).
632 Easy
632 of 995
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.
633 Medium
633 of 995
Application +4 | -1
If [t][t] stands for the integral part of tt, then ∫05π/12[tan⁡x]dx=\int\limits_{0}^{5 \pi / 12}[\tan x] d x=
View answer and explanation
Correct option: C
π4\frac{\pi}{4}
Option C (π4\frac{\pi}{4}) is correct because after using the properties of the greatest integer function and trigonometric functions, the integral evaluates to π4\frac{\pi}{4}.
634 Hard
634 of 995
Analyzing +4 | -1
The value of ∫−11  sin⁡−1  [x2  +  12]  dx  +  ∫−11  cos⁡−1  [x2−  12]  dx\int \limits_{ - 1}^1 \,\,{\sin ^{ - 1}}\,\,\left[ {{x^2}\,\, + \,\,\frac{1}{2}} \right]\,\,dx\,\, + \,\,\int \limits_{ - 1}^1 \,\,{\cos ^{ - 1}}\,\,\left[ {{x^2} - \,\,\frac{1}{2}} \right]\,\,dx, where [ ⋅ ]\left[ {\, \cdot \,} \right] denotes the greatest integer function, is
View answer and explanation
Correct option: B
2 π2\,\pi
Option B (2π) is correct because the sum of the two integrals simplifies to 2π after evaluating the greatest integer functions and using trigonometric identities.
635 Easy
635 of 995
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.
636 Easy
636 of 995
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.
637 Easy
637 of 995
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.
638 Easy
638 of 995
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.
639 Easy
639 of 995
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.
640 Easy
640 of 995
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.
641 Easy
641 of 995
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.
642 Easy
642 of 995
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.
643 Medium
643 of 995
Application +4 | -1
∫01  tan⁡−1  (1x2−x+1)  dx\int \nolimits_0^1 \,\,{\tan ^{ - 1}}\,\,\left( {\frac{1}{{{x^2} - x + 1}}} \right)\,\,dx is equal to
View answer and explanation
Correct option: D
π2  −  log⁡  2\frac{\mathrm{\pi }}{2}\,\, - \,\,\log \,\,2
The correct answer is D (π2−log⁡,2\frac{\mathrm{\pi }}{2}\\ - \log \\,2) because the integral can be simplified using trigonometric identities and integration by parts.
644 Easy
644 of 995
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.
645 Medium
645 of 995
Application +4 | -1
The value of the integral ∫−π2π2sin⁡4x(1+log⁡(2+sin⁡x2−sin⁡x))dx\int\limits_ {-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^4 x\left(1+\log \left(\frac{2+\sin x}{2-\sin x}\right)\right) d x is
View answer and explanation
Correct option: C
38π\frac{3}{8} \pi
Option C (38π\frac{3}{8} \pi) is correct because it is the result obtained after using the property of definite integrals to eliminate the logarithmic term and then evaluating the remaining integral.

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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
646 Easy
646 of 995
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}}.
647 Easy
647 of 995
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.
648 Easy
648 of 995
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.
649 Easy
649 of 995
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.
650 Easy
650 of 995
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.
651 Easy
651 of 995
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.
652 Easy
652 of 995
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.
653 Easy
653 of 995
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.
654 Easy
654 of 995
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.
655 Medium
655 of 995
Application +4 | -1
The value of the ∫  sin⁡  x  +  cos⁡  x3  +  sin⁡  2x  dx\int \,\,\frac{{\sin \,\,x\,\, + \,\,\cos \,\,x}}{{3\,\, + \,\,\sin \,\,2x}}\,\,dx is
View answer and explanation
Correct option: A
14  log⁡  (2−sin⁡  x  +  cos⁡  x2+sin⁡  x  −  cos⁡  x)+c\frac{1}{4}\,\,\log \,\,\left( {\frac{{2 - \sin \,\,x\,\, + \,\,\cos \,\,x}}{{2 + \sin \,\,x\,\, - \,\,\cos \,\,x}}} \right) + c
Option A (14log⁡(2−sin⁡,x,+,cos⁡,x2+sin⁡,x,−,cos⁡,x)+c\frac{1}{4}\log \left( {\frac{{2 - \sin \\,x\\, + \\,\cos \\,x}}{{2 + \sin \\,x\\, - \\,\cos \\,x}}} \right) + c) is correct because it is the result of the integration after a suitable substitution.
656 Easy
656 of 995
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.
657 Medium
657 of 995
Application +4 | -1
The volume of the solid obtained by revolving about yy-axis the area enclosed between the ellipse x2+9y2=9{x^2} + 9{y^2} = 9 and the straight line x+3y=3,  x + 3y = 3,\; in the first quadrant is
View answer and explanation
Correct option: A
3π3\mathrm{\pi }
Option A, 3π3\mathrm{\pi }, is the correct volume because it is found by subtracting the volume generated by revolving the line from the volume generated by revolving the ellipse.
658 Easy
658 of 995
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.
659 Medium
659 of 995
Application +4 | -1
Let f(x)=∫1x2−t2dtf(x)=\int_1^x \sqrt{2-t^2} d t. Then the real roots of the equation x2−f′(x)=0x^2-f^{\prime}(x)=0 are
View answer and explanation
Correct option: A
±1\pm 1
Option A (±1\pm 1) is the correct answer because it satisfies the equation derived from the given integral and its derivative.
660 Medium
660 of 995
Application +4 | -1
∫02π  (sin⁡  x  +  ∣sin⁡  x)  dx\int \nolimits_0^{2\pi } \,\,(\sin \,\,x\,\, + \,\,|\sin \,\,x)\,\,dx is equal to
View answer and explanation
Correct option: B
4\mathrm{4}
Option B (4) is correct because the integral of |sin x| from 0 to 2π is 4, and the integral of sin x from 0 to 2π is 0.
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