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JEE • MATHEMATICS • DEFINITE INTEGRATION

Definite integration practice · Medium

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451 Medium
451 of 596
Application +4 | -1
The volume of the solid generated by revolving the region bounded by y=x2+1y = {x^2} + 1 and y=2x+1y = 2x + 1 about x−x -axis is
View answer and explanation
Correct option: A
104π15  cu  units  \frac{{104\mathrm{\pi }}}{{15}}\mathrm{\;cu\;units\;}
Option A (104π15  cu  units  \frac{{104\mathrm{\pi }}}{{15}}\mathrm{\;cu\;units\;}) is correct because it represents the volume of the solid generated by revolving the region between the two curves about the x-axis, calculated using the disk method.
452 Medium
452 of 596
Application +4 | -1
Let f(x)f\left( x \right) be a non-negative continuous function such that the area bounded by the curve y=f(x),  xy = f\left( x \right),\;x-axis and the ordinates x=π4  x = \frac{\mathrm{\pi }}{4}\mathrm{\;} and x=β>π4 is πx = \mathrm{\beta } > \frac{\mathrm{\pi }}{4}\,\mathrm{is}\,\mathrm{\pi }
(βsin⁡β+π4cos⁡β+2β)  the  f(π2)s\left( {\mathrm{\beta }\sin \mathrm{\beta } + \frac{\mathrm{\pi }}{4}\cos \mathrm{\beta } + \sqrt 2 \mathrm{\beta }} \right)\;\mathrm{the\;}f\left( {\frac{\mathrm{\pi }}{2}} \right)\mathrm{s}
View answer and explanation
Correct option: A
(1−π4+2)\left( {1 - \frac{\mathrm{\pi }}{4} + \sqrt 2 } \right)
Option A ((1−π4+2)\left( {1 - \frac{\mathrm{\pi }}{4} + \sqrt 2 } \right)) is correct because it is the result of differentiating the given area function and substituting β=π2\beta = \frac{{\pi }}{2}.
453 Medium
453 of 596
Application +4 | -1
Let aa and bb be real constants such that the function ff defined by f(x)={x2+3x+a,x≤1bx+2,x>1f(x)=\left\{\begin{array}{cc}x^2+3 x+a, & x \leq 1 \\ b x+2, & x > 1\end{array}\right. be differentiable on R.\mathbb{R}. Then, the value of ∫−22f(x)dx\int\limits_{{-2}}^2 f(x) d x equals
View answer and explanation
Correct option: B
17
The correct answer is B (17) because by ensuring differentiability and continuity, we can find the values of a and b, and then evaluate the definite integral.
454 Medium
454 of 596
Application +4 | -1
If g(x)=∫0xcos⁡4t dtg(x)=\int\limits_{0}^x \cos ^4 t \,d t, then g(x+π)g(x+\pi) equals
View answer and explanation
Correct option: A
g(x)+g(π)g(x)+g(\pi)
Option A (g(x)+g(π)g(x)+g(\pi)) is correct because g(x+π)g(x+\pi) can be expressed as the sum of two integrals, which simplifies to g(x)+g(π)g(x)+g(\pi).
455 Medium
455 of 596
Analyzing +4 | -1
The value of ∫01ln⁡(x+1)1+x2dx\int\limits_{0}^1 \frac{\ln (x+1)}{1+x^2} d x is
View answer and explanation
Correct option: B
πln⁡28\frac{\pi \ln 2}{8}
Option B (πln⁡28\frac{\pi \ln 2}{8}) is correct because the integral can be solved using a substitution that simplifies the integrand.
456 Medium
456 of 596
Application +4 | -1
∫π/43π/4dx1+cos⁡x\int\limits_{{\pi / 4}}^{3 \pi / 4} \frac{d x}{1+\cos x} is equal to
View answer and explanation
Correct option: A
2
Option A (2) is correct because the integral simplifies to an expression involving tangent functions, which evaluates to 2.
457 Medium
457 of 596
Analyzing +4 | -1
Let R→R\mathbb{R} \rightarrow \mathbb{R} be function defined as f(x)=a sin⁡(π[x]2)+[2−x],a∈Rf(x)=\operatorname{a\,sin}\left(\frac{\pi[x]}{2}\right)+[2-x], a \in \mathbb{R}, where [t][t] is the greatest integer less than or equal to tt. If lim⁡x→−1f(x)\lim _{x \rightarrow-1} f(x) exists, then the value of ∫04f(x)dx\int_0^4 f(x) d x is equal to
View answer and explanation
Correct option: B
−2-2
Option D (2) is correct because by finding the value of 'a' using the limit condition and then integrating the function from 0 to 4, we get the result as -2.
458 Medium
458 of 596
Application +4 | -1
∫0ax2(a2−x2)3/2dx=\int\limits_{0}^a x^2\left(a^2-x^2\right)^{3 / 2} d x=
View answer and explanation
Correct option: A
πa632\frac{\pi a^6}{32}
Option A (πa632\frac{{\pi a^6}}{{32}}) is correct because the integral is solved using trigonometric substitution and Gamma functions.
459 Medium
459 of 596
Application +4 | -1
∫π/43π/4dx1+cos⁡x\int_{\pi /4}^{3\pi /4} {\frac{{dx}}{{1 + \cos x}}} is equal to
View answer and explanation
Correct option: B
2
Option C (1.1×10−9M1.1 \times {10^{ - 9}}M) is correct because it accounts for the dilution effect when calculating the original concentration of Ba2+Ba^{2+}.
460 Medium
460 of 596
Application +4 | -1
The value of the definite integral ∫a+2πa+5π/2  (sin⁡−1  (cos⁡  x)+cos⁡−1  (sin⁡ x))  dx\int \nolimits_{a + 2\pi }^{a + 5\pi /2} \,\,({\sin ^{ - 1}}\,\,(\cos \,\,x) + {\cos ^{ - 1}}\,\,(\sin \,x))\;dx is equal to
View answer and explanation
Correct option: B
π24\frac{{{\mathrm{\pi }^2}}}{4}
Option B (π24\frac{{{\mathrm{\pi }^2}}}{4}) is correct because after simplifying the integrand using inverse trigonometric identities and evaluating the integral, the result is this value.
461 Medium
461 of 596
Application +4 | -1
∫02π11+tan⁡4xdx\int_0^{2 \pi} \frac{1}{1+\tan ^4 x} d x is equal to
View answer and explanation
Correct option: A
π\pi
Option A (π) is correct because using properties of definite integrals and symmetry, the integral simplifies to π.
462 Medium
462 of 596
Analyzing +4 | -1
If f(x)=∫5x8+7x6(x2+1+2x7)2dx,(x≥0)f(x)=\int \frac{5 x^8+7 x^6}{\left(x^2+1+2 x^7\right)^2} d x,(x \geq 0), and f(0)=0f(0)=0, then the value of f(1)f(1) is:
View answer and explanation
Correct option: A
−12-\frac{1}{2}
Option D (14\frac{1}{4}) is correct because after performing the integration and applying the initial condition f(0)=0, we find f(1) = 1/4.
463 Medium
463 of 596
Application +4 | -1
∫0[x]2x2[x]dx\int\limits_{0}^{[x]} \frac{2^x}{2^{[x]}} d x equals (where [.] denotes the greatest integer function)
View answer and explanation
Correct option: B
[x]ln⁡2\frac{[x]}{\ln 2}
Option B ([x]ln⁡2\frac{{[x]}}{{\ln 2}}) is correct because the integral simplifies to a sum of integrals over unit intervals, each of which evaluates to 1ln⁡2\frac{{1}}{{\ln 2}}.
464 Medium
464 of 596
Application +4 | -1
The value of the integral ∫−π/2π/2{x2+log⁡(π+xπ−x)}cos⁡x dx, is\int_{ - \pi /2}^{\pi /2} {\left\{ {{x^2} + \log \left( {\frac{{\pi + x}}{{\pi - x}}} \right)} \right\}} \cos x\,dx,\,\mathrm{ is }
View answer and explanation
Correct option: B
π22−4\frac{\pi^2}{2}-4
The correct answer is B (π22−4\frac{\pi^2}{2}-4) because after splitting the integral and evaluating, the result is π22−4\frac{\pi^2}{2}-4.
465 Medium
465 of 596
Application +4 | -1
∫0π  x  sin⁡4  x dx\int \nolimits_0^\pi \,\,x\,\,{\sin ^4}\,\,x\,dx is equal to
View answer and explanation
Correct option: B
3 π216\frac{{3\,{\pi ^2}}}{{16}}
Option B (3 π216\frac{{3\,{\pi ^2}}}{{16}}) is correct because the integral is solved using properties of definite integrals and trigonometric identities.

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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
466 Medium
466 of 596
Application +4 | -1
The value of ∫12  dx(x+1)  x2−1\int \nolimits_1^2 \,\,\frac{{dx}}{{\left( {x + 1} \right)\,\,\sqrt {{x^2} - 1} }} is
View answer and explanation
Correct option: B
13\frac{1}{{\sqrt 3 }}
The correct answer is 1/√(3) because after substituting x = sec θ and evaluating the definite integral, we obtain this value.
467 Medium
467 of 596
Application +4 | -1
The value of the definite integral ∫37cos⁡x2cos⁡x2+cos⁡(10−x)2dx\int\limits_{3}^7 \frac{\cos x^2}{\cos x^2+\cos (10-x)^2} d x is:
View answer and explanation
Correct option: A
2
Option A (2) is correct because the integral can be simplified using the property ∫abf(x)dx=∫abf(a+b−x)dx\int_a^b f(x) dx = \int_a^b f(a+b-x) dx and then adding the two integrals.
468 Medium
468 of 596
Application +4 | -1
If ∫(x9+x6+x3)(2x6+3x3+6)13dx=1A(2x9+3x6+6x3)B+C\int\left(x^9+x^6+x^3\right)\left(2 x^6+3 x^3+6\right)^{\frac{1}{3}} d x=\frac{1}{A}\left(2 x^9+3 x^6+6 x^3\right)^B+C, where CC is integration constant then AB{AB} is equal to
View answer and explanation
Correct option: A
32
Option A (32) is correct because it is the product of A and B after evaluating the integral and comparing with the given form.
469 Medium
469 of 596
Analyzing +4 | -1
Let ff be a differentiable function from R\mathbf{R} to R\mathbf{R} such that [f(x)[f(x) −f(y)∣≤2∣x−y∣3/2-f(y)|\leq 2| x-\left.y\right|^{3 / 2}, for all x,y,∈Rx, y, \in \mathbf{R}. If f(0)=1f(0)=1 then ∫01f2(x)dx\int\limits_{0}^1 f^2(x) d x is equal to:
View answer and explanation
Correct option: A
1
Option A (1) is correct because the given condition implies that f(x)f(x) is a constant function, and since f(0)=1f(0) = 1, f(x)=1f(x) = 1 for all xx.
470 Medium
470 of 596
Application +4 | -1
The integral ∫dx(x+1)34(x−2)54\int \frac{\mathrm{dx}}{(x+1)^{\frac{3}{4}}(x-2)^{\frac{5}{4}}} is equal to:
View answer and explanation
Correct option: B
4(x+1x−2)14+C4\left(\frac{x+1}{x-2}\right)^{\frac{1}{4}}+C
Option B (4(x+1x−2)14+C4\left(\frac{{x+1}}{{x-2}}\right)^{\frac{{1}}{{4}}}+C) is correct because the integral can be solved by substitution, letting t=x+1x−2t = \frac{{x+1}}{{x-2}}.
471 Medium
471 of 596
Understanding +4 | -1
The value of ∫−ππ(1−x2)  sin⁡  x  cos⁡2  x  dx\int \limits_{ - \pi }^\pi \left( {1 - {x^2}} \right)\,\,\sin \,\,x\,\,{\cos ^2}\,\,x\,\,dx, is
View answer and explanation
Correct option: A
0
Option A is correct because the integrand is an odd function, and the integral of an odd function over a symmetric interval is zero.
472 Medium
472 of 596
Application +4 | -1
∫−22∣x∣dx=\int\limits_{{-2}}^2|x| d x=
View answer and explanation
Correct option: D
4
Option D (4) is correct because the integral of the absolute value function ∣x∣|x| from -2 to 2 is the sum of the areas of two triangles, each with area 2, resulting in a total area of 4.
473 Medium
473 of 596
Analyzing +4 | -1
The integral ∫01/alog⁡(1+ax)1+a2x2 dx(a>0)\int\limits_{0}^{1 / a} \frac{\log (1+a x)}{1+a^2 x^2} \mathrm{~d} x(a>0) is equal to
View answer and explanation
Correct option: C
1a(log⁡2)π8\frac{1}{a}(\log 2) \frac{\pi}{8}
Option C (1a(log⁡2)π8\frac{1}{a}(\log 2) \frac{\pi}{8}) is correct because after substitution and integration, the integral evaluates to 1a(log⁡2)π8\frac{1}{a}(\log 2) \frac{\pi}{8}.
474 Medium
474 of 596
Application +4 | -1
∫0π/4  log⁡  (1+tan⁡  x)  dx\int \nolimits_0^{\pi /4} \,\,\log \,\,(1 + \tan \,\,x)\,\,dx is equal to
View answer and explanation
Correct option: A
π8  log⁡e  2\frac{\pi }{8}\,\,{\log _e}\,\,2
Option A is correct because the integral is solved by using properties of definite integrals and logarithmic functions.
475 Medium
475 of 596
Application +4 | -1
If I=∫01  cos⁡  {2cot⁡−1  1−x1+x}  dxI = \int \limits_0^1 \,\,\cos \,\,\left\{ {2{{\cot }^{ - 1}}\,\,\sqrt {\frac{{1 - x}}{{1 + x}}} } \right\}\,\,dx , then
View answer and explanation
Correct option: B
I=−  12I = - \,\,\frac{1}{2}
Option C (π\pi) is correct because the integral of the odd functions over the symmetric interval is zero, and only the integral of 1 remains.
476 Medium
476 of 596
Application +4 | -1
∫02af(x)f(x)+f(2a−x)dx=\int\limits_{0}^{2 a} \frac{f(x)}{f(x)+f(2 a-x)} d x=
View answer and explanation
Correct option: B
aa
Option B is correct because the integral can be solved using the property ∫abf(x)f(x)+f(a+b−x)dx=b−a2\int_a^b \frac{f(x)}{f(x)+f(a+b-x)} dx = \frac{b-a}{2}.
477 Medium
477 of 596
Application +4 | -1
lim⁡n→∞1n∑r=12nrn2+r2\lim \limits_{{n \rightarrow \infty}} \frac{1}{n} \sum\limits_{{r=1}}^{2 n} \frac{r}{\sqrt{n^2+r^2}} equals
View answer and explanation
Correct option: B
5−1\sqrt{5}-1
Option D (5−1\sqrt{5}-1) is correct because the limit of the Riemann sum converts to the integral ∫02x1+x2dx\int_0^2 \frac{x}{\sqrt{1+x^2}} dx, which evaluates to 5−1\sqrt{5}-1.
478 Medium
478 of 596
Application +4 | -1
∫π/6π/3  dx1+cot⁡  x\int \nolimits_{\pi /6}^{\pi /3} \,\,\frac{{dx}}{{1 + \sqrt {\cot \,\,x} }} is equal to
View answer and explanation
Correct option: C
π12\frac{\mathrm{\pi }}{12}
Option C (π12\frac{\mathrm{\pi }}{{12}}) is correct because the definite integral is evaluated using the property ∫ab f  (x)  dx=∫ab f  (a+b−x)  dx\int \nolimits_a^b \,f\,\,\left( x \right)\,\,dx = \int \nolimits_a^b \,f\,\,\left( {a + b - x} \right)\,\,dx.
479 Medium
479 of 596
Application +4 | -1
Let g(x)=cos⁡x2,f(x)=xg(x)=\cos x^2, f(x)=\sqrt{x}, and α,β(α<β)\alpha, \beta(\alpha < \beta) be the roots of the quadratic equation 18x2−9πx+π2=018 x^2-9 \pi x+\pi^2=0. Then the area (in sq. units) bounded by the curve y=gof(x)y=g o f(x) and the lines x=αx=\alpha and x=βx=\beta, is
View answer and explanation
Correct option: D
12(3−1)\frac{1}{2}(\sqrt{3}-1)
Option D (12(3−1)\frac{1}{2}(\sqrt{3}-1)) is correct because it is the area under the cosine curve between the roots of the given quadratic equation.
480 Medium
480 of 596
Application +4 | -1
∫  1  +  x2x   1  +  x4  dx\int \,\,\frac{{1\,\, + \,\,{x^2}}}{{x\;\,\sqrt {1\,\, + \,\,{x^4}} }}\,\,dx is equal to
View answer and explanation
Correct option: B
log⁡  ∣x  −  1x  +  (x  −  1x)2  +  2∣  +  C\log \,\,\left| {x\,\, - \,\,\frac{1}{x}\,\, + \,\,\sqrt {{{\left( {x\,\, - \,\,\frac{1}{x}} \right)}^2}\,\, + \,\,2} } \right|\,\, + \,\,C
Option B is correct. The integral simplifies by dividing numerator and denominator by x2 and then substituting to get a standard logarithmic form.
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