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JEE • MATHEMATICS • DEFINITE INTEGRATION

Definite integration practice · Mixed difficulty

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721 Easy
721 of 995
Analyzing +4 | -1
If I1=∫03π  f(cos⁡2  x) dx{I_1} = \int \limits_0^{3\pi } \,\,f\left( {{{\cos }^2}\,\,x} \right)\,dx and I2=∫0π  f(cos⁡2  x)  dx{I_2} = \int \limits_0^\pi \,\,f\left( {{{\cos }^2}\,\,x} \right)\,\,dx then
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because I1=3I2{I_1} = 3{I_2}, which is not represented by options A, B, or C.
722 Easy
722 of 995
Application +4 | -1
∫−22  (x  −  ∣ x ∣ )  dx\int \nolimits_{ - 2}^2 \,\,\left( {x\,\, - \,\,\left| {\,x\,} \right|\,} \right)\,\,dx is equal to
View answer and explanation
Correct option: D
− 4- \,4
Option D ( − 4\ - \,4) is correct because the integral evaluates to -4 after considering the definition of the absolute value function.
723 Easy
723 of 995
Application +4 | -1
If F(x)=∫3x(2+d dtcos⁡t)dtF(x)=\int\limits_{3}^x\left(2+\frac{d}{\mathrm{~d} t} \cos t\right) \mathrm{d} t then F′(π6)F^{\prime}\left(\frac{\pi}{6}\right) is equal to
View answer and explanation
Correct option: D
3/23 / 2
Option D (3/2) is correct because the derivative of the integral is found using the Fundamental Theorem of Calculus and evaluating at x = π/6.
724 Easy
724 of 995
Application +4 | -1
The value of ∫14  ∣ x−3 ∣dx\int \nolimits_1^4 \,\,\left| {\,x - 3\,} \right|dx is equal to
View answer and explanation
Correct option: B
52\frac{5}{2}
Option B is correct because the absolute value function requires splitting the integral at x=3, and then integrating each part separately, resulting in 5/2.
725 Easy
725 of 995
Application +4 | -1
∫0π/4log⁡(1+tan⁡2θ+2tan⁡θ)dθ=\int\limits_{0}^{\pi / 4} \log \left(1+\tan ^2 \theta+2 \tan \theta\right) d \theta=
View answer and explanation
Correct option: C
(πlog⁡2)/4(\pi \log 2) / 4
Option C is correct because the integral is solved using properties of logarithms and definite integrals, specifically using substitution and the property of integrals with limits 0 to a.
726 Medium
726 of 995
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.
727 Easy
727 of 995
Application +4 | -1
The value of ∫01 x31+x8  dx\int \limits_0^1 \,\frac{{{x^3}}}{{1 + {x^8}}}\,\,dx, is
View answer and explanation
Correct option: C
π16\frac{\pi }{16}
Option C (π16\frac{\pi }{16}) is correct because it is the result of substituting x4=tx^4 = t and evaluating the resulting integral.
728 Easy
728 of 995
Application +4 | -1
The value of I=∫−π/2π/2cos⁡x−cos⁡3x dxI=\int\limits_{{-\pi / 2}}^{\pi / 2} \sqrt{\cos x-\cos ^3 x} \mathrm{~d} x is
View answer and explanation
Correct option: C
4/34 / 3
Option C (4/3) is correct because the integral simplifies to 2∫0π/2cos⁡xsin⁡xdx2 \int_0^{\pi/2} \sqrt{{\cos x}} \sin x dx, which can be easily evaluated using substitution.
729 Easy
729 of 995
Application +4 | -1
Let g(x)={1,0≤x<1x3,1≤x<4x,4≤x<9g(x)=\left\{\begin{array}{cl}1, & 0 \leq x < 1 \\ x^3, & 1 \leq x < 4 \\ \sqrt{x}, & 4 \leq x < 9\end{array}\right. then ∫09g(x)dx\int\limits_{0}^9 g(x) \mathrm{d} x is
View answer and explanation
Correct option: C
929/12929 / 12
Option C is correct because the integral can be solved by splitting the integral based on the piecewise function definition.
730 Easy
730 of 995
Application +4 | -1
∫0π/2  sin⁡2  xsin⁡  x  +  cos⁡  x  dx\int \limits_0^{\pi /2} \,\,\frac{{{{\sin }^2}\,\,x}}{{\sin \,\,x\,\, + \,\,\cos \,\,x}}\,\,dx is equal to
View answer and explanation
Correct option: C
12  log⁡  (2+1)\frac{1}{{\sqrt 2 }}\,\,\log \,\,\left( {\sqrt 2 + 1} \right)
Option C (12  log⁡  (2+1)\frac{1}{{\sqrt 2 }}\,\,\log \,\,\left( {\sqrt 2 + 1} \right)) is correct because using the property of definite integrals and simplifying leads to the given answer.
731 Easy
731 of 995
Application +4 | -1
For x>0x > 0, let f(x)=∫1xlog⁡t1+tdtf(x)=\int\limits_{1}^x \frac{\log t}{1+\mathrm{t}} \mathrm{dt}. Then f(x)+f(1x)f(x)+f\left(\frac{1}{x}\right) is equal to
View answer and explanation
Correct option: C
12(log⁡x)2\frac{1}{2}(\log x)^2
Option C (12(log⁡x)2\frac{1}{2}(\log x)^2) is correct because it is the result obtained after simplifying the expression f(x)+f(1x)f(x) + f(\frac{1}{x}) using a suitable substitution and evaluating the resulting integral.
732 Easy
732 of 995
Application +4 | -1
Evaluate ∫0π/2  sin⁡  xsin⁡  x  +  cos⁡  x  dx  is\int \nolimits_0^{\pi /2} \,\,\frac{{\sqrt {\sin \,\,x} }}{{\sqrt {\sin \,\,x\,\, + \,\,} \sqrt {\cos \,\,x} }}\,\,dx\,\,\mathrm{is} is〗
View answer and explanation
Correct option: A
π4\frac{\pi }{4}
Option A (π4\frac{\pi}{4}) is correct because using the property of definite integrals ∫0af(x)dx=∫0af(a−x)dx\int \nolimits_0^a f(x) dx = \int \nolimits_0^a f(a-x) dx simplifies the integral.
733 Easy
733 of 995
Application +4 | -1
The value of lim⁡x→0∫0xcos⁡t2 dtx\lim \limits_{{x \rightarrow 0}} \frac{\int\limits_{0}^x \cos t^2 \mathrm{~d} t}{x} is
View answer and explanation
Correct option: A
1
Option A (1) is correct because L'Hopital's rule can be applied to evaluate the limit of the indeterminate form 0/0.
734 Medium
734 of 995
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+}.
735 Medium
735 of 995
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.
736 Easy
736 of 995
Application +4 | -1
The value of the integral ∫αβdx(x−α)(β−x)\int\limits_ \alpha^\beta \frac{d x}{\sqrt{(x-\alpha)(\beta-x)}} for β>α\beta>\alpha, is
View answer and explanation
Correct option: D
π\pi.
Option D (π\pi) is correct because the integral represents the area under a semicircle, which evaluates to π\pi.
737 Medium
737 of 995
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.
738 Hard
738 of 995
Analyzing +4 | -1
Consider the integrals
I1=∫01 e−xcos⁡2  x dx,    I2=∫01 e−x2cos⁡2  x dx,    I3=∫01 e−x2  dx{I_1} = \int \limits_0^1 \,{e^{ - x}}{\cos ^2}\,\,x\,dx,\;\,\,{I_2} = \int \limits_0^1 \,{e^{ - {x^2}}}{\cos ^2}\,\,x\,dx,\;\,\,{I_3} = \int \limits_0^1 \,{e^{ - {x^2}}}\;dx and I4=∫01 e−(1/2)x2  dx{I_4} = \int \limits_0^1 \,{e^{ - \left( {1/2} \right){x^2}}}\,\,dx The greatest of these integral is
View answer and explanation
Correct option: D
I4{I_4}
Option D (I4{I_4}) is correct because by comparing the integrands over the interval [0,1], e−(1/2)x2{e^{ - (1/2){x^2}}} is the greatest, leading to the largest integral value.
739 Easy
739 of 995
Application +4 | -1
If f(x)f(x) is differentiable and ∫0t2xf(x)dx=25t5\int_0^{t^2} x f(x) d x=\frac{2}{5} t^5, then f(425)f\left(\frac{4}{25}\right) equals
View answer and explanation
Correct option: A
2/52 / 5
Option A (2/5) is correct because differentiating both sides of the equation using Leibniz's rule and solving for f(4/25) leads to the answer 2/5.
740 Easy
740 of 995
Analyzing +4 | -1
If ∫0100 πsin⁡2xe(xπ−[xπ]) dx=απ31+4π2,  α∈R\int\limits_0^{100\,\pi } {\frac{{{{\sin }^2}x}}{{{e^{\left( {\frac{x}{\pi } - \left[ {\frac{x}{\pi }} \right]} \right)}}}}} \,dx = \frac{{\alpha {\pi ^3}}}{{1 + 4{\pi ^2}}},\,\,\alpha \in R, where [x][x] is the greatest integer less than or equal to xx then the value of α\alpha is
View answer and explanation
Correct option: A
200 (1−e−1)200\,(1 - {e^{ - 1}})
Option A (200 (1−e−1)200\,(1 - {e^{ - 1}})) is correct because the integral can be simplified using the periodicity of the sine function and then evaluated, leading to the given result.
741 Easy
741 of 995
Analyzing +4 | -1
If ∫0xf(t)dt=x+∫x1tf(t)dt\int\limits_{0}^x f(t) d t=x+\int_x^1 t f(t) d t then f(1)f(1) is
View answer and explanation
Correct option: A
12\frac{1}{2}
Option A (12\frac{{1}}{{2}}) is correct because by differentiating the given equation and substituting x=1, we find that f(1) equals 12\frac{{1}}{{2}}.
742 Easy
742 of 995
Application +4 | -1
The value of integral ∫−24  x  [x]  dx\int \limits_{ - 2}^4 \,\,x\,\,\left[ x \right]\,\,dx, is
View answer and explanation
Correct option: A
41  / 2\mathrm{41}\,\,\mathrm{/}\,\mathrm{2}
Option A (41/2) is correct because the integral is evaluated by splitting it into intervals where the greatest integer function is constant, and then summing the results.
743 Easy
743 of 995
Application +4 | -1
The value of the definite integral ∫π245π24dx1+tan⁡2x3\int\limits_{\frac{\pi }{{24}}}^{\frac{{5\pi }}{{24}}} {\frac{{dx}}{{1 + \sqrt[3]{{\tan 2x}}}}} is
View answer and explanation
Correct option: C
π12\frac{\pi }{{12}}
Option C (π12\frac{π }{{12}}) is correct because it is the value of the definite integral after applying the property of definite integrals and simplifying.
744 Easy
744 of 995
Application +4 | -1
The equation of the tangent to the curve
y=∫x4x6dt1+t2 at x=1 is y=\int\limits_{{x^4}}^{x^6} \frac{\mathrm{d} t}{\sqrt{1+t^2}} \text { at } x=1 \text { is }
View answer and explanation
Correct option: D
None of these
Option C (2) is correct because the distance between the parallel lines is calculated by rewriting the equation and using the formula for the distance between parallel lines.
745 Medium
745 of 995
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.
746 Easy
746 of 995
Application +4 | -1
The value of ∫0αdx1−cos⁡αcos⁡x(0<α<π/2)\int\limits_{0}^\alpha \frac{\mathrm{d} x}{1-\cos \alpha \cos x}(0 < \alpha < \pi / 2) is
View answer and explanation
Correct option: A
πsin⁡α\frac{\pi}{\sin \alpha}
Option A (πsin⁡α\frac{\pi}{\sin \alpha}) is correct because after substitution and integration, the result matches this expression.
747 Easy
747 of 995
Analyzing +4 | -1
The value of the integral ∫−1/21/2cos⁡xlog⁡(1+x1−x)dx\int\limits_{{-1 / 2}}^{1 / 2} \cos x \log \left(\frac{1+x}{1-x}\right) d x, is
View answer and explanation
Correct option: A
0
Option A (0) is correct because the integrand is an odd function, and the integral is evaluated over a symmetric interval.
748 Easy
748 of 995
Application +4 | -1
The value of ∫1e2  dxx (1  +  log⁡  x)2\int \nolimits_1^{{e^2}} \,\,\frac{{dx}}{{x\,{{(1\,\, + \,\,\log \,\,x)}^2}}} is
View answer and explanation
Correct option: A
23\frac{2}{3}
Option A (23\frac{2}{3}) is correct because the definite integral evaluates to 2/3 after a simple substitution.
749 Easy
749 of 995
Application +4 | -1
Let f(x)=∫1x  2  −  t2 dtf\left( x \right) = \int \nolimits_1^x \,\,\sqrt {2\,\, - \,\,{t^2}} \,dt.〗 Then, real roots of the equation x2−f′(x)=0{x^2} - f'\left( x \right) = 0 are
View answer and explanation
Correct option: A
±  1\pm \,\,1
Option A (±  1\pm \,\,1) is correct because these are the real roots obtained by solving the equation x2−f′(x)=0{x^2} - f'\left( x \right) = 0 after finding f′(x)f'(x).
750 Medium
750 of 995
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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