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JEE • MATHEMATICS • DEFINITE INTEGRATION

Definite integration practice · Mixed difficulty

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781 Easy
781 of 995
Application +4 | -1
The value of α\alpha in (−π,0)(-\pi, 0) satisfying sin⁡α+∫α2αcos⁡2x dx=0\sin \alpha+\int\limits_ \alpha^{2 \alpha} \cos 2 x \mathrm{~d} x=0 is
View answer and explanation
Correct option: C
−π/3-\pi / 3
Option C is correct because the equation can be solved by evaluating the integral and solving the resulting trigonometric equation.
782 Medium
782 of 995
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.
783 Easy
783 of 995
Analyzing +4 | -1
The value of the integral ∫0π/2  log⁡  ∣tan⁡  x∣dx  is\int \nolimits_0^{\pi /2} \,\,\log \,\,|\tan \,\,x|dx\,\,\mathrm{is}
View answer and explanation
Correct option: B
0\mathrm{0}
Option B (0) is correct because the integral of log|tan x| from 0 to π/2 is 0 due to the symmetry of the function.
784 Medium
784 of 995
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.
785 Easy
785 of 995
Analyzing +4 | -1
If af(x)+bf(1x)=1x−5,x≠0,a≠ba f(x)+b f\left(\frac{1}{x}\right)=\frac{1}{x}-5, x \neq 0, a \neq b, then ∫12f(x)dx\int\limits_{1}^2 f(x) d x equals
View answer and explanation
Correct option: B
(ln⁡2−5)a+7b2a2−b2\frac{(\ln 2-5) a+\frac{7 b}{2}}{a^2-b^2}
The correct answer is A ((ln⁡2−5)a+72ba2−b2\frac{(\ln 2-5) a+\frac{7}{2} b}{a^2-b^2}) because by solving the system of equations for f(x)f(x) and then integrating, the result matches this option.
786 Medium
786 of 995
Application +4 | -1
The value of ∫0π∣cos⁡x∣3 dx\int\limits_{0}^\pi|\cos x|^3 \mathrm{~d} x is:
View answer and explanation
Correct option: B
43\frac{4}{3}
Option B (43\frac{4}{3}) is correct because it is the value obtained after evaluating the definite integral of the absolute value of cosine cubed over the given interval.
787 Easy
787 of 995
Analyzing +4 | -1
If [t][t] denotes the greatest integer ≤t\leq t, then the value of ∫01[2x−∣3x2−5x+2∣+1]dx\int_0^1\left[2 x-\left|3 x^2-5 x+2\right|+1\right] d x is
View answer and explanation
Correct option: A
37+13−46\frac{\sqrt{37}+\sqrt{13}-4}{6}
Option A (37+13−46\frac{\sqrt{37}+\sqrt{13}-4}{6}) is correct because it is the value of the definite integral calculated by splitting the integral based on the absolute value function and evaluating each part.
788 Easy
788 of 995
Understanding +4 | -1
If ff is integrable on [a,b][a, b] then ∫ab[f(x)−fave ]dx\int\limits_{a}^b\left[f(x)-f_{\text {ave }}\right] \mathrm{d} x is equal to
View answer and explanation
Correct option: D
0
Option D (0) is correct because the integral of a function minus its average value over an interval is always zero.
789 Easy
789 of 995
Application +4 | -1
∫14log⁡e[x]dx\int_1^4 \log _e[x] d x equals
View answer and explanation
Correct option: A
log⁡e6\log _e 6
Option A (log⁡e6\log _e 6) is correct because the integral is evaluated by breaking it into intervals where the greatest integer function is constant.
790 Medium
790 of 995
Application +4 | -1
The area between y=xy=x and y=x2,0≤x≤2y=x^2, 0 \leq x \leq 2 is
View answer and explanation
Correct option: A
1
Option C (1/3) is incorrect, the correct answer is 1 because it is the sum of two integrals, one from 0 to 1 and another from 1 to 2, representing the area between the curves.
791 Hard
791 of 995
Application +4 | -1
∫−32  {∣ x+1 ∣  +  ∣ x+2 ∣  +  ∣ x−1 ∣}dx\int \nolimits_{ - 3}^2 \,\,\left\{ {\left| {\,x + 1\,} \right|\,\, + \,\,\left| {\,x + 2\,} \right|\,\, + \,\,\left| {\,x - 1\,} \right|} \right\}dx is equal to
View answer and explanation
Correct option: C
472\frac{{47}}{2}
Option C (472\frac{{47}}{2}) is correct because it is the result of correctly evaluating the definite integral by breaking it into intervals based on the absolute value function.
792 Hard
792 of 995
Application +4 | -1
∫−11  [x+ [x+ [x]]]  dx\int \nolimits_{ - 1}^1 \,\,\left[ {x + \,} \right[x + \,[x]]]\,\,dx where [ ⋅ ]\left[ {\, \cdot \,} \right] denotes greatest integer function, is equal to
View answer and explanation
Correct option: C
−3- 3
Option D (-4) is correct because the integral of the greatest integer function is evaluated by splitting the integral into intervals where the greatest integer function is constant.
793 Easy
793 of 995
Application +4 | -1
If f(0)=2,f′(x)=f(x),ϕ(x)=x+f(x)f(0)=2, f^{\prime}(x)=f(x), \phi(x)=x+f(x) then ∫01f(x)ϕ(x)dx\int\limits_{0}^1 f(x) \phi(x) d x is
View answer and explanation
Correct option: B
2e22 e^2
Option B (2e22 e^2) is correct because the integral can be solved by recognizing that f(x) = 2e^x and then using integration by parts.
794 Medium
794 of 995
Application +4 | -1
∫01  ddx  [sin⁡−1  (2x1+x2)]    dx\int \nolimits_0^1 \,\,\frac{d}{{dx}}\,\,\left[ {{{\sin }^{ - 1}}\,\,\left( {\frac{{2x}}{{1 + {x^2}}}} \right)} \right]\,\,\;dx is equal to
View answer and explanation
Correct option: C
π2\frac{\mathrm{\pi }}{2}
Option C (π2\frac{\mathrm{\pi }}{2}) is correct because the integral of a derivative is the original function evaluated at the limits of integration.
795 Easy
795 of 995
Application +4 | -1
∫02nπ  {∣sin⁡x∣  −   ∣ 12sin⁡  x ∣}dx\int \nolimits_0^{2n\pi } \,\,\left\{ {|\sin x|\,\, - \,\,\,\left| {\,\frac{1}{2}\sin \,\,x\,} \right|} \right\}dx equals
View answer and explanation
Correct option: B
2n2n
Option B (2n2n) is correct because the integral of the absolute value of sine over one period is 2, and there are 2n such half-periods in the interval [0, 2nπ].

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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
796 Medium
796 of 995
Analyzing +4 | -1
If f(x)=∣ sin⁡  x  +  sin⁡  2x  +  sin⁡  3xsin⁡  2xsin⁡  3x3  +  4sin⁡  x34  sin⁡  x1  +  sin⁡  xsin⁡  x1 ∣f\left( x \right) = \left| {\,\begin{array}{cccccccccccccccccccc} {\sin \,\,x\,\, + \,\,\sin \,\,2x\,\, + \,\,\sin \,\,3x}&{\sin \,\,2x}&{\sin \,\,3x}\\ {3\,\, + \,\,4\sin \,\,x}&3&{4\,\,\sin \,\,x}\\ {1\,\, + \,\,\sin \,\,x}&{\sin \,\,x}&1 \end{array}\,} \right|, then the value of ∫0π/2 f(x)dx\int \nolimits_0^{\pi /2} \,f\left( x \right)dx is
View answer and explanation
Correct option: C
13\frac{1}{3}
Option C (13\frac{1}{3}) is correct because after simplifying the determinant, the resulting integral evaluates to 13\frac{1}{3}.
797 Easy
797 of 995
Application +4 | -1
For any n∈Nn \in N and x∈R+x \in {R^ + }, the value of the integral ∫0n[x](x−[x])dx\int \limits_0^{n\left[ x \right]} \left( {x - \left[ x \right]} \right)dx, is
View answer and explanation
Correct option: C
n2  [x]\frac{n}{2}\,\,\left[ x \right]
Option C (n2  [x]\frac{n}{2}\,\,[ x ]) is correct because the integral of the fractional part of x over one period is 1/2, and we have n[x] such periods.
798 Easy
798 of 995
Application +4 | -1
The value of the integral ∫01  1(1+x2)3/2  dx\int \limits_0^1 \,\,\frac{1}{{{{\left( {1 + {x^2}} \right)}^{3/2}}}}\,\,dx, is
View answer and explanation
Correct option: B
1/21/\sqrt 2
Option B (1/21/\sqrt 2) is correct because it results from a trigonometric substitution and evaluating the resulting integral.
799 Easy
799 of 995
Application +4 | -1
∫0π/2  sin⁡2/3  xsin⁡2/3  x  +  cos⁡2/3  x  dx\int \nolimits_0^{\pi /2} \,\,\frac{{{{\sin }^{2/3}}\,\,x}}{{{{\sin }^{2/3}}\,\,x\,\, + \,\,{{\cos }^{2/3}}\,\,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 using the property of definite integrals, ∫0af(x)dx=∫0af(a−x)dx\int_0^a f(x) dx = \int_0^a f(a-x) dx, simplifies the integral.
800 Easy
800 of 995
Application +4 | -1
The value of ∫π/4π/2cot⁡θcosec⁡2θdθ\int\limits_{{\pi / 4}}^{\pi / 2} \cot \theta \operatorname{cosec}^2 \theta \mathrm{d} \theta is
View answer and explanation
Correct option: B
12\frac{1}{2}
Option B (12\frac{1}{2}) is correct. The integral is solved by using a simple substitution to simplify the expression and then evaluating the resulting integral.
801 Easy
801 of 995
Analyzing +4 | -1
The value of the integral ∫−aa  xex21+x2  dx\int \nolimits_{ - a}^a \,\,\frac{{x{e^x}^2}}{{1 + {x^2}}}\,\,dx is
View answer and explanation
Correct option: B
0
Option B (0) is correct because the integrand is an odd function, and the integral is evaluated over a symmetric interval.
802 Medium
802 of 995
Understanding +4 | -1
If I=1π∫02πecos⁡θcos⁡(sin⁡θ)dθI=\frac{1}{\pi} \int\limits_{0}^{2 \pi} e^{\cos \theta} \cos (\sin \theta) d \theta, then find the value of II.
View answer and explanation
Correct option: A
2
Option A (2) is correct because the integral represents the real part of the power series expansion of eze^z integrated over a circle, which simplifies to 2.
803 Medium
803 of 995
Application +4 | -1
∫0πecos⁡xecos⁡x+e−cos⁡xdx\int\limits_{0}^\pi \frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} d x is equal to
View answer and explanation
Correct option: B
π2\frac{\pi}{2}
Option B (π/2) is correct because by using the property of definite integrals and adding the original integral to a transformed version, the integral simplifies to π/2.
804 Medium
804 of 995
Application +4 | -1
The value of ∫−π/2π/2    (x3  +  x  cos⁡  x  +  tan⁡5  x  +  1)  dx\int \limits_{ - \pi /2}^{\pi /2} \,\,\,\,\left( {{x^3}\,\, + \,\,x\,\,\cos \,\,x\,\, + \,\,{{\tan }^5}\,\,x\,\, + \,\,1} \right)\,\,dx is equal to
View answer and explanation
Correct option: C
π\pi
Option C (π\pi) is correct because the integral of the odd functions over the symmetric interval is zero, leaving only the integral of 1 from -π\pi/2 to π\pi/2.
805 Easy
805 of 995
Understanding +4 | -1
If I=∫0π/4sin⁡2xdxI=\int\limits_{0}^{\pi / 4} \sin ^2 x d x and J=∫0π/4cos⁡2xdxJ=\int\limits_{0}^{\pi / 4} \cos ^2 x d x, then I=I=
View answer and explanation
Correct option: A
π4−J\frac{\pi}{4}-J
Option C is correct because the integrals of sin²(x) and cos²(x) from 0 to π/4 are related by I + J = π/4.
806 Easy
806 of 995
Application +4 | -1
∫0π/2  dx1+  tan⁡3  x\int \nolimits_0^{\pi /2} \,\,\frac{{dx}}{{1 + \,\,{{\tan }^3}\,\,x}} is equal to
View answer and explanation
Correct option: C
π4\frac{\pi }{4}
Option C (π4\frac{\pi }{4}) is correct because the integral is evaluated using properties of definite integrals and trigonometric identities.
807 Medium
807 of 995
Application +4 | -1
A function f(x)f(x) satisfying ∫01f(tx)dt=nf(x)\int\limits_{0}^1 f(t x) d t=n f(x), where x>0x>0, is
View answer and explanation
Correct option: A
f(x)=C⋅x1−nnf(x)=C \cdot x^{\frac{1-n}{n}}
Option A is correct because the function satisfies the given integral equation, which leads to a differential equation that can be solved.
808 Medium
808 of 995
Analyzing +4 | -1
If the value of the integral ∫05X+[X]ex−[X]dx=αe−1×β\int_0^5 {\frac{{X + \left[ X \right]}}{{{e^{x - \left[ X \right]}}}}} dx = \alpha {e^{ - 1}} \times \beta, where α,  β∈R,  5α+6β=0\alpha ,\,\,\beta \in R,\,\,5\alpha + 6\beta = 0 and [X][X] denotes the greatest integer less than or equal to xx, then the value of (α+β)2{\left( {\alpha + \beta } \right)^2} is equal to:
View answer and explanation
Correct option: B
25
Option B (25) is correct because after evaluating the integral and using the given condition 5α+6β=05\alpha + 6\beta = 0, we find that (α+β)2=25(\alpha + \beta)^2 = 25.
809 Easy
809 of 995
Application +4 | -1
\lim \limits_\limits {n \rightarrow \infty} \frac{1+2^4+3^4+\ldots n^4}{n^5}-\lim \limits_{{n \rightarrow \infty}} \frac{1+2^3+3^3+\ldots n^3}{n^5}\lim \limits_\limits {n \rightarrow \infty} \frac{1+2^4+3^4+\ldots n^4}{n^5}-\lim \limits_{{n \rightarrow \infty}} \frac{1+2^3+3^3+\ldots n^3}{n^5}
View answer and explanation
Correct option: A
15\frac{1}{5}
Option A (15\frac{1}{5}) is correct because the limit is evaluated by converting the sums to integrals and subtracting.
810 Easy
810 of 995
Application +4 | -1
∫0π/2  x  sin⁡  x  dx\int \nolimits_0^{\pi /2} \,\,x\,\,\sin \,\,x\,\,dx is equal to
View answer and explanation
Correct option: B
1\mathrm{1}
Option B (1) is correct because it is the result of integrating x*sin(x) by parts over the interval [0, pi/2].
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