Definite Integration84 questions10 PYQ

Definite Integration — JEE Maths Practice Questions & Solutions

84 questions on Definite Integration with full step-by-step solutions, including past-year (PYQ) problems. Free to practice.

hardPYQ · JEE Main 2026
If α=∫023log⁡2(x2+4) dx+∫242x−4  dx\alpha=\displaystyle\int_0^{2\sqrt3}\log_2\left(x^2+4\right)\,dx+\int_2^{4}\sqrt{2^x-4}\,\,dx, then α2\alpha^2 is equal to
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mediumPYQ · JEE Main 2026
Let (21−a+21+a), f(a), (3a+3−a)(2^{1-a}+2^{1+a}),\ f(a),\ (3^a+3^{-a}) be in A.P. and α\alpha be the minimum value of f(a)f(a). Then the value of ∫ln⁡(α−1)ln⁡αdxe2x−e−2x\displaystyle\int_{\ln(\alpha-1)}^{\ln\alpha}\dfrac{dx}{e^{2x}-e^{-2x}} is:
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mediumPYQ · JEE Main 2026
Let ∫−22(∣sin⁡x∣+[xsin⁡x])dx=2(3−cos⁡2)+β\displaystyle\int_{-2}^{2}\big(|\sin x|+[x\sin x]\big)dx=2(3-\cos2)+\beta, where [⋅][\cdot] is the greatest integer function. Then βsin⁡(β2)\beta\sin\left(\dfrac{\beta}{2}\right) equals
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mediumPYQ · JEE Main 2026
The value of the integral ∫−π/4π/4(32cos⁡4x1+esin⁡x)dx\displaystyle\int_{-\pi/4}^{\pi/4}\left(\dfrac{32\cos^4 x}{1+e^{\sin x}}\right)dx is
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mediumPYQ · JEE Main 2026
The value of ∫020π(sin⁡4x+cos⁡4x) dx\displaystyle\int_0^{20\pi}(\sin^4x+\cos^4x)\,dx is equal to
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mediumPYQ · JEE Main 2026
The value of the integral ∫π/6π/34−cosec⁡2xcos⁡4x dx\displaystyle\int_{\pi/6}^{\pi/3}\dfrac{4-\operatorname{cosec}^2 x}{\cos^4 x}\,dx is
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mediumPYQ · JEE Main 2026
The value of the integral ∫0∞log⁡e(x)x2+4 dx\displaystyle\int_0^{\infty}\dfrac{\log_e(x)}{x^2+4}\,dx is
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mediumPYQ · JEE Main 2026
The integral ∫01cot⁡−1(1+x+x2)dx\displaystyle\int_0^{1}\cot^{-1}\left(1+x+x^2\right)dx is equal to
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mediumPYQ · JEE Main 2026
If α=1\alpha=1 and β=1+i2\beta=1+i\sqrt2, where i=−1i=\sqrt{-1}, are two roots of the equation x3+ax2+bx+c=0x^3+ax^2+bx+c=0, a,b,c∈Ra,b,c\in\mathbb{R}, then ∫−11(x3+ax2+bx+c)dx\displaystyle\int_{-1}^{1}\left(x^3+ax^2+bx+c\right)dx is equal to
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mediumPYQ · JEE Main 2026
Let [⋅][\cdot] denote the greatest integer function. Then the value of ∫03ex+e−x[x]! dx\displaystyle\int_0^{3}\dfrac{e^x+e^{-x}}{[x]!}\,dx is
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hard
The value of ∫0n2[x ] dx\displaystyle\int_{0}^{n^{2}}\big[\sqrt{x}\,\big]\,dx, where [ ⋅ ][\,\cdot\,] denotes the greatest integer function and n∈Nn\in\mathbb{N}, is
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hard
The value of ∫−11{x2+x−3} dx\displaystyle\int_{-1}^{1}\{x^{2}+x-3\}\,dx, where { ⋅ }\{\,\cdot\,\} denotes the fractional part of xx, is
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hard
The value of ∫−2n 2n+12{x2}sin⁡πx dx\displaystyle\int_{-2n}^{\,2n+\frac12}\left\{\dfrac{x}{2}\right\}\sin\pi x\,dx, where { ⋅ }\{\,\cdot\,\} denotes the fractional part, is
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hard
Let In=∫0∞e−xsin⁡nx dxI_{n}=\displaystyle\int_{0}^{\infty}e^{-x}\sin^{n}x\,dx, n∈Nn\in\mathbb{N}, n>1n>1. Then I2008I2006\dfrac{I_{2008}}{I_{2006}} equals
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hard
The value of the integral ∣ ∫02π[2sin⁡x] dx ∣\left|\,\displaystyle\int_{0}^{2\pi}[2\sin x]\,dx\,\right|, where [ ⋅ ][\,\cdot\,] denotes the greatest integer function, is
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hard
If I=∫0π/2e−αsin⁡x dxI=\displaystyle\int_{0}^{\pi/2}e^{-\alpha\sin x}\,dx, α∈(0,∞)\alpha\in(0,\infty), then
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hard
Let f(x)=∫1xett2 dtf(x)=\displaystyle\int_{1}^{x}\dfrac{e^{t}}{t^{2}}\,dt for all x>0x>0. Then the value of ∫1xet(t+a)2 dt\displaystyle\int_{1}^{x}\dfrac{e^{t}}{(t+a)^{2}}\,dt is
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hard
If y=x2∫1xln⁡t dty=x^{2}\displaystyle\int_{1}^{x}\sqrt{\ln t}\,dt, then the value of d3ydx3\dfrac{d^{3}y}{dx^{3}} at x=ex=e is
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hard
Let In=∫0π/2cos⁡nx cos⁡(nx) dxI_{n}=\displaystyle\int_{0}^{\pi/2}\cos^{n}x\,\cos(nx)\,dx, n∈Nn\in\mathbb{N}. Then I2001:I2002\sqrt{I_{2001}:I_{2002}} can be the eccentricity of
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hard
Let In=∫01xntan⁡−1x dxI_{n}=\displaystyle\int_{0}^{1}x^{n}\tan^{-1}x\,dx. If anIn+2+bnIn=cna_{n}I_{n+2}+b_{n}I_{n}=c_{n} for all n∈N, n≥1n\in\mathbb{N},\ n\ge 1, then
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hard
Let an=∫0π/21−cos⁡2nx1−cos⁡2x dxa_{n}=\displaystyle\int_{0}^{\pi/2}\dfrac{1-\cos 2nx}{1-\cos 2x}\,dx. Then the value of ∣a1a2a3a4a5a6a7a8a9∣\begin{vmatrix}a_{1} & a_{2} & a_{3}\\[2pt] a_{4} & a_{5} & a_{6}\\[2pt] a_{7} & a_{8} & a_{9}\end{vmatrix} equals
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hard
The value of the integral
∫−11dx(x2+x+1)+x4+3x2+1\int_{-1}^{1}\frac{dx}{\left(x^2+x+1\right)+\sqrt{x^4+3x^2+1}}
 is
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hard
Let α\alpha be a real number, then the integral, (n∈I)\left(n \in I\right)
I(α)=∫−11sin⁡α dx1−2xcos⁡α+x2=I\left(\alpha\right) = \int_{-1}^{1}\frac{\sin\alpha\,dx}{1-2x\cos\alpha+x^2} =
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hard
If ff is a continuous and differentiable function in x∈(0,1)x \in \left(0,1\right) such that
∑r=010(f(x+r)−∣ex−r−1∣)=0,\sum_{r=0}^{10}\left(f\left(x+r\right)-\left|e^x-r-1\right|\right) = 0 ,
 then ∫011f(x)dx\displaystyle\int_0^{11}f\left(x\right)dx is, where e=2.78e = 2.78, ln⁡2=0.693\ln2 = 0.693 and ln⁡3=1.098\ln3 = 1.098.
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hard
In=∫n/2(n+1)/2sin⁡(πsin⁡2πx)(2)xdx,n∈II_n = \int_{n/2}^{\left(n+1\right)/2}\frac{\sin\left(\pi\sin^2\pi x\right)}{\left(\sqrt2\right)^x}dx , \qquad n \in I
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hard
Let I=∫01(x9−x5+x)(3x8−4x4+6)3/4dxI = \displaystyle\int_0^1\frac{\left(x^9-x^5+x\right)}{\left(3x^8-4x^4+6\right)^{3/4}}dx, then the value of (12 I)4\left(12\,I\right)^4 is
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hard
Let f:[0,4]→Rf:\left[0,4\right]\to\mathbb R be a continuous function such that∣f(x)∣=6∣(x−n+1)(x−n)∣\left|f(x)\right| = 6\left|\left(x-n+1\right)\left(x-n\right)\right|, where n∈Nn \in \mathbb N ∀ x∈[n−1,n]\forall\,x \in \left[n-1,n\right], and g(x)=∫0xf(t) dt−∫x4f(t) dtg(x) = \displaystyle\int_0^xf(t)\,dt-\int_x^4f(t)\,dt has its maximum at x=2x = 2. Then ∫1/24f(t) dt\displaystyle\int_{1/2}^{4}f(t)\,dt is/are
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medium
The value of ∫eπ2[log⁡πx] d(log⁡ex)\displaystyle\int_{e}^{\pi^{2}}\big[\log_{\pi}x\big]\,d(\log_{e}x), where [ ⋅ ][\,\cdot\,] denotes the greatest integer function, is
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medium
∫0π/4(πx−4x2)ln⁡(1+tan⁡x) dx\displaystyle\int_{0}^{\pi/4}\big(\pi x-4x^{2}\big)\ln(1+\tan x)\,dx equals
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medium
∫0π/2sin⁡x sin⁡2x sin⁡3x sin⁡4x dx\displaystyle\int_{0}^{\pi/2}\sin x\,\sin 2x\,\sin 3x\,\sin 4x\,dx equals
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medium
If f(x)=∫01dt1+∣x−t∣f(x)=\displaystyle\int_{0}^{1}\dfrac{dt}{1+|x-t|}, then the value of f′ ⁣(12)f'\!\left(\dfrac12\right) is
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medium
Let f(x)f(x) be the maximum and g(x)g(x) be the minimum of {x∣x∣, x2∣x∣}\big\{x|x|,\,x^{2}|x|\big\}. Then ∫−11[f(x)−g(x)] dx\displaystyle\int_{-1}^{1}\big[f(x)-g(x)\big]\,dx equals
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medium
Let T>0T>0 be a fixed real number. Suppose ff is continuous with f(x+T)=f(x)f(x+T)=f(x) for all x∈Rx\in\mathbb{R}. If I=∫0Tf(x) dxI=\displaystyle\int_{0}^{T}f(x)\,dx, then ∫33+3Tf(2x) dx=KI\displaystyle\int_{3}^{3+3T}f(2x)\,dx=KI, where KK is
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medium
If [ ⋅ ][\,\cdot\,] denotes the greatest integer function, ∫02π[ ∣sin⁡x∣−∣cos⁡x∣ ] dx\displaystyle\int_{0}^{2\pi}\big[\,|\sin x|-|\cos x|\,\big]\,dx equals
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medium
∫−22max⁡(x+∣x∣, x−[x]) dx\displaystyle\int_{-2}^{2}\max\big(x+|x|,\ x-[x]\big)\,dx, where [x][x] denotes the greatest integer ≤x\le x, equals
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medium
∫02πx ln⁡ ⁣(3+cos⁡x3−cos⁡x)dx\displaystyle\int_{0}^{2\pi}x\,\ln\!\left(\dfrac{3+\cos x}{3-\cos x}\right)dx equals
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medium
The range of the function f(x)=∫1x∣t∣ dtf(x)=\displaystyle\int_{1}^{x}|t|\,dt, x∈[−12,12]x\in\left[-\dfrac12,\dfrac12\right], is
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medium
The value of the definite integral ∫−1/21/2(sin⁡−1(3x−4x3)−cos⁡−1(4x3−3x))dx\displaystyle\int_{-1/2}^{1/2}\Big(\sin^{-1}(3x-4x^{3})-\cos^{-1}(4x^{3}-3x)\Big)dx is
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medium
A function g(x)g(x) is given by g(x)=x1+(log⁡ex)(log⁡ex)⋯∞g(x)=\dfrac{x}{1+(\log_{e}x)(\log_{e}x)\cdots\infty}, where x∈[1,∞)x\in[1,\infty). Then ∫12eg(x) dx\displaystyle\int_{1}^{2e}g(x)\,dx is equal to
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medium
Suppose f(x)=∫0x1+t4 dtf(x)=\displaystyle\int_{0}^{x}\sqrt{1+t^{4}}\,dt for all real xx and let f(1)=cf(1)=c. Then the value of (f−1)′(c)\big(f^{-1}\big)'(c) is
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medium
If xx satisfies (∫01dtt2+2tcos⁡α+1)x2−(∫−33t2sin⁡2tt2+1 dt)x−2=0\left(\displaystyle\int_{0}^{1}\dfrac{dt}{t^{2}+2t\cos\alpha+1}\right)x^{2}-\left(\displaystyle\int_{-3}^{3}\dfrac{t^{2}\sin 2t}{t^{2}+1}\,dt\right)x-2=0 for 0<α<π0<\alpha<\pi, then the value of xx is
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medium
A cubic f(x)f(x) vanishes at x=−2x=-2 and has local minimum/maximum at x=−1x=-1 and x=13x=\dfrac13. If ∫−11f(x) dx=143\displaystyle\int_{-1}^{1}f(x)\,dx=\dfrac{14}{3}, then f(1)=f(1)=
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medium
If A=∫1sin⁡θt dt1+t2A=\displaystyle\int_{1}^{\sin\theta}\dfrac{t\,dt}{1+t^{2}} and B=∫1cosec⁡θdtt (1+t2)B=\displaystyle\int_{1}^{\operatorname{cosec}\theta}\dfrac{dt}{t\,(1+t^{2})}, then ∣AA2BeA+BB2−11A2+B2−1∣\begin{vmatrix}A & A^{2} & B\\[2pt] e^{A+B} & B^{2} & -1\\[2pt] 1 & A^{2}+B^{2} & -1\end{vmatrix} equals
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medium
If f(x)f(x) is a monotonic and differentiable function, then ∫f(a)f(b)2x(b−f−1(x)) dx\displaystyle\int_{f(a)}^{f(b)}2x\big(b-f^{-1}(x)\big)\,dx equals
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medium
If A=∫0πcos⁡x(x+2)2 dxA=\displaystyle\int_{0}^{\pi}\dfrac{\cos x}{(x+2)^{2}}\,dx, then ∫0π/2sin⁡2xx+1 dx\displaystyle\int_{0}^{\pi/2}\dfrac{\sin 2x}{x+1}\,dx is equal to
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medium
The value of ∫0nπ+t(∣cos⁡x∣+∣sin⁡x∣) dx\displaystyle\int_{0}^{n\pi+t}\big(|\cos x|+|\sin x|\big)\,dx, where 0<t<π20<t<\dfrac{\pi}{2}, is
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medium
If ∫01e5x6(x−α) dx=0\displaystyle\int_{0}^{1}e^{5x^{6}}(x-\alpha)\,dx=0, then
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medium
The value of the integral ∫01/3dx(1+x2)1−x2\displaystyle\int_{0}^{1/\sqrt3}\dfrac{dx}{(1+x^{2})\sqrt{1-x^{2}}} must be
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medium
∫0∞f ⁣(xn+x−n)ln⁡x dxx\displaystyle\int_{0}^{\infty}f\!\big(x^{n}+x^{-n}\big)\ln x\,\dfrac{dx}{x} equals
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medium
The value of ∫−13(tan⁡−1xx2+1+tan⁡−1x2+1x)dx\displaystyle\int_{-1}^{3}\left(\tan^{-1}\dfrac{x}{x^{2}+1}+\tan^{-1}\dfrac{x^{2}+1}{x}\right)dx is
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medium
If f(x)=∫1xdt2+t4f(x)=\displaystyle\int_{1}^{x}\dfrac{dt}{2+t^{4}}, then
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medium
∫0π/4(xxsin⁡x+cos⁡x)2dx\displaystyle\int_{0}^{\pi/4}\left(\dfrac{x}{x\sin x+\cos x}\right)^{2}dx equals
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medium
The value of ∫−π/4π/41+sin⁡xcos⁡x cos⁡2x dx\displaystyle\int_{-\pi/4}^{\pi/4}\dfrac{1+\sin x}{\cos x\,\sqrt{\cos 2x}}\,dx is
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medium
If a function f(x)f(x) satisfies the relation f(x)=ex+∫01exf(t) dtf(x)=e^{x}+\displaystyle\int_{0}^{1}e^{x}f(t)\,dt, then f(x)=f(x)=
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medium
Consider the real-valued function y=f(x)=∫01x2+t22−t dty=f(x)=\displaystyle\int_{0}^{1}\dfrac{x^{2}+t^{2}}{2-t}\,dt. It represents
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medium
If I1=∫01x2(2−x3) ex3 dxI_{1}=\displaystyle\int_{0}^{1}\dfrac{x^{2}}{\big(2-x^{3}\big)\,e^{x^{3}}}\,dx and I2=∫01ex1+x dxI_{2}=\displaystyle\int_{0}^{1}\dfrac{e^{x}}{1+x}\,dx, then I1:I2I_{1}:I_{2} equals
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medium
Let f(x)f(x) satisfy f′(x)=f(x)f'(x)=f(x) with f(0)=1f(0)=1, and let gg satisfy f(x)+g(x)=x2f(x)+g(x)=x^{2}. The value of ∫01f(x) g(x) dx\displaystyle\int_{0}^{1}f(x)\,g(x)\,dx is
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medium
∫−2π2πsin⁡6x(sin⁡6x+cos⁡6x)(1+e−x) dx\displaystyle\int_{-2\pi}^{2\pi}\dfrac{\sin^{6}x}{\big(\sin^{6}x+\cos^{6}x\big)\big(1+e^{-x}\big)}\,dx equals
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medium
lim⁡n→∞(cos⁡π2n cos⁡2π2n cos⁡3π2n⋯cos⁡(n−1)π2n)1/n\displaystyle\lim_{n\to\infty}\left(\cos\dfrac{\pi}{2n}\,\cos\dfrac{2\pi}{2n}\,\cos\dfrac{3\pi}{2n}\cdots\cos\dfrac{(n-1)\pi}{2n}\right)^{1/n} equals
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medium
f(x)f(x) is a differentiable function such that
f(x)=x2+∫0xe−tf(x−t) dt,f(x) = x^2+\int_0^x e^{-t}f(x-t)\,dt ,
 then
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medium
If
I1=∫0π/2sin⁡(sin⁡x)sin⁡x dx,I2=∫0π/2sin⁡xx dx,I3=∫0π/2sin⁡(tan⁡x)tan⁡x dx,I_1 = \int_0^{\pi/2}\frac{\sin\left(\sin x\right)}{\sin x}\,dx ,\qquad I_2 = \int_0^{\pi/2}\frac{\sin x}{x}\,dx ,\qquad I_3 = \int_0^{\pi/2}\frac{\sin\left(\tan x\right)}{\tan x}\,dx ,
 then which of the following is true
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medium
If 3f(x)+4f(3−x)=3x2−83f(x)+4f(3-x) = 3x^2-8 and ∫αβf(x) dx≤0\displaystyle\int_{\alpha}^{\beta}f(x)\,dx \le 0, where α\alpha, β\beta (α<β)\left(\alpha<\beta\right) are roots of x2−3x+a=0x^2-3x+a = 0, then 'aa' can be
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medium
If
1e−1∑n=1501(n+1)!∫−nne{∣x∣} dx1+esin⁡x=1−1(ab)!\frac{1}{e-1}\sum_{n=1}^{50}\frac{1}{(n+1)!}\int_{-n}^{n}e^{\{|x|\}}\,\frac{dx}{1+e^{\sin x}} = 1-\frac{1}{(ab)!}
 where aa and bb are prime, then ∣a−b∣\left|a-b\right| is equal to ({ ⋅ }\{\,\cdot\,\} denotes the fractional part)
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medium
If 3x2+1x4+x2≥max⁡f(x)\dfrac{3x^2+1}{\sqrt{x^4+x^2}} \ge \max f(x), where f(x)f(x) is constant for 0<x≤20<x\le\sqrt2, then the minimum value of
∫023x2+1x4+x2 dx\int_{0}^{\sqrt2}\frac{3x^2+1}{\sqrt{x^4+x^2}}\,dx
 is
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medium
If f(n)=∑k=1k=n∫kk+1(k+1)dxx(x+1)f(n) = \displaystyle\sum_{k=1}^{k=n}\int_{k}^{k+1}\frac{\left(k+1\right)dx}{x\left(x+1\right)} and g(x)=(x+1)ln⁡(x+2x+1)+f(x)g(x) = \left(x+1\right)\ln\left(\dfrac{x+2}{x+1}\right)+f(x), then
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medium
∫48(16x−64+x+x−16x−64)dx=\int_4^8\left(\sqrt{\sqrt{16x-64}+x}+\sqrt{x-\sqrt{16x-64}}\right)dx =
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medium
If f(0)=1f\left(0\right) = 1, f(2)=3f\left(2\right) = 3, f′(2)=5f'\left(2\right) = 5 then the value of the definite integral ∫01x f′′(2x)dx\displaystyle\int_0^1x\,f''\left(2x\right)dx is
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medium
The value of
[π⋅21⋅23⋅43⋅45⋅65⋅67⋅87⋅89⋯∞]\left[\pi\cdot\frac21\cdot\frac23\cdot\frac43\cdot\frac45\cdot\frac65\cdot\frac67\cdot\frac87\cdot\frac89\cdots\infty\right]
 is Where [ ⋅ ]\left[\ \cdot\ \right] is G.I.F.
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medium
Let f(x)=∫x1ln⁡(1−t2)dtf(x) = \displaystyle\int_x^1\ln\left(1-t^2\right)dt, x∈(0,1)x \in \left(0,1\right) and α=∫01f(x) dx\alpha = \displaystyle\int_0^1f(x)\,dx, then the value of 4α+54\alpha+5 is
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medium
f(x)f(x) is a continuous and increasing function satisfying f(x)=f(x−3)+4f(x) = f\left(x-3\right)+4 and ∫06f(x) dx=0\displaystyle\int_0^6f(x)\,dx = 0. Choose the CORRECT option(s)
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medium
f:R→Rf:\mathbb R\to\mathbb R is a continuous function satisfying
2f(x)+f(x2+2)=x2+2x+3,2f(x)+f\left(\frac x2+2\right) = x^2+2x+3 ,
 then ∫03f(x) dx\displaystyle\int_0^3f(x)\,dx is
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medium
Let f(x)f(x) be an even function such that f(x)+f(x−2)=x2+(1−2x)f(x)+f\left(x-2\right) = x^2+\left(1-2x\right) ∀ x∈R\forall\,x \in \mathbb R and
∫02(f(x)x2−2x−1)dx=1+aln⁡(b),\int_0^2\left(\frac{f(x)}{x^2-2x-1}\right)dx = 1+a\ln\left(b\right),
 then the value of [a2]+[b]\left[a^2\right]+\left[b\right] is (where [ . ][\,.\,] represents the greatest integer function)
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easy
∫0π/2sin⁡x1+cos⁡x+sin⁡x dx\displaystyle\int_{0}^{\pi/2}\dfrac{\sin x}{1+\cos x+\sin x}\,dx equals
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easy
If f(x)=∫1/xxcos⁡t2 dtf(x)=\displaystyle\int_{1/x}^{\sqrt{x}}\cos t^{2}\,dt for x>0x>0, then df(x)dx\dfrac{df(x)}{dx} is
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easy
If ∫0π/2sin⁡10x cos⁡10x dx=2−n∫0π/2(sin⁡10x+cos⁡10x) dx\displaystyle\int_{0}^{\pi/2}\sin^{10}x\,\cos^{10}x\,dx=2^{-n}\displaystyle\int_{0}^{\pi/2}\big(\sin^{10}x+\cos^{10}x\big)\,dx, n∈Nn\in\mathbb{N}, then n=n=
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easy
If ∫sin⁡x1t2f(t) dt=1−sin⁡x\displaystyle\int_{\sin x}^{1}t^{2}f(t)\,dt=1-\sin x for all x∈(0,π2]x\in\left(0,\dfrac{\pi}{2}\right], then the value of f ⁣(13)f\!\left(\dfrac{1}{\sqrt3}\right) is
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easy
Let f:(0,∞)→Rf:(0,\infty)\to\mathbb{R} and F(x)=∫0xf(t) dtF(x)=\displaystyle\int_{0}^{x}f(t)\,dt. If F ⁣(x2)=x2(1+x)F\!\big(x^{2}\big)=x^{2}(1+x), then f(4)f(4) equals
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easy
If f(x)=∫0x(1+t3)−1/2 dtf(x)=\displaystyle\int_{0}^{x}\big(1+t^{3}\big)^{-1/2}\,dt and g(x)g(x) is the inverse of ff, then the value of g′′(x)g2(x)\dfrac{g''(x)}{g^{2}(x)} is
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easy
The value of ∫36(x+12x−36+x−12x−36)dx\displaystyle\int_{3}^{6}\left(\sqrt{x+\sqrt{12x-36}}+\sqrt{x-\sqrt{12x-36}}\right)dx is
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easy
If z≠0z\ne 0, then ∫x=0100[arg⁡∣z∣] dx\displaystyle\int_{x=0}^{100}\big[\arg|z|\big]\,dx, where [ ⋅ ][\,\cdot\,] denotes the greatest integer function, is
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easy
If f(x)f(x) is a differentiable function and ∫0x3t2f(t) dt=313x13+5\displaystyle\int_{0}^{x^{3}}t^{2}f(t)\,dt=\dfrac{3}{13}x^{13}+5, then f ⁣(827)f\!\left(\dfrac{8}{27}\right) equals
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easy
Let f(x), g(x), h(x)f(x),\,g(x),\,h(x) be continuous on [0,2a][0,2a] with f(2a−x)=f(x)f(2a-x)=f(x), g(2a−x)=g(x)g(2a-x)=g(x) and h(x)+h(2a−x)=3h(x)+h(2a-x)=3. Then ∫02af(x) g(x) h(x) dx\displaystyle\int_{0}^{2a}f(x)\,g(x)\,h(x)\,dx equals
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easy
Let aa and bb be positive real numbers and f(a,b)=∫abexa−ebxx dx\displaystyle f\left(a,b\right) = \int_a^b\frac{e^{\frac xa}-e^{\frac bx}}{x}\,dx, then
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easy
For positive integers k=1,2,3,…,nk = 1,2,3,\dots,n, let SkS_k denote the area of △AOBk\triangle AOB_k (where OO is the origin) such that ∠AOBk=kπ2n\angle AOB_k = \dfrac{k\pi}{2n}, OA=1OA = 1 and OBk=kOB_k = k. Then the value of lim⁡n→∞1n2∑k=1nSk\displaystyle\lim_{n\to\infty}\frac1{n^2}\sum_{k=1}^{n}S_k is
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