Recently Active Calculus Questions

Find ∫ ( 1/x6 + 1/x8 )1/3dx i.e integrate cubic root of (1/x6+1/x8) ...

\hspace{-16}\bf{\int_{0}^{\frac{\pi}{4}}\ln \left(\frac{1+\sin^2 2x}{\sin^4 x+\cos^4 x}\right)dx} ...

\hspace{-16}\bf{\int_{-1}^{1}\frac{2x^{1004}+x^{3014}+x^{2008}.\sin(x)^{2007}}{1+x^{2010}}dx} ...

\hspace{-16}\bf{\int_{0}^{4\pi}\ln\left|13.\sin (x)+3\sqrt{3}.\cos (x)\right|dx}= ...

\hspace{-16}\bf{\int\frac{x^2.\cos^{-1}\big(x\sqrt{x}\big)}{\big(1-x^3\big)^2}dx} ...

\hspace{-16}\bf{\int_{\frac{25\pi}{4}}^{\frac{53\pi}{4}}\frac{1}{(1+2^{\sin x}).(1+2^{\cos x})}dx} ...

\hspace{-16}\bf{\lim_{n\rightarrow \infty}\sum_{i=0}^{n}\;\sum_{j=0}^{n-i}\frac{x^j}{i!\;.\;j!}=} ...

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\hspace{-16}\bf{(1)\;\; \int_{0}^{\infty}\frac{\ln (x)}{x^2+4}dx}$\\\\\\ $\bf{(2)}\;\;$ Find Max. value of $\bf{\int_{0}^{1}f^3(x)dx}$\\\\\\ Given $\bf{\mid f(x)\mid \leq 1}$ and $\bf{\int_{0}^{1}f(x)dx=0}$\\\\\\ $\bf{(3)\;\; ...

\int_{\pi /2}^{5\pi /2}{ \frac{e^{tan^{-1}(sinx)}}{e^{tan^{-1}(sinx) }+ e^{tan^{-1}(cosx)}}} dx A)1 B)Ï€ C)e D)none of these ...

\hspace{-16}$Let $\bf{a\in \mathbb{R}}$. If the value of $\bf{\int_{-\pi+a}^{3\pi+a}\mid x-a-\pi\mid\sin \left(\frac{x}{2}\right)dx=-16}$\\\\\\ Then Sum of all Integer value of $\bf{a}$ in $\bf{\left[0,314\right]}$ is $\bf{k\ ...

\hspace{-16}\bf{\int \frac{\sin^2(2012\; x)}{\sin \; (x)}dx} ...

consider a function f on non-negative integers such that f(0)=1 , f(1)=0 and f(n) + f(n-1) = nf(n-1) + (n-1)f(n-2) for n≥2 . Show that , \frac{f(n)}{n!} = \sum_{k=0}^{n}{\frac{(-1)^{k}}{k!}} ...

\hspace{-16}\mathbf{\lim_{n\rightarrow \infty}\sum_{r=0}^{n}\frac{1}{\binom{n}{r}}=} ...

\hspace{-16}\bf{\int_{-\pi}^{\pi}\frac{\sin (0.5+n)x}{2\sin (0.5)x}dx=}$\\\\\\ Where $\mathbf{n\in\mathbb{N}}$ and $\bf{0.5=\frac{1}{2}}$ ...

find ∫ x3-2/ (x3+1)3 dx ...

\hspace{-16}\mathbf{\int_{1}^{\infty}\frac{(x^3+3)}{x^6.(x^2+1)}dx=\frac{a+b\pi}{c}}$.\\\\\\ Then $\mathbf{(a,b,c)=}$ ...

∫ x2+x-1/x3+x2 -6x dx ...

Please suggest a good book for solving quality calculus problems for JEE preparation. ...

\hspace{-16}$Let $\bf{L=\lim_{n\rightarrow \infty}\frac{1}{\frac{1}{\sqrt{2}}.\sqrt{\big(\frac{1}{2}}+\frac{1}{2}\sqrt{\frac{1}{2}}\big).\sqrt{\big(\frac{1}{2}}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}...... ...

∫l log x/x1/2 upper limit 1 lower limit 0 form the positive side ...

lim (1+3x)10/x as x→0+ ...

\hspace{-16}\bf{\int \big(x-\sqrt{x^2+1}\big)^{2012}dx} ...

\hspace{-16}$If $\bf{x,y,z\in \mathbb{R}}$ and $\bf{f(x).f(y).f(z) = 12f(xyz) ô€€€ -16xyz}$\\\\ Then No. of function which satisy $\bf{f:\mathbb{R}\rightarrow \mathbb{R}}$ is ...

∫sin2(log x) dx also ∫ log x/(1 + log x)2 dx ...

∫[ ( x2 + 1 ) / ( x4 + 1) ] dx ...

\hspace{-16}$If $\bf{f(1)=1}$ and $\bf{f(x+5) > f(x) +5 f(x+1) < f(x) + 1}$\\\\ and $\bf{g(x)=f(x)-x+1}$.Then $\bf{g(2012)}$ is ...

if x*tan(y) + 1 = (1+x^2)^(1/2) , find dy/dx. ...

\int_{0}^{x}{f(t)d(t)} ----> 5 as |x| ---> 1, then value of 'a' so that the equation 2x + \int_{0}^{x}{f(t)d(t)} = a has at least two roots of opposite signs in (-1,1) is a) 0<a<1 b) 0<a<3 c) -1<a<∞ ...

let f be a one to one continuous function such that f(2) =3and f(5)=7. given (2to5)∫ f(x) dx =17 then value of (3to7)∫ f-1(x) dx is (a) 10 (b) 11 (c)12 (d) 13 ans c 2 if nCk is cmbination of n diff. thing taking k at a ti ...