Definite Integration
$ \text { The value of } 5050 \frac{\int_0^1\left(1-x^{50}\right)^{100} d x}{\int_0^{\frac{1}{1}}\left(1-x^{50}\right)^{101} d x} \text { is : } $
Explanation:
$ \begin{aligned} \mathrm{I} & =\frac{5050 \int_0^1\left(1-x^{50}\right)^{100} d x}{\int_0^1\left(1-x^{50}\right)^{100} d x} \\ & =5050 \frac{\mathrm{I}_{100}}{\mathrm{I}_{101}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{aligned} $
Now, $\mathrm{I}_{101}=\int_0^1\left(1-x^{50}\right)^{101} d x$
Using integration by part
$ \begin{aligned} & \begin{aligned} = & {\left[\left(1-x^{50}\right)^{101} x\right]_0^1 } \end{aligned}+\int_0^1 101 \times 50\left(1-x^{50)^{100}} x^{49} x d x\right. \\ & =0+5050 \int_0^1 x^{50}\left(1-x^{50)^{100}} d x\right. \\ & =-5050 \int_0^1\left[\left(1-x^{50}\right)-1\right]\left(1-x^{50}\right)^{100} d x \\ & =-5050\left[\int_0^1\left(1-x^{50}\right)^{101} d x-\int_0^1\left(1-x^{50}\right)^{100} d x\right] \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $
Using (ii) in (i), we have
$ \begin{aligned} \mathrm{I} & =\frac{5050 \mathrm{I}_{100}}{5050 \mathrm{I}_{100}} \times 5051 \\ & =5051 \end{aligned} $
If $a_n=\frac{3}{4}-\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^3+\cdots \cdots(-1)^{n-1}\left(\frac{3}{4}\right)^n$ and $b_n=1-a_n$, then find the minimum natural number $n_0$ such that $b_n>a_n \forall n>n_0$
Explanation:
$ \begin{array}{ll} & a_n=\frac{3}{4}-\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^3+\ldots(-1)^{n-1}\left(\frac{3}{4}\right)^n \\ & =\frac{3}{4}\left[\frac{1-\left(\frac{-3}{4}\right)^n}{1+\frac{3}{4}}\right]=\frac{3}{7}\left[1-\left(\frac{-3}{4}\right)^n\right] \\ & b_n>a_n \Rightarrow 2 a_n<1 \\ \Rightarrow & \frac{6}{7}\left[1-\left(\frac{-3}{4}\right)^n\right]<1 \\ \Rightarrow & 1-\left(\frac{-3}{4}\right)^n<\frac{7}{6} \\ \Rightarrow & \frac{-1}{6}<\left(\frac{-3}{4}\right)^n \\ \Rightarrow & \frac{1}{6}>\frac{-(-3)^n}{2^{2 n}} \\ \Rightarrow & 2^{2 n-1}>-(-3)^{n+1} \end{array} $
For $n$ to be even, inequality always holds.
For $n$ to be odd, it holds for $n \geq 7$.
Thus minimum natural number
$ n_0=6 $
Evatuate:
$\int_\limits{0}^{\pi} e^{|\cos x|}\left[2 \sin \left(\frac{1}{2} \cos x\right)+3 \cos \left(\frac{1}{2} \cos x\right)\right] \sin x ~d x$
$\int\limits_0^{\pi /2} {f\left( {\cos 2x} \right)\cos x\,dx = \sqrt 2 } \int\limits_0^{\pi /4} {f\left( {\sin 2x} \right)\cos x\,dx.} $
function such that for all $x \in R$, $f\left( {x + T} \right) = f\left( x \right)$.
If $I = \int\limits_0^T {f\left( x \right)dx} $ then the value of $\int\limits_3^{3 + 3T} {f\left( {2x} \right)dx} $ is
function such that for all $x \in R$, $f\left( {x + T} \right) = f\left( x \right)$.
If $I = \int\limits_0^T {f\left( x \right)dx} $ then the value of $\int\limits_3^{3 + 3T} {f\left( {2x} \right)dx} $ is
$f\left( x \right) + f\left( {{1 \over x}} \right)$ and show that $f\left( e \right) + f\left( {{1 \over e}} \right) = {1 \over 2}.$
Here, $\ln t = {\log _e}t$.
${1 \over 2} \le f\left( t \right) \le 1,$ for $t \in \left[ {0,1} \right]$ and $\,0 \le f\left( t \right) \le {1 \over 2},$ for $t \in \left[ {1,2} \right]$.
Then $g(2)$ satisfies the inequality
equal to $y$, then the value of the integral $\int\limits_{\pi /2}^{3\pi /2} {\left[ {2\sin x} \right]dx} $ is
Hence or otherwise, evaluate the integral
$\int_0^1 {{{\tan }^{ - 1}}\left( {1 - x + {x^2}} \right)dx.} $
then one of the possible values of $k$ is ............
$\int_1^2 {f\left( x \right)dx} = .......$
$\int\limits_0^1 {f\left( x \right)dx = {{2A} \over \pi },} $ then constants $A$ and $B$ are
${{\sin 2kx} \over {\sin x}} = 2\left[ {\cos x + \cos 3x + ......... + \cos \left( {2k - 1} \right)x} \right]$
Hence prove that $\int\limits_0^{\pi /2} {\sin 2kx\,\cot \,x\,dx = {\pi \over 2}} $
$\int\limits_{ - \pi /2}^{\pi /2} {\left[ {f\left( x \right) + f\left( { - x} \right)} \right]\left[ {g\left( x \right) - g\left( { - x} \right)} \right]dx} $ is
$f\left( x \right) = f\left( {a - x} \right)$ and $g\left( x \right) + g\left( {a - x} \right) = 2,$
then show that $\int\limits_0^a {f\left( x \right)g\left( x \right)dx = \int\limits_0^a {f\left( x \right)dx} } $
Where [ ] denotes the greatest integer function, equals .............