Definite Integration

$ \begin{aligned} \operatorname{Minimum}\left\{\mathrm{x}^2,\{\mathrm{x}\}\right\} & =\mathrm{x}^2 ; \mathrm{x} \in[0,1) \\\\ {\left[\mathrm{x}-\log _{\mathrm{e}} \mathrm{x}\right] } & =1 ; \mathrm{x} \in[1,2) \end{aligned} $

$ \therefore \mathrm{f}(\mathrm{x})=\left\{\begin{array}{l} \mathrm{e}^{\mathrm{x}^2} ; \mathrm{x} \in[0,1) \\\\ \mathrm{e} ; \mathrm{x} \in[1,2) \end{array}\right. $

$ \begin{aligned} & \int\limits_0^2 x f(x)=\int\limits_0^1 x \cdot e^{x^2} d x+\int\limits_1^2 x \cdot e d x \\\\ & x^2=t \Rightarrow 2 x d x=d t \end{aligned} $

$ =\frac{1}{2} \int\limits_0^1 e^t d t+\left.e \frac{x^2}{2}\right|_1 ^2 $

$ \begin{aligned} & =\frac{1}{2}(\mathrm{e}-1)+\frac{1}{2}(4-1) \mathrm{e} \\\\ & =2 \mathrm{e}-\frac{1}{2} \end{aligned} $

$\int_\limits{-\log _{e} 2}^{\log _{e} 2} e^{x}\left(\log _{e}\left(e^{x}+\sqrt{1+e^{2 x}}\right)\right) d x$

Let $e^x=t \Rightarrow e^x d x=d t$

When, $x \rightarrow-\log _e 2$, then $t \rightarrow \frac{1}{2}$

When, $x \rightarrow \log _e 2$, then $t \rightarrow 2$

$ I=\int_\limits{\frac{1}{2}}^2\left[\log _e\left(t+\sqrt{1+t^2}\right)\right] d t $ ...........(i)

On applying integration by part method in Eq. (i), we get

$ \begin{aligned} & I=\left[t \log _e\left(t+\sqrt{1+t^2}\right)\right]_{1 / 2}^2-\int_{1 / 2}^2 \frac{t}{t+\sqrt{1+t^2}}\left(1+\frac{2 t}{2 \sqrt{1+t^2}}\right) d t \\\\ & =2 \log _e(2+\sqrt{5})-\frac{1}{2} \log _e\left(\frac{1+\sqrt{5}}{2}\right)-\int_{1 / 2}^2 \frac{t}{\sqrt{1+t^2}} d t \end{aligned} $

$ =\log _e\left(\frac{(2+\sqrt{5})^2}{\left(\frac{1+\sqrt{5}}{2}\right)^{1 / 2}}\right)-\frac{1}{2} \int_{1 / 2}^2 \frac{2 t}{\sqrt{1+t^2}} d t $ .............(ii)

Let $\quad I_1=\int_{1 / 2}^2 \frac{2 t}{\sqrt{1+t^2}} d t$

Let $1+t^2=w$

$2 t d t=d w$

When, $t \rightarrow \frac{1}{2}$, then $w=\frac{5}{4}$

When, $t \rightarrow 2$, then $w=5$

$ \begin{aligned} I_1 & =\int_{5 / 4}^5 \frac{1}{\sqrt{w}} d w \\\\ & =[2 \sqrt{w}]_{5 / 4}^5 \\\\ & =2\left[\sqrt{5}-\frac{\sqrt{5}}{2}\right]=\sqrt{5} \end{aligned} $

On substitute value of $I_1$ in Eq. (ii), we get

$ I=\log _e\left(\frac{\sqrt{2}(2+\sqrt{5})^2}{\sqrt{1+\sqrt{5}}}\right)-\frac{\sqrt{5}}{2} $

Given that

$ \int\limits_0^{t^2}\left(f(x)+x^2\right) d x=\frac{4}{3} t^3, \forall t>0 $

On differentiating using Leibnitz rule, we get

$ \begin{aligned} & \left(f\left(t^2\right)+t^4\right) \times 2 t=\frac{4}{3} \times 3 t^2 \\\\ & \Rightarrow f\left(t^2\right)+t^4=2 t \\\\ & \Rightarrow f\left(t^2\right)=2 t-t^4 \end{aligned} $

On substituting $\frac{\pi}{2}$ for $t$, we get

$ f\left(\frac{\pi^2}{4}\right)=2\left(\frac{\pi}{2}\right)-\frac{\pi^4}{16}=\pi\left(1-\frac{\pi^3}{16}\right) $

Let $I=\int\limits_0^\pi f(x) \sin x d x$ ..........(i)

$ \begin{aligned} & =\int\limits_0^\pi f(\pi-x) \sin (\pi-x) d x \\\\ & =\int\limits_0^\pi f(\pi-x) \sin x d x ........(ii) \end{aligned} $

On adding Equations (i) and (ii), we get

$ \begin{aligned} & 2 I=\int\limits_0^\pi[f(x)+f(\pi-x)] \sin x d x \\\\ & \Rightarrow 2 I=\int\limits_0^\pi \pi^2 \sin x d x=\pi^2 \int\limits_0^\pi \sin x d x \\\\ & =\pi^2(-\cos x)_0^\pi=-\pi^2(-1-1)=2 \pi^2 \\\\ & \Rightarrow I=\pi^2 \end{aligned} $

Let $L=\lim\limits_{n \rightarrow \infty}\left\{\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)\left(2^{\frac{1}{2}}-2^{\frac{1}{5}}\right) \ldots\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n-1}}\right)\right\}$

Here, $\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)^n<\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)\left(2^{\frac{1}{3}}-2^{\frac{1}{5}}\right)$ $ \ldots\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n-1}}\right)<\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n-1}}\right)^n $

$ \Rightarrow \lim\limits_{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)^n < L < \lim\limits_{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n-1}}\right)^n $

Since, $\lim\limits_{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)^n=0$ and $\lim\limits_{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n-1}}\right)^n=0$

$\therefore L = 0$

We have,

$ 5 f(x)+4 f\left(\frac{1}{x}\right)=\frac{1}{x}+3, x>0 $ ..........(i)

On replacing $x$ by $\frac{1}{x}$ in (i), we get

$ 5 f\left(\frac{1}{x}\right)+4 f(x)=x+3 $ ..........(ii)

Now, using Eq. (i) $\times 5-$ (ii) $\times 4$, we get

$ \begin{aligned} & 25 f(x)-16 f(x) =\left(\frac{5}{x}+15\right)-(4 x+12) \\\\ &\Rightarrow 9 f(x) =\frac{5}{x}-4 x+3 \\\\ &\Rightarrow f(x) =\frac{1}{9}\left(\frac{5}{x}-4 x+3\right) ...........(iii) \end{aligned} $

$ \begin{aligned} \therefore \quad & 18 \int_1^2 f(x) d x=18 \int_1^2 \frac{1}{9}\left(\frac{5}{x}-4 x+3\right) d x \text { [Using Eq. (iii)] }\\\\ = & 2 \int_1^2\left(\frac{5}{x}-4 x+3\right) d x \end{aligned} $

$ \begin{aligned} & =2\left[5 \log _e x-4\left(\frac{x^2}{2}\right)+3 x\right]_1^2 \\\\ & =2\left[\left(5 \log _e 2-2(2)^2+3(2)\right)-\left(5 \log 1-2(1)^2+3(1)\right)\right] \\\\ & =2\left[5 \log _e 2-8+6+2-3\right] [\because \log 1=0]\\\\ & =10 \log _e 2-6 \end{aligned} $

$ I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{(2-\cos 2 x)} d x $

Using $\int_a^b f(x) d x=\int_a^b f(a+b-x) d x$

$ I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\frac{-x+\frac{\pi}{4}}{2-\cos 2 x}\right) d x $

$\begin{aligned} & \therefore 2 I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\pi d x}{2(2-\cos 2 x)} \\\\ & \Rightarrow I=\frac{2 \pi}{4} \int_0^{\frac{\pi}{4}}\left(\frac{d x}{\left.\frac{2-1-\tan ^2 x}{1+\tan ^2 x}\right)}\right. \\\\ & \Rightarrow I=\frac{\pi}{2} \int_0^{\frac{\pi}{4}}\left(\frac{1+\tan ^2 x}{1+3 \tan ^2 x}\right) d x\end{aligned}$

Now,

$ \begin{aligned} & \tan x=t \\\\ & =\frac{\pi}{2} \int_0^1 \frac{d t}{1+3 t^2} \\\\ & =\frac{\pi}{2}\left[\frac{\tan ^{-1}(\sqrt{3} t)}{\sqrt{3}}\right]_0^1 \\\\ & =\frac{\pi}{2 \sqrt{3}}\left(\frac{\pi}{3}\right)=\frac{\pi^2}{6 \sqrt{3}} \end{aligned} $

$ \begin{aligned} & \lim _{n \rightarrow \infty}\left[\frac{1}{1+n}+\frac{1}{2+n}+\frac{1}{3+n}+\ldots \ldots+\frac{1}{2 n}\right] \\\\ & =\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{r+n} \\\\ & =\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{n}\left(\frac{1}{\frac{r}{n}+1}\right) \\\\ & =\int_0^1 \frac{d x}{x+1} \\\\ & =\left.\log _e(1+\mathrm{x})\right|_0 ^1 \\\\ & =\log _e^2 \end{aligned} $

$I=\int\limits_{0}^{\alpha} \frac{x}{\sqrt{x+\alpha}-\sqrt{x}} d x, \alpha>0$

$ \begin{aligned} & =\frac{1}{\alpha} \int\limits_{0}^{\alpha} x(\sqrt{x+\alpha}+\sqrt{x}) d x \end{aligned} $

$\eqalign{ & = {1 \over \alpha }\int\limits_0^\alpha {\left( {x\sqrt {x + \alpha } + x\sqrt x } \right)} dx \cr & = {1 \over \alpha }\int\limits_0^\alpha {\left[ {\left( {x + \alpha - \alpha } \right)\sqrt {x + \alpha } + {x^{{3 \over 2}}}} \right]} dx \cr} $

$ \begin{aligned} & =\frac{1}{\alpha}\int\limits_0^\alpha \left[(x+\alpha)^{3 / 2}-\alpha(x+\alpha)^{1 / 2}+x^{3 / 2}\right] d x \\\\ & =\frac{1}{\alpha}\left[\frac{2}{5}(x+\alpha)^{5 / 2}-\alpha \frac{2}{3}(x+\alpha)^{3 / 2}+\frac{2}{5} x^{5 / 2}\right]_0^\alpha \\\\ & =\frac{1}{\alpha}\left(\frac{2}{5}(2 \alpha)^{5 / 2}-\frac{2 \alpha}{3}(2 \alpha)^{3 / 2}+\frac{2}{5} \alpha^{5 / 2}-\frac{2}{5} \alpha^{5 / 2}+\frac{2}{3} \alpha^{5 / 2}\right) \\\\ & =\frac{1}{\alpha}\left(\frac{2^{7 / 2} \alpha^{5 / 2}}{5}-\frac{2^{5 / 2} \alpha^{5 / 2}}{3}+\frac{2}{3} \alpha^{5 / 2}\right)=\alpha^{3 / 2}\left(\frac{2^{7 / 2}}{5}-\frac{2^{5 / 2}}{3}+\frac{2}{3}\right) \\\\ & =\frac{\alpha^{3 / 2}}{15}(24 \sqrt{2}-20 \sqrt{2}+10)=\frac{\alpha^{3 / 2}}{15}(4 \sqrt{2}+10) \end{aligned} $

Now, $\frac{\alpha^{3 / 2}}{15}(4 \sqrt{2}+10)=\frac{16+20 \sqrt{2}}{15}=\frac{2 \sqrt{2}(4 \sqrt{2}+10)}{15}$

$ \Rightarrow \alpha^{3 / 2}=2 \sqrt{2}=2^{3 / 2} $

$\Rightarrow \alpha=2 $

$\phi(x)=\frac{1}{\sqrt{x}} \int_{\pi / 4}^{x}\left(4 \sqrt{2} \sin t-3 \phi^{\prime}(t)\right) d t$

$\Rightarrow \phi^{\prime}(x)=\frac{-1}{2 x^{3 / 2}} \int_{\pi / 4}^{x}\left(4 \sqrt{2} \sin t-3 \phi^{\prime}(t)\right) d t$

$+\frac{1}{\sqrt{x}}\left(4 \sqrt{2} \sin (x)-3 \phi^{\prime}(x)\right)$

At, $x=\frac{\pi}{4}$

$\phi^{\prime}\left(\frac{\pi}{4}\right)=\frac{-1}{2\left(\frac{\pi}{4}\right)^{3 / 2}} \times 0+\sqrt{\frac{4}{\pi}}\left(4 \sqrt{2} \times \frac{1}{\sqrt{2}}-3 \phi^{\prime}\left(\frac{\pi}{4}\right)\right)$

$\Rightarrow \phi^{\prime}\left(\frac{\pi}{4}\right)\left[1+\frac{6}{\sqrt{\pi}}\right]=\frac{2}{\sqrt{\pi}} \times 4$

$\Rightarrow \phi^{\prime}\left(\frac{\pi}{4}\right)=\frac{8}{6+\sqrt{x}}$

$\int_{0}^{\alpha} \frac{t^{50}}{1-t} d t$

$ \begin{aligned} & = \int_{0}^{\alpha} \frac{t^{50}-1+1}{1-t}\\\\ & =-\int_{0}^{\alpha}\left(\frac{1-t^{50}}{1-t}-\frac{1}{1-t}\right) d t\\\\ & =-\int_{0}^{\alpha}\left(1+t+\ldots . .+t^{49}\right)+\int_{0}^{\alpha} \frac{1}{1-t} d t \end{aligned} $

$=-\left(\frac{\alpha^{50}}{50}+\frac{\alpha^{49}}{49}+\ldots . .+\frac{\alpha^{1}}{1}\right)+\left(\frac{\ln (1-\mathrm{f})}{-1}\right)_{0}^{\alpha}$

$=-\mathrm{P}_{50}(\alpha)-\ln (1-\alpha)$

$=-\mathrm{P}_{50}(\alpha)-\beta$

Let I = $\int_\limits{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{(2+3 \sin x)}{\sin x(1+\cos x)} d x$

$ =\int\limits_{\pi / 3}^{\pi / 2} \frac{2}{\sin x(1+\cos x)} d x+\int\limits_{\pi / 3}^{\pi / 2} \frac{3}{1+\cos x} d x $

$ \begin{aligned} =\int\limits_{\pi / 3}^{\pi / 2} \frac{2(1-\cos x)}{\sin x\left(1-\cos ^2 x\right)} & d x +\int\limits_{\pi / 3}^{\pi / 2} \frac{3}{1+\cos x} d x \end{aligned} $

$ =\int\limits_{\pi / 3}^{\pi / 2} \frac{2(1-\cos x)}{\sin ^3 x} d x+\int\limits_{\pi / 3}^{\pi / 2} \frac{3}{2 \cos ^2 \frac{x}{2}} d x $

$ \begin{array}{r} =\int\limits_{\pi / 3}^{\pi / 2} \frac{2}{\sin ^3 x} d x-\int\limits_{\pi / 3}^{\pi / 2} 2 \cot x \operatorname{cosec}^2 x d x +\int\limits_{\pi / 3}^{\pi / 2} \frac{3}{2} \sec ^2 \frac{x}{2} d x \end{array} $

$ =2 I_1-2 I_2+\frac{3}{2} I_3 $

Now,

$ \begin{aligned} I_1 & =\int\limits_{\pi / 3}^{\pi / 2} \operatorname{cosec}^3 x d x \\\\ & =\int\limits_{\pi / 3}^{\pi / 2} \sqrt{1+\cot ^2 x} \operatorname{cosec}^2 x d x \end{aligned} $

Put $\cot x=t \Rightarrow-\operatorname{cosec}^2 x d x=d t$

When, $x=\frac{\pi}{3} \Rightarrow t=\frac{1}{\sqrt{3}} \text { and } x=\frac{\pi}{2} \Rightarrow t=0$

$ \begin{aligned} \therefore & I_1=-\int\limits_{\frac{1}{\sqrt{3}}}^0 \sqrt{1+t^2} d t \\\\ = & \int\limits_0^{\frac{1}{\sqrt{3}}} \sqrt{1+t^2} d t \\\\ = & {\left[\frac{t}{2} \sqrt{1+t^2}+\frac{1}{2} \log \left[t+\sqrt{1+t^2}\right]\right]_0^{\frac{1}{\sqrt{3}}} } \\\\ = & \frac{1}{3}+\frac{1}{2} \log \sqrt{3} \end{aligned} $

$ I_2=\int\limits_{\pi / 3}^{\pi / 2} \cot x \operatorname{cosec}^2 x d x $

Put $\cot x=t \Rightarrow \operatorname{cosec}^2 x d x=-d t$

When, $x=\frac{\pi}{3}$ $\Rightarrow t=\frac{1}{\sqrt{3}}$ and $x=\frac{\pi}{2} \Rightarrow t=0$

$ \therefore \begin{aligned} I_2 & =-\int\limits_{\frac{1}{\sqrt{3}}}^0 t d t \\\\ & =\int\limits_0^{\frac{1}{\sqrt{3}} t} t d t=\left[\frac{t^2}{2}\right]_0^{\frac{1}{\sqrt{3}}}=\frac{1}{6} \end{aligned} $

$ \begin{aligned} & I_3=\int\limits_{\frac{\pi}{3}}^{\frac{\pi}{2}} \sec ^2 \frac{x}{2} d x \\\\ = & 2\left(\tan \frac{\pi}{4}-\tan \frac{\pi}{6}\right)=2\left(1-\frac{1}{\sqrt{3}}\right) \end{aligned} $

$ \begin{aligned} & \therefore \quad I=2 I_1-2 I_2+\frac{3}{2} I_3 \\\\ & =2\left(\frac{1}{3}+\frac{1}{2} \log \sqrt{3}\right)-2\left(\frac{1}{6}\right) +\frac{3}{2}\left(2\left(1-\frac{1}{\sqrt{3}}\right)\right) \end{aligned} $

$ \begin{aligned} & =\frac{2}{3}+\log \sqrt{3}-\frac{1}{3}+3-\sqrt{3} \\\\ & =\frac{10}{3}-\sqrt{3}+\log _e \sqrt{3} \end{aligned} $

$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 0}^{n - 1} {{3 \over n}{{\left( {2 + {r \over n}} \right)}^2}} $

$ = \int_0^1 {3{{(2 + x)}^2}\,dx} $

$ = \left. {3\,.\,{{{{(2 + x)}^3}} \over 3}} \right|_0^1$

$ = {3^3} - {2^3} = 19$

$I = {{3(e - 1)} \over e}\int_1^2 {{x^2}{e^{[x] + [{x^3}]}}dx} $

$ = {{3(e - 1)} \over e}\int_1^2 {{x^2}{e^{1 + [{x^3}]}}dx} $ ($\because$ $[x] = 1$ when $x \in (12)$)

$ = 3(e - 1)\int_1^2 {{x^2}{e^{[{x^3}]}}dx} $

Let ${x^3} = t$

$I = (e - 1)\int_1^8 {{e^{[t]}}dt} $

$ = ({e^{ - 1}})({e^1} + {e^2} + {e^3}\, + \,...\, + \,{e^7})$

$ = (e - 1)e{{({e^7} - 1)} \over {e - 1}}$

$ = {e^8} - e$

$\int_1^2 {{{{t^4} + 1} \over {{t^6} + 1}}dt} $

$ = \int_1^2 {{{{{({t^2} + 1)}^2}} \over {{t^6} + 1}}dt - 2\int_1^2 {{{{t^2}} \over {{t^6} + 1}}dt} } $

$ = \int_1^2 {{{{t^2} + 1} \over {{t^4} - {t^2} + 1}}dt - 2\int_1^2 {{{{t^2}} \over {{{({t^3})}^2} + 1}}dt} } $

$ = \left. {{{\tan }^{ - 1}}(2t + \sqrt 3 ) + {{\tan }^{ - 1}}(2t - \sqrt 3 )} \right|_1^2 - \left. {{2 \over 3}{{\tan }^{ - 1}}({t^3})} \right|_1^2$

$ = {\tan ^{ - 1}}(4 + \sqrt 3 ) + {\tan ^{ - 1}}(4 - \sqrt 3 ) - {\tan ^{ - 1}}(2 + \sqrt 3 ) - {\tan ^{ - 1}}(2 + \sqrt 3 ) - {\tan ^{ - 1}}(2\sqrt 3 ) - {2 \over 3}({\tan ^{ - 1}}8 - {\tan ^{ - 1}}1)$

$ = {\tan ^{ - 1}}2 + {1 \over 3}{\tan ^{ - 1}}8 - {\pi \over 3}$

$I = \int\limits_{{1 \over 2}}^2 {{{{{\tan }^{ - 1}}x} \over x}dx} $ ..... (i)

$x \to {1 \over x}$

$I = \int\limits_{{1 \over 2}}^2 {{1 \over x}{{\tan }^{ - 1}}{1 \over x}dx} $ ..... (ii)

$2I = \int\limits_{{1 \over 2}}^2 {{1 \over x}\,.\,{\pi \over 2}dx} $

$ = \left. {{\pi \over 2}\ln x} \right|_{{1 \over 2}}^2 = \pi \ln 2$

$ \Rightarrow I = {\pi \over 2}\ln 2$

$f(x)=x+\int\limits_{0}^{\pi / 2}(\sin x \cos y+\cos x \sin y) f(y) d y$

$f(x)=x+\int\limits_{0}^{\pi / 2}((\cos y f(y) d y) \sin x+(\sin y f(y) d y) \cos x)\quad....(1)$

On comparing with

$f(x)=x+\frac{a}{\pi^{2}-4} \sin x+\frac{b}{\pi^{2}-4} \cos x, x \in \mathbb{R}$ then

$\Rightarrow \frac{a}{\pi^{2}-4}=\int_{0}^{\pi / 2} \cos y f(y) d y \quad....(2)$

$\Rightarrow \frac{b}{\pi^{2}-4}=\int_{0}^{\pi / 2} \sin y f(y) d y \quad....(3)$

Add (2) and (3)

$\frac{a+b}{\pi^{2}-4}=\int_{0}^{\pi / 2}(\sin y+\cos y) f(y) d y \quad....(4)$

$\frac{a+b}{\pi^{2}-4}=\int_{0}^{\pi / 2}(\sin y+\cos y) f\left(\frac{\pi}{2}-y\right) d y\quad....(5)$

Add (4) and (5)

$\frac{2(a+b)}{\pi^{2}-4}=\int_{0}^{\pi / 2}(\sin y+\cos y)\left(\frac{\pi}{2}+\frac{(a+b)}{\pi^{2}-4}(\sin y+\cos y)\right) d y$

$=\pi+\frac{a+b}{\pi^{2}-4}\left(\frac{\pi}{2}+1\right)$

$(a+b)=-2 \pi(\pi+2)$

$I=\int \frac{d x}{x^{3}\left(x^{2}+2\right)^{2}}$

$=\frac{1}{4} \int \frac{x}{x^{2}+2} d x+\frac{1}{4} \int \frac{x}{\left(x^{2}+2\right)^{2}}-\frac{1}{4} \int \frac{d x}{x}+\frac{1}{4} \int \frac{d x}{x^{3}}$

$=\frac{1}{8} \ln \left(x^{2}+2\right)-\frac{\ln x}{4}-\frac{1}{8\left(x^{2}+2\right)}-\frac{1}{8 x^{3}}$

Now, $16 \int_{1}^{2} \frac{d x}{x^{3}\left(x^{2}+2\right)^{2}}=2 \ln 6-2 \ln 3-4 \ln 2+\frac{11}{6}$

$=\frac{11}{6}-\ln 4$

$ f(x)=\int_0^2 e^{|x-t|} d t $

For $x>2$

$ f(x)=\int_0^2 e^{x-t} d t=e^x\left(1-e^{-2}\right) $

For $x<0$

$ f(x)=\int_0^2 e^{t-x} d t=e^{-x}\left(e^2-1\right) $

For $x \in[0,2]$

$ \begin{aligned} & f(x)=\int_0^x e^{x-t} d t + \int_x^2 e^{t-x} d t \\\\ & =e^{2-x}+e^x-2 \end{aligned} $

For $x>2$

$ \left.f(x)\right|_{\min =e^2-1} $

For $\mathrm{x}<0$

$ \left.f(x)\right|_{\min =e^2-1} $

For $x \in[0,2]$

$ \left.f(x)\right|_{\min }=2(e-1) $

$ \int_{\frac{3 \sqrt{2}}{4}}^{\frac{3 \sqrt{3}}{4}} \frac{48}{\sqrt{9-4 x^2}} d x $

We have $\int \frac{d x}{\sqrt{a^2-x^2}}=\sin ^{-1} \frac{x}{a}+C$

Hence $\int_{\frac{3 \sqrt{2}}{4}}^{\frac{3 \sqrt{3}}{4}} \frac{48}{\sqrt{9-4 x^2}} d x=\frac{48}{2} \times\left[\sin ^{-1} \frac{2 x}{3}\right]_{\frac{3 \sqrt{2}}{4}}^{\frac{3 \sqrt{3}}{4}}$

$=24 \times\left[\sin ^{-1}\left(\frac{2}{3} \times \frac{3 \sqrt{3}}{4}\right)-\sin ^{-1}\left(\frac{2}{3} \times \frac{3 \sqrt{2}}{4}\right)\right]$ $=24 \times\left[\sin ^{-1} \frac{\sqrt{3}}{2}-\sin ^{-1} \frac{1}{\sqrt{2}}\right]$

$=24 \times\left(\frac{\pi}{3}-\frac{\pi}{4}\right)$

$=24 \times \frac{\pi}{12}=2 \pi$

$\begin{aligned} & f:(0,1) \rightarrow R, \\\\ & f(x)=\sqrt{n}, x \in\left[\frac{1}{n+1}, \frac{1}{n}\right), n \in N\end{aligned}$

$g:(0,1) \rightarrow R$ where

$ \int_{x^2}^x \sqrt{\frac{1-t}{t}} d t < g(x) < 2 \sqrt{x}, x \in(0,1) $

Now (According to the question)

$ \begin{aligned} & \lim _{x \rightarrow \infty} f(x) \cdot g(x) \\\\ & \Rightarrow \text { Put } x=\frac{1}{n} \\\\ & \lim _{n \rightarrow \infty} f\left(\frac{1}{n}\right) g\left(\frac{1}{n}\right) \end{aligned} $

$\begin{aligned} \lim _{n \rightarrow \infty} \sqrt{n-1} \int_{\frac{1}{n^2}}^{\frac{1}{n}} \sqrt{\frac{1-t}{t} d t} \leq \lim _{n \rightarrow \infty} & f\left(\frac{1}{n}\right) g\left(\frac{1}{n}\right) \leq \lim _{n \rightarrow \infty} \sqrt{n}-1 \frac{2}{\sqrt{n}}\end{aligned}$

$\Rightarrow \lim\limits_{n \rightarrow \infty} \frac{\int_{\frac{1}{n^2}}^{\frac{1}{n}} \sqrt{\frac{1-t}{t}} d t}{\frac{1}{\sqrt{n-1}}} \leq \lim\limits_{n \rightarrow \infty} f\left(\frac{1}{n}\right) g\left(\frac{1}{n}\right) \leq 2$

$\begin{gathered}\Rightarrow \frac{\lim\limits_{n \rightarrow \infty} \frac{-1}{n^2} \sqrt{n-1}+\frac{2}{n^3} \sqrt{n^2-1}}{\frac{1}{2(n-1)^{\frac{3}{2}}}} \leq \lim _{n \rightarrow \infty} f\left(\frac{1}{n}\right) g\left(\frac{1}{n}\right) \leq 2\end{gathered}$

$\begin{aligned} & \therefore \lim _{n \rightarrow \infty} \frac{2(n-1)^2}{n^2}-\frac{4(n-1)^{\frac{3}{2}} \sqrt{n^2-1}}{n^3}=2 \\\\ & \therefore \lim _{n \rightarrow \infty} f\left(\frac{1}{n}\right) g\left(\frac{1}{n}\right)=2 \text { (Using Sandwich Theorem) }\end{aligned}$

$ \text { } \begin{aligned} \int_0^{\pi / 2} \frac{x \tan x \sec ^2 x}{\tan ^4 x+1} d x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\ =\int_0^{\pi / 2} \frac{x \sin x}{\frac{\cos ^3 x}{\sin ^4 x+\cos ^4 x} x} d x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\ =\int_0^{\pi / 2} \frac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ =\int_0^{\pi / 2} \frac{\left(\frac{\pi}{2}-x\right) \sin \left(\frac{\pi}{2}-x\right) \cos \left(\frac{\pi}{2}-x\right)}{\sin ^4\left(\frac{\pi}{2}-x\right)+\cos ^4\left(\frac{\pi}{2}-x\right)} d x \\ =\int_0^{\pi / 2} \frac{\left(\frac{\pi}{2}-x\right) \cos x \sin x}{\sin ^4 x+\cos ^4 x} d x \end{aligned} $

On adding Eqs. (i) and (ii), we get

$ 2 I=\frac{\pi}{2} \int_0^{\pi / 2} \frac{\sin x \cos x}{\sin ^4 x+\cos ^4 x} d x $

Put, $\sin ^2 x=t$

$ \Rightarrow 2 \sin x \cos x=d t $

When, $x=0, t=0$

and $x=\frac{\pi}{2}, t=1$

$ \begin{aligned} I & =\frac{\pi}{4} \int_0^1 \frac{\frac{1}{2} d t}{2 t^2-2 t+1} \\ \Rightarrow I & =\frac{\pi}{8} \int_0^1 \frac{d t}{\left(t-\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2} \\ & =\frac{\pi}{8}\left[\tan ^{-1}\left(\frac{t-\frac{1}{2}}{\frac{1}{2}}\right)\right]_0^1 \\ & =\frac{\pi}{8}\left[\tan ^{-1}(1)-\tan ^{-1}(-1)\right] \\ & =\frac{\pi}{8}\left(\frac{\pi}{4}+\frac{\pi}{4}\right)=\frac{\pi^2}{16} \end{aligned} $

Let $I=\int_3^6 \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Using properties

$ \begin{aligned} \int_a^b f(x) d x & =\int_a^b f(a+b-x) d x \\ I & =\int_3^6 \frac{\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}} d x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

On adding Eqs. (i) and (ii), we get

$ 2 I=\int_3^6 d x \Rightarrow I=\frac{1}{2}[x]_3^6=\frac{3}{2} $

$\ln P=$

$ \begin{aligned} & \frac{1}{n}\left[\begin{array}{r} \ln \left(1+\frac{1}{n^2}\right)+\ln \left(1+\frac{2^2}{n^2}\right)+\ln \left(1+\frac{3^2}{n^2}\right) \\ \ldots \ln \left(1+\frac{n^2}{n^2}\right) \end{array}\right] \\ & \Rightarrow \ln P=\sum_{r=1}^n \ln \left(1+\frac{r^2}{n^2}\right) \frac{1}{n} \\ & \Rightarrow \ln P=\int_0^1 \ln \left(1+x^2\right) d x\left\{\begin{array}{l} r / n=x \\ 1 / n=d x \end{array}\right. \end{aligned} $

Using by parts,

$ \begin{aligned} & \ln P=\left[\ln \left(1+x^2\right) \cdot x-\int \frac{1}{1+x^2} \cdot 2 x \cdot x d x\right]_0^1 \\ & \Rightarrow \ln P=\ln 2-0-2 \int_0^1 \frac{1+x^2-1}{1+x^2} d x \\ & =\ln 2-2\left[x-\tan ^{-1} x\right]_0^1=\ln 2-2+\frac{\pi}{2} \\ & \Rightarrow P=e^{\left(\ln 2+\frac{\pi}{2}-2\right)}=2 e^{\frac{\pi-4}{2}} \end{aligned} $

$ \begin{aligned} & \text { Let } I=\int_{-1}^1 x|x| d x \\ & |x|=\left\{\begin{array}{cc} -x & ; x<0 \\ x & ; x>0 \end{array}\right. \\ & I=\int_{-1}^0-x^2 d x+\int_0^1 x^2 d x=\left[\frac{-x^3}{3}\right]_{-1}^0+\left[\frac{x^3}{3}\right]_0^1 \\ & =0-\left[\frac{-(-1)^3}{3}\right]+\left[\frac{1}{3}\right] \\ & I=\frac{-1}{3}+\frac{1}{3}=0 \end{aligned} $

Let

$ \begin{aligned} I & =\int_{-\pi / 2}^{\pi / 2} \sin ^2 x \cos ^2 x(\sin x+\cos x) d x \\ & =\int_{-\pi / 2}^{\pi / 2} \sin ^3 x \cos ^2 x d x+\int_{-\pi / 2}^{\pi / 2} \sin ^2 x \cos ^3 x d x \end{aligned} $

Now, we know that for an odd function

$ f(x)=\int_{-a}^a f(x) d x=0 $

And for an even function $g(x)$,

$ \int_{-a}^a g(x) d x=2 \int_0^a g(x) d x $

Now, $f(x)=\sin ^3 x \cdot \cos ^2 x$ is an odd function and $g(x)=\sin ^2 x \cdot \cos ^3 x$ is an even function.

Hence, first integration is 0 .

For second part, we get

$ \begin{aligned} & =2 \int_0^{\pi / 2} \sin ^2 x \cos ^3 x d x \\ & =2 \int_0^{\pi / 2} \sin ^2 x\left(1-\sin ^2 x\right) \cos x d x \end{aligned} $

Now, let $t=\sin x$

$ d t=\cos x d x $

And for $x=0, t=0$ and for $x=\frac{\pi}{2}, t=1$

Hence, integration becomes,

$ \begin{aligned} 2 \int_0^1 t^2\left(1-t^2\right) & d t=2 \int_0^1\left(t^2-t^4\right) d t \\ = & 2\left[\frac{t^3}{3}-\frac{t^5}{5}\right]_0^1=2\left[\frac{1}{3}-\frac{1}{5}\right]=\frac{4}{15} \end{aligned} $

$ \begin{aligned} &\\ &\text { } \begin{aligned} & \int_0^3\left(3 x^2-4 x+2\right) d x=k \\ & \quad\left\{\because \int x^n d x=\frac{x^{n+1}}{n+1}+C\right\} \end{aligned} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad 3 \int_0^3 x^2 d x-4 \int_0^3 x d x+2 \int_0^3 d x=k \\ & \Rightarrow \quad 3\left[\frac{x^3}{3}\right]_0^3-4\left[\frac{x^2}{2}\right]_0^3+2[x]_0^3=k \\ & \Rightarrow \quad 3\left[\frac{3^3}{3}-0\right]-4\left[\frac{3^2}{2}-0\right]+2[3-0]=k \\ & \Rightarrow \quad 27-18+6=k \Rightarrow k=15 \\ & \Rightarrow \quad 3 x^2-4 x+2=3 k / 5 \\ & \Rightarrow \quad 3 x^2-4 x+2=\frac{3 \times 15}{5} \\ & \Rightarrow \quad 3 x^2-4 x+2=9 \\ & \Rightarrow \quad 3 x^2-4 x-7=0 \\ & \Rightarrow \quad 3 x^2-7 x+3 x-7=0 \\ & \Rightarrow \quad(3 x-7)(x+1)=0 \\ & \Rightarrow \quad x=\frac{7}{3}, x=-1 \end{aligned} $

$ \begin{aligned} & \text { Let } I=\int_0^\pi \frac{x \cos ^2 x}{1+\sin x} d x \\ & =\int_0^\pi \frac{x\left(1-\sin ^2 x\right)}{1+\sin x} d x \\ & =\int_0^\pi \frac{x(1-\sin x)(1+\sin x)}{1+\sin x} d x \\ & =\int_0^\pi x(1-\sin x) d x \\ & =\int_0^\pi x d x+\int_0^\pi x(-\sin x) d x \\ & =\left|\frac{x^2}{2}\right|_0^\pi+\left|x \cos x-\int \cos x d x\right|_0^\pi \\ & =\left|\frac{x^2}{2}\right|_0^\pi+|x \cos x-\sin x d x|_0^\pi \\ & =\frac{\pi^2}{2}+\pi \cos \pi-\sin \pi-0+\sin 0 \\ & =\frac{\pi^2}{2}-\pi-0+0=\frac{\pi^2-2 \pi}{2}=\frac{\pi(\pi-2)}{2} \end{aligned} $

$ \begin{aligned} & \text { Let } I=\int_{-2}^2[2-x] d x \\ & =\int_{-2}^{-1} 3 d x+\int_{-1}^0 2 d x+\int_0^1 1 d x+\int_1^2 0 d x \end{aligned} $

$ \begin{aligned} & =|3 x|_{-2}^1+|2 x|_{-1}^0+|x|_0^1+0 \\ & =3(-1+2)+2(0+1)+(1-0) \\ & =3 \times 1+2 \times 1+1=3+2+1=6 \end{aligned} $

Let $I=\int_0^2 \frac{x}{(2-x)^{3 / 4}} d x$

Let $2-x=v$

$ \Rightarrow-d x=d v \Rightarrow d x=-d v $

If $\quad x=0 \Rightarrow v=2$

and if $x=2 \Rightarrow v=0$

$ \begin{aligned} & \therefore \quad I=-\int_2^0 \frac{2-v}{v^{3 / 4}} d v \\ & =\int_0^2\left[\frac{2}{v^{3 / 4}}-\frac{v}{v^{3 / 4}}\right] d v \\ & =2 \int_0^2 v^{-3 / 4} d v-\int_0^2 v^{1 / 4} d v \\ & =2\left|\frac{v^{1 / 4}}{1 / 4}\right|_0^2-\left|\frac{v^{5 / 4}}{5 / 4}\right|_0^2=8(2)^{1 / 4}-\frac{4}{5}(2)^{5 / 4} \\ & =2^{1 / 4}\left[8-\frac{4 \times 2}{5}\right]=2^{1 / 4}\left[\frac{40-8}{5}\right]=(32 / 5) 2^{1 / 4} \end{aligned} $

Given, $\int_0^2 x^3(2-x)^4 d x$

Let $2-x=z \Rightarrow-d x=d z \Rightarrow d x=-d z$

If $\quad 2-0=z$

$ \Rightarrow \quad z=2 \text { and } 2-2=z \Rightarrow z=0 $

$ \therefore-\int_2^0(2-z)^3 z^4 d z=\int_0^2(2-z)^3 z^4 d z $

$ =\int_0^2\left(8-z^3-12 z+6 z^2\right) z^4 d z $

$ =\int_0^2\left(8 z^4-z^7-12 z^5+6 z^6\right) d z $

$ =\left|\frac{8 z^5}{5}-\frac{z^8}{8}-\frac{12 z^6}{6}+\frac{6 z^7}{7}\right|_0^2 $

$ =\left|\frac{8}{5} z^5-\frac{z^8}{8}-2 z^6+\frac{6}{7} z^7\right|_0^2 $

$ =\frac{8}{5}(2)^5-\frac{(2)^8}{8}-2(2)^6+\frac{6}{7}(2)^7 $

$ =\frac{256}{5}-32-128+\frac{6}{7} \times 128 $

$ =\frac{7 \times 256-160 \times 5 \times 7+5 \times 6 \times 128}{35}=\frac{32}{35} $

$ \begin{aligned} & \text {Let } I=\int_0^3\left|x^2-3 x+2\right| d x \\ & \begin{array}{r} x^2-3 x+2=\left\{\begin{array}{cc} x^2-3 x+2, & 0 < x < 1 \\ -\left(x^2-3 x+2\right), & 1 < x < 2 \\ \left(x^2-3 x+2\right), & 2 < x < 3 \end{array}\right. \\ =\int_0^1\left(x^2-3 x+2\right) d x-\int_1^2\left(x^2-3 x+2\right) \\ d x+\int_2^3\left(x^2-3 x+2\right) d x \end{array} \\ & =\left(\frac{x^3}{3}-\frac{3 x^2}{2}+2 x\right)_0^1-\left(\frac{x^3}{3}-\frac{3 x^2}{2}+2 x\right)_1^2 +\left(\frac{x^3}{3}-\frac{3 x^2}{2}+2 x\right)_2^3 \end{aligned} $

$ \begin{aligned} & =\left(\frac{1}{3}-\frac{3}{2}+2\right)-\left(\frac{8}{3}-\frac{12}{2}+4\right)+ \left(\frac{1}{3}-\frac{3}{2}+2\right)+\left(\frac{27}{3}-\frac{27}{2}+6\right)-\left(\frac{8}{3}-\frac{12}{2}+4\right) \\ & =11 / 6 \end{aligned} $

Let

$ \begin{aligned} & I=\int_{\frac{-\pi}{8092}}^{\frac{\pi}{8092}} \frac{\sec (2023) x}{1+(2023)^{(2023 x)}} d x \\ & \int_a^b f(x) d x=\int_a^b f(a+b-x) d x \\ & I=\int_{\frac{\pi}{8092}}^{\frac{\pi}{8092}} \frac{(2023)^{(2023 x)} \sec (2023) x}{1+(2023)^{2023 x}} d x \ldots \end{aligned} $

On adding Eqs. (i) and (ii), we get

$ \begin{aligned} &2I=\int_{-\frac{\pi}{8092}}^{\frac{\pi}{8092}} \sec (2023 x) d x \\ & I=\int_0^{\frac{\pi}{8092}} \sec (2023 x) d x \\ = & {\left[\frac{\log / \sec (2023 x)+\tan (2023 x)}{2023}\right]_0^{\frac{\pi}{8092}}+C } \\ = & \frac{\log (\sqrt{2}+1)}{2023}+C \end{aligned} $

$ \text { } \begin{aligned} I & =\int_0^2 x^{5 / 2} \sqrt{2-x} d x \\ \text { Let } x & =2 \sin ^2 \theta \\ d x & =4 \sin \theta \cos \theta d \theta \end{aligned} $

$ \begin{aligned} & \text { At } x=0, \theta=0 \\ & \text { At } x=2, \theta=\pi / 2 \\ & \begin{aligned} \therefore \quad & =\int_0^{\pi / 2}\left(2 \sin ^2 \theta\right)^{5 / 2} \sqrt{2-2 \sin ^2 \theta} \\ & 4 \sin \theta \cos \theta d \theta \\ = & \int_0^{\pi / 2} 2^{5 / 2} \sin ^5 \theta \cdot 2^{1 / 2} \cos \theta 4 \sin \theta \cos \theta d \theta \\ = & 32 \int_0^{\pi / 2} \sin ^6 \theta \cos ^2 \theta d \theta \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Using Walli's formula, }\\ &\begin{aligned} \int_0^{\pi / 2} & \sin ^m x \cos ^n x d x \\ & =\frac{[(m-1)(m-3)(m-5) \ldots \ldots][(n-1)(n-3) \ldots .]}{[(m+n)(m+n-2)(m+n-4) \ldots . . k]} \cdot k \\ k & =\left\{\begin{array}{l} \frac{\pi}{2}, m \text { and } n \text { are even } \\ 1, \text { otherwise } \end{array}\right. \\ & =32 \cdot \frac{(5 \cdot 3 \cdot 1)(1)}{8 \cdot 6 \cdot 4 \cdot 2} \times \frac{\pi}{2}=\frac{5 \pi}{8} \end{aligned} \end{aligned} $