(1)

Given $f\left(x\right)=\max \{x^2,1+[x\left]\right\}$

So, ${\int }_0^{2}f\left(x\right) dx={\int }_0^{1}1 dx+{\int }_1^{\sqrt{2}}2 dx+{\int }_{\sqrt{2}}^2{x}^2 dx$

               $=1+2(\sqrt{2}-1)+\frac{8-2\sqrt{2}}{3}=\frac{5+4\sqrt{2}}{3}$

(3)

$f\left(x\right)=x+{\int }_0^{\pi /2}(\sin x\cos y+\cos x\sin y)f\left(y\right) dy$

$=x+{\int }_0^{\pi /2}\left(\right(\cos y f\left(y\right) dy)\sin x+(\sin y f\left(y\right) dy\left)\cos x\right)$                 ...(i)

On comparing with $f\left(x\right)=x+\frac{a}{{\pi }^2-4}\sin x+\frac{b}{{\pi }^2-4}\cos x$

$\Rightarrow \frac{a}{{\pi }^2-4}={\int }_0^{\pi /2}\cos y f\left(y\right) dy$                          ...(ii)

$\Rightarrow \frac{b}{{\pi }^2-4}={\int }_0^{\pi /2}\sin y f\left(y\right) dy$                              ...(iii)

Adding (ii) and (iii), we get

$\frac{a+b}{{\pi }^2-4}={\int }_0^{\pi /2}(\sin y+\cos y)f\left(y\right) dy$                 ...(iv)

$\frac{a+b}{{\pi }^2-4}={\int }_0^{\pi /2}(\sin y+\cos y)f(\frac{{\displaystyle \pi }}{{\displaystyle 2}}-y)dy$         ...(v)

Adding (iv) and (v), we get

$\frac{2(a+b)}{{\pi }^2-4}={\int }_0^{\pi /2}(\sin y+\cos y)[\frac{{\displaystyle \pi }}{{\displaystyle 2}}+\frac{{\displaystyle (a+b)}}{{\displaystyle {\pi }^2-4}}(\sin y+\cos y\left)\right]dy$

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

(4)

We have the integral as follows  

$I={\int }_{1/2}^2\frac{{\tan }^{-1}x}{x}dx \dots \left(1\right)$

By substituting $x=\frac{1}{t}$ and $dx=-\frac{1}{t^2}dt$ in equation (1),

$I=-{\int }_2^{1/2}\frac{{\tan }^{-1}\frac{1}{t}}{\frac{1}{t}}·\frac{1}{t^2}·dt$

$I=-{\int }_2^{1/2}\frac{{\tan }^{-1}\frac{1}{t}}{t}dt={\int }_{1/2}^2\frac{{\cot }^{-1}t}{t}dt={\int }_{1/2}^2\frac{{\cot }^{-1}x}{x}dx \dots \left(2\right)$

On adding equation (1) and (2), we have  

$2I={\int }_{1/2}^2\frac{{\tan }^{-1}x+{\cot }^{-1}x}{x}dx;$   $2I=\frac{\pi }{2}{\int }_{1/2}^2\frac{dx}{x}=\frac{\pi }{2}{\left[\log x\right]}_{1/2}^2$

      $2I=\frac{\pi }{2}[{\log }_{e}2-{\log }_e\frac{1}{2}]=\pi {\log }_{e}2$

$\Rightarrow$ $2I={\mathrm{\pi log}}_{\mathrm{e}}2$     $\therefore$$I=\frac{\pi }{2}{\log }_{e}2$

Hence, option (4) is the correct answer.

(2)

Let $I={\int }_1^2\left[\frac{{\displaystyle t^4+1}}{{\displaystyle t^6+1}}\right]dt$

$={\int }_1^2\frac{1}{t^2+1}dt+\frac{1}{3}{\int }_1^2\frac{3t^2}{{\left(t^3\right)}^2+1}dt$

$={\left[{\tan }^{-1}t\right]}_1^2+{\left[\frac{{\displaystyle 1}}{{\displaystyle 3}}\right({\tan }^{-1}{t}^3\left)\right]}_1^2$

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

(4)

Let $I=\frac{3(e-1)}{e}{\int }_1^2{x}^2{e}^{\left[x\right]+\left[x^3\right]} dx$

Between 1 and 2, $\left[x\right]$= 1, so $e^{\left[x\right]}=e$.

$\therefore$   $I=3(e-1){\int }_1^2{x}^2·e^{\left[x^3\right]} dx$

Let $x^3=t\Rightarrow x^{2}dx=\frac{dt}{3}$

$I=\frac{3(e-1)}{3}{\int }_1^8{e}^{\left[t\right]} dt=(e-1){\int }_1^8{e}^{\left[t\right]} dt$

$=(e-1)[e+e^2+e^3+e^4+e^5+e^6+e^7]$$=(e-1)\left[\frac{{\displaystyle e(e^7-1)}}{{\displaystyle e-1}}\right]$

$\Rightarrow e(e^7-1)=e^8-e$

(4)

We have, $\underset{n\to \infty }{\lim }\frac{3}{n}\{4+{(2+\frac{1}{n})}^2+{(2+\frac{2}{n})}^2+\cdots +{(3-\frac{1}{n})}^2\}$

$=\underset{n\to \infty }{\lim }\frac{3}{n}\{{(2+\frac{{\displaystyle 0}}{{\displaystyle n}})}^2+{(2+\frac{{\displaystyle 1}}{{\displaystyle n}})}^2+\cdots +(2+{\left(\frac{{\displaystyle n-1}}{{\displaystyle n}}\right))}^2\}$

$=\underset{n\to \infty }{\lim }\frac{3}{n}\sum _{r=0}^{n-1}{(2+\frac{{\displaystyle r}}{{\displaystyle n}})}^2=3{\int }_0^1{(2+x)}^{2}dx$

Let $2+x=tdx=dt$$\Rightarrow 3{\int }_2^3{t}^2 dt=3\times {\left[\frac{{\displaystyle t^3}}{{\displaystyle 2}}\right]}_2^3=\frac{3}{3}[27-8]=19$

(2)

We have, $P_n\left(x\right)=x+\frac{x^2}{2}+\frac{x^3}{3}+\cdots +\frac{x^n}{n}, x\in (0,1)$                       ...(i)

Let $I={\int }_0^{\alpha }\frac{t^{50}}{1-t}dt={\int }_0^{\alpha }\frac{t^{50}+1-1}{1-t}dt={\int }_0^{\alpha }(\frac{{\displaystyle -(1-t^{50})}}{{\displaystyle 1-t}}+\frac{{\displaystyle 1}}{{\displaystyle 1-t}})dt$

$={\int }_0^{\alpha }-(1+t+t^2+\dots +t^{49})dt+{\int }_0^{\alpha }\frac{{\displaystyle 1}}{{\displaystyle 1-t}}dt$

$=-{[t+\frac{{\displaystyle t^2}}{{\displaystyle 2}}+\frac{{\displaystyle t^3}}{{\displaystyle 3}}+\cdots +\frac{{\displaystyle t^{50}}}{{\displaystyle 50}}]}_0^{\alpha }-{\left[\log \right(1-t\left)\right]}_0^{\alpha }$

$=-\log (1-\alpha )-[\alpha +\frac{{\displaystyle {\alpha }^2}}{{\displaystyle 2}}+\frac{{\displaystyle {\alpha }^3}}{{\displaystyle 3}}+\cdots +\frac{{\displaystyle {\alpha }^{50}}}{{\displaystyle 50}}]$

$=-\beta -P_{50}\left(\alpha \right)$                   $[\because \beta ={\log }_e(1-\alpha ) \text{and from (i)}]$

$=-(\beta +P_{50}(\alpha \left)\right)$

(3)

Let $I={\int }_{\pi /3}^{\pi /2}\frac{2+3\sin x}{\sin x(1+\cos x)}dx$

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

$={\int }_{\pi /3}^{\pi /2}\frac{2}{{\sin }^{3}x}dx-{\int }_{\pi /3}^{\pi /2}\frac{2\cos x}{{\sin }^{3}x}dx+\frac{3}{2}{\int }_{\pi /3}^{\pi /2}\frac{1}{{\cos }^2\frac{x}{2}}dx$

$=2{\int }_{\pi /3}^{\pi /2}\sqrt{cose{c}^{2}x} cose{c}^{2}xdx-2{\int }_{\pi /3}^{\pi /2}\cot x·cose{c}^{2}xdx+\frac{3}{2}{\left[2\tan \frac{{\displaystyle x}}{{\displaystyle 2}}\right]}_{\pi /3}^{\pi /2}$

$=2{\int }_{\pi /3}^{\pi /2}\sqrt{1+{\cot }^{2}x} cose{c}^2 dx-2{\int }_{\pi /3}^{\pi /2}\cot x·cose{c}^{2}dx+3[\tan \frac{{\displaystyle \pi }}{{\displaystyle 4}}-\tan \frac{{\displaystyle \pi }}{{\displaystyle 6}}]$

Let $\cot x=t\Rightarrow cose{c}^{2}x dx=-dt$ and  

For $x=\frac{\pi }{2}; t=0$ and for $x=\frac{\pi }{3}; t=\frac{1}{\sqrt{3}}$

$\therefore$   $I=-2{\int }_{1/\sqrt{3}}^0\sqrt{1+t^2} dt+2{\int }_{1/\sqrt{3}}^{0}t dt+3[1-\frac{1}{\sqrt{3}}]$

$=-2{[\frac{{\displaystyle t}}{{\displaystyle 2}}\sqrt{1+t^2}+\frac{{\displaystyle 1}}{{\displaystyle 2}}{\log }_e(t+\sqrt{1+t^2}\left)\right]}_{1/\sqrt{3}}^0 +{\left[t^2\right]}_{1/\sqrt{3}}^0+(3-\sqrt{3})$

$=\frac{2}{3}+\log \sqrt{3}-\frac{1}{3}+3-\sqrt{3}=\frac{10}{3}-\sqrt{3}+{\log }_e\sqrt{3}$

(2)

By Leibnitz rule,

$\varphi \text{'}\left(x\right)=\frac{1}{\sqrt{x}}\left[\right(4\sqrt{2}\sin x-3\varphi \text{'}\left(x\right))·1-0]-\frac{1}{2}{x}^{-3/2}{\int }_{\pi /4}^x(4\sqrt{2}\sin t-3\varphi \text{'}(t\left)\right)dt$

Put $x=\frac{\pi }{4}$, we get $\varphi \text{'}\left(\frac{{\displaystyle \pi }}{{\displaystyle 4}}\right)=\frac{2}{\sqrt{\pi }}[4-3\varphi \text{'}(\frac{{\displaystyle \pi }}{{\displaystyle 4}}\left)\right]+0$

$\Rightarrow (1+\frac{{\displaystyle 6}}{{\displaystyle \sqrt{\pi }}})\varphi \text{'}\left(\frac{{\displaystyle \pi }}{{\displaystyle 4}}\right)=\frac{8}{\sqrt{\pi }}$$=\frac{8}{\sqrt{\pi }}\times \left(\frac{{\displaystyle \sqrt{\pi }}}{{\displaystyle 6+\sqrt{\pi }}}\right)=\frac{8}{6+\sqrt{\pi }}$

(1)

After rationalising, $I={\int }_0^{\alpha }\frac{x}{\alpha }(\sqrt{x+\alpha }+\sqrt{x})dx$

$={\int }_0^{\alpha }\frac{1}{\alpha }[{(x+\alpha )}^{3/2}-\alpha {(x+\alpha )}^{1/2}+x^{3/2}]dx$

$=\frac{1}{\alpha }{[\frac{{\displaystyle 2}}{{\displaystyle 5}}{(x+\alpha )}^{5/2}-\alpha \frac{{\displaystyle 2}}{{\displaystyle 3}}{(x+\alpha )}^{3/2}+\frac{{\displaystyle 2}}{{\displaystyle 5}}{x}^{5/2}]}_0^{\alpha }$

$=\frac{1}{\alpha }(\frac{{\displaystyle 2}}{{\displaystyle 5}}{\left(2\alpha \right)}^{5/2}-\frac{{\displaystyle 2\alpha }}{{\displaystyle 3}}{\left(2\alpha \right)}^{3/2}+\frac{2}{5}{\alpha }^{5/2}-\frac{{\displaystyle 2}}{{\displaystyle 5}}{\alpha }^{5/2}+\frac{{\displaystyle 2}}{{\displaystyle 3}}{\alpha }^{5/2})$

$=\frac{{\displaystyle 1}}{{\displaystyle \alpha }}(\frac{{\displaystyle 2^{7/2}{\alpha }^{5/2}}}{{\displaystyle 5}}-\frac{{\displaystyle 2^{5/2}{\alpha }^{5/2}}}{{\displaystyle 3}}+\frac{{\displaystyle 2}}{{\displaystyle 3}}{\alpha }^{5/2})$$={\alpha }^{3/2}(\frac{{\displaystyle 2^{7/2}}}{{\displaystyle 5}}-\frac{{\displaystyle 2^{5/2}}}{{\displaystyle 3}}+\frac{{\displaystyle 2}}{{\displaystyle 3}})$

$=\frac{{\alpha }^{3/2}}{15}(24\sqrt{2}-20\sqrt{2}+10)=\frac{{\alpha }^{3/2}}{15}(4\sqrt{2}+10)$

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

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