Integrals

$l={\displaystyle \underset{1}{\overset{3}{\int }}\left[x^2-2x+1-3\right]dx=}{\displaystyle \underset{1}{\overset{3}{\int }}\left[{\left(x-1\right)}^2-3\right]dx={\displaystyle \underset{1}{\overset{3}{\int }}\left[{\left(x-1\right)}^2\right]dx-3{\displaystyle \underset{1}{\overset{3}{\int }}dx}}}$  .(A)

$l_1={\displaystyle \underset{1}{\overset{3}{\int }}\left[{\left(x-1\right)}^2\right]dxPut{\left(x-1\right)}^2=t}$

$l_1=\frac{1}{2}\left[0{\displaystyle \underset{0}{\overset{1}{\int }}\frac{dt}{\sqrt[]{t}}}{\displaystyle \underset{1}{\overset{2}{\int }}\frac{dt}{\sqrt[]{t}}}+2{\displaystyle \underset{2}{\overset{3}{\int }}\frac{dt}{\sqrt[]{t}}}+3{\displaystyle \underset{3}{\overset{4}{\int }}\frac{dt}{\sqrt[]{t}}}\right]=\frac{1}{2}\left\{{\left|\frac{t^{\frac{-1}{2}+1}}{-\frac{1}{2}+1}\right|}_1^2{\left|+2\frac{t^{\frac{-1}{2}+1}}{-\frac{1}{2}+1}\right|}_2^3{+3\frac{t^{\frac{-1}{2}+1}}{-\frac{1}{2}+1}]}_3^4\right\}$      = $5-\sqrt[]{2}-\sqrt[]{3}$  

Hence from (A)

= $5-\sqrt[]{2}-\sqrt[]{3}-\sqrt[]{3}-6=-1-\sqrt[]{2}-\sqrt[]{3}$

2^nd method       

${\displaystyle \underset{1}{\overset{3}{\int }}\left[{\left(x-1\right)}^2\right]dx-6..............\left(A\right)}$

From (A), $l=5-\sqrt[]{2}-\sqrt[]{3}-6=-1-\sqrt[]{2}-\sqrt[]{3}$  

Let sinθ= t ${\displaystyle \int \frac{sin\theta \left(2sin\theta .cos\theta \right)\left(si{n}^6\theta +si{n}^4\theta +si{n}^2\theta \right)\sqrt[]{2si{n}^4\theta +3si{n}^2\theta +6}}{2si{n}^2\theta }}$ dθ

sinθ = t

cos θ. dθ = dt

${\displaystyle \int \frac{u^{1/2}}{12}}du=\frac{u^{3/2}}{18}+C=\frac{{\left(2t^6+3t^4+6t^2\right)}^{3/2}}{18}+C=\frac{{\left(2si{n}^6\theta +3si{n}^4\theta +6si{n}^2\theta \right)}^{3/2}}{18}+C$

$\underset{n\to \infty }{lim}{\left(1+\frac{1+\frac{1}{2}+.......+\frac{1}{n}}{n^2}\right)}^n$ limit is in the form of $1^{\infty }$  

$\mathcal{l}=exp\left(\underset{n\to \infty }{lim}\frac{1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{n}}{n^2}\right)$             

$0\le 1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{n}\le 1+\frac{1}{\sqrt[]{2}}+\frac{1}{\sqrt[]{3}}+....+\frac{1}{\sqrt[]{n}}\le 2\sqrt[]{n}-1$        

Taking limit   $\left(n\to \infty \right)$

$\mathcal{l}$ = exp (0) (from sandwich)

  $\mathcal{l}=1$          

Second Method :

$1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{n}\ge ln\left(n+1\right).......\left(i\right)$

$1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{n}\le 1+{\displaystyle {\int }_1^2\frac{1}{2}dx\le 1+lnn..........\left(ii\right)}$

From (i) & (ii)

$ln\left(n+1\right)\le 1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{n}\le 1+Inn,\forall n\in N,n\ge 2$           

As $\underset{n\to \infty }{lim}\frac{ln\left(n+1\right)}{n}=0$  

and $\underset{n\to \infty }{lim}\frac{1+ln\left(n\right)}{n}=0$  

$\therefore$ from sandwich theorem

$\underset{n\to \infty }{lim}\frac{1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{n}}{n}=0$  

$\therefore e^0=1$   

$l={\displaystyle \int \frac{8x^3+20x^2}{x^4+5x^3-7x^2}}dx=4{\displaystyle \int \frac{2x+5}{x^2+5x-7}dx=4ln\left|x^2+5x-7\right|+c}$

$l_n={\displaystyle \underset{\pi /4}{\overset{\pi /2}{\int }}co{t}^{n}xdx=}{\displaystyle \underset{\pi /4}{\overset{\pi /2}{\int }}co{t}^{n-2}x\left(cose{c}^{2}x-1\right)dx}$

$={\left[\frac{-co{t}^{n-1}x}{n-1}\right]}_{\pi /4}^{\pi /2}-l_{n-2}=\frac{1}{n-1}-l_{n-2}$

$l_n+l_{n-2}=\frac{1}{n-1}$

$\begin{array}{l}n=4\Rightarrow l_4+l_2=\frac{1}{3}\\ n=5\Rightarrow l_5+l_3=\frac{1}{4}\\ n=6\Rightarrow l_6+l_4=\frac{1}{5}\end{array}\}\Rightarrow \frac{1}{l_2+l_4},\frac{1}{l_3+l_5},\frac{1}{l_4+l_6}$ are in A.P.

$x^2=-2\left(y-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$  

y² = -4 (x – 1)

Required area = $2{\displaystyle {\int }_{-1}^0\left[2\sqrt[]{x+1}-\frac{1-x^2}{2}\right]dx}$  

$=2{\left[\frac{4}{3}{\left(x+1\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{1}{2}\left(x-\frac{x^3}{3}\right)\right]}_{-1}^0$          

= 2

${\displaystyle \underset{}{\overset{}{\int }}\frac{si{n}^{2}xdx}{e^{\left\{\frac{x}{\pi }\right\}}}=l}$

$f\left(x\right)=\frac{si{n}^{2}x}{e^{\left\{\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$\pi $}\right.\right\}}}$ is periodic with period =

$l=100{\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{si{n}^{2}x}{e^{x/\pi }}dx.}$

$=50{\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{1-cos2x}{e^{x/\pi }}}dx=50{\displaystyle \underset{0}{\overset{\pi }{\int }}{e}^{-x/\pi }dx-50}{\displaystyle \underset{0}{\overset{\pi }{\int }}{e}^{-x/\pi }}cos2xdx$

$=50\pi \left(1-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$e$}\right.\right)\left[1-\frac{1}{1+4{\pi }^2}\right]=\frac{50\left(1-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$e$}\right.\right)4{\pi }^3}{1+4{\pi }^2}$

$\therefore \alpha =200\left(1-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$e$}\right.\right)$

$\frac{1}{\alpha \left(\alpha +1\right)\left(\alpha +2\right)..........\left(\alpha +20\right)}=\frac{A_0}{\alpha }+\frac{A_1}{\alpha +1}+\frac{A_2}{\alpha +2}+......+\frac{A_{20}}{\alpha +20}$

Solving by partial fraction, we get

$A_{13}=\frac{-1}{14!7!},A_{14}=\frac{-1}{14!6!}and{A}_{15}=\frac{-1}{14!5!}$

$\therefore \frac{A_{14}+A_{15}}{A_{13}}=\frac{3}{10}\Rightarrow {\left(\frac{A_{14}+A_{15}}{A_{13}}\right)}^2={\left(\frac{3}{10}\right)}^2\Rightarrow 100{\left(\frac{A_{14}+A_{15}}{A_{13}}\right)}^2=9$

A = $-\frac{y}{x}+2sinx+2$  

$\therefore \frac{dy}{dx}+\frac{y}{x}=2sinx+2$ . (i)

$\Rightarrow l.F=e^{{\displaystyle \int !lnxdx}}$ = x from (i) d (xy) = 2 ${\displaystyle \int xsinxdx+2{\displaystyle \int xdx}}$

xy = 2 [-xcos x + sin x] + x² + c       . (ii)

according to question, we get c = 0 $\therefore y\left(\frac{\pi }{2}\right)=\frac{4}{\pi }+\frac{\pi }{2}$

$\underset{x\to 1}{lim}\frac{x^{n}f\left(1\right)-f\left(x\right)}{x-1}$  

$=\underset{x\to 1}{lim}\frac{9x^n-\left(x^6+2x^4+x^3+2x+3\right)}{x-1}$                

= 9n – 19 = 44 -> n = 7