Integrals

$A=\left\{\left(x,y\right):x^2\le y\le min\{x+2,4-3x\right\}$

So, area of the required region

$A={\displaystyle \underset{-1}{\overset{\frac{1}{2}1}{\int }}\left(x+2-x^2\right)dx+{\displaystyle \underset{\frac{1}{2}}{\overset{1}{\int }}\left(4-3x-x^2\right)}dx}$

$={\left[\frac{x^2}{2}+2x-\frac{x^3}{3}\right]}^{\frac{1}{2}}+{\left[4x-\frac{3x^2}{2}-\frac{x^3}{3}\right]}^1$

= $\frac{17}{6}$

$\left|f\left(x\right)\right|\le 800\Rightarrow 2n^2-n-1\le 800$  

$\Rightarrow 2n^2-n-801\le 0$                

${\displaystyle \sum _{x\in S}^{}f\left(x\right)={\displaystyle \sum _{}^{}\left(2x^2-x-1\right)}}$                

$=2\left(19^2+18^2+........1^2+0^2+1^2+.....+20^2\right)$                

= 10620

Let f(x) = 8 sin x - sin 2x

$f\text{'}\left(x\right)=8cosx-2cos2x$              

$f\text{'}\text{'}\left(x\right)=-8sinx+4sin2x$               

= -8 (sin x) (1 – cos x)

f’’(x) < 0, f’(x) is a decreasing function

  ^{$f\text{'}\left(\frac{\pi }{3}\right)<f\text{'}\left(x\right)<f\text{'}\left(\frac{\pi }{a}\right)$}               

  $5<f\text{'}\left(x\right)<\frac{8}{\sqrt[]{2}}$

${\displaystyle \underset{\pi /4}{\overset{\pi /3}{\int }}dx<{\displaystyle \underset{\pi /4}{\overset{\pi /3}{\int }}\frac{f\left(x\right)}{x}<{\displaystyle \underset{\pi /4}{\overset{\pi /3}{\int }}\frac{8}{\sqrt[]{2}}}}}$

$5\pi 12<l<\frac{8}{\sqrt[]{2}}.\frac{\pi }{\sqrt[]{2}}$        

  $l={\displaystyle \underset{-2}{\overset{2}{\int }}\frac{\left|x^3+x\right|}{e^{x\left|x\right|}+1}dx}$ ……. (i)

$l={\displaystyle \underset{-2}{\overset{2}{\int }}\frac{\left|x^3+x\right|}{e^{-x\left|x\right|}+1}dx}$ …. (ii)

$=\left(\frac{16}{4}+\frac{4}{2}\right)$ - 0

= 4 + 2 = 6

I(x) = ${\displaystyle \int \frac{se{c}^{2}x-2022}{si{n}^{2022}x}dx}$  

$={\displaystyle \int si{n}^{-2022}x\cdot se{c}^{2}xdx-{\displaystyle \int 2022si{n}^{-2022}xdx}}$

$=si{n}^{-2022}x\cdot tanx-{\displaystyle \int \left(-2022\right)si{n}^{-2023}x\cdot cosx\cdot tanxdx-{\displaystyle \int 2022si{n}^{-2022}xdx}}$

$=tanx\cdot si{n}^{-2022}x+2022{\displaystyle \int si{n}^{-2022}-2022{\displaystyle \int si{n}^{-2022}xdx}}$

 I(x) = $tanx\cdot si{n}^{-2022}x+c$  

Given,   $I\left(\frac{\pi }{4}\right)=2^{1011}$

-> $2^{1011}=\frac{1}{{\left(\frac{1}{\sqrt[]{2}}\right)}^{2022}}+c\Rightarrow c=0$

$I\left(x\right)=\frac{tanx}{si{n}^{2022}x},I\left(\frac{\pi }{3}\right)=\frac{\sqrt[]{3}}{\left(\frac{\sqrt[]{3}}{2}\right)}=\frac{2^{2022}}{{\left(\sqrt[]{3}\right)}^{2021}}$  

$l={\displaystyle {\int }_{-3}^{103}\left(\left[sin\left(\pi x\right)\right]+e^{\left[cos\left(2\pi x\right)\right]}\right)}dx$

$\left[sin\pi x\right]$ is periodic with period 2 and

$e^{\left[cos2\pi x\right]}$ is periodic with period 1.

So,

$I=52{\displaystyle {\int }_0^2\left(\left[sin\left(\pi x\right)\right]+e^{\left[cos2\pi x\right]}\right)}dx$

= $\frac{52}{e}$

$r{\cdot }^n\text{?}{C}_r=n{\cdot }^{n-1}\text{?}{C}_{r-1}$

${\displaystyle \sum _{k=1}^{10}{\left(k{\cdot }^{10}\text{?}{C}_k\right)}^2}={\displaystyle \sum _{k=1}^{10}{\left(10{\cdot }^9\text{?}{C}_{K-1}\right)}^2}$

= $100{\cdot }^{18}\text{?}{C}_9$  

22000L =   $100{\cdot }^{18}\text{?}{C}_9$

 L =

$18!=2^{16}\cdot 3^8\cdot 5^3\cdot 7^2\cdot 11^1\cdot 13\cdot 17$

$9+4+2+1$ $9!=2^7\cdot 3^4\cdot 5^1\cdot 7^1$

6 + 2                    4 + 2 + 1   $\frac{18!}{{\left(9!\right)}^2}=2^2\cdot 5\cdot 11\cdot 13\cdot 17$            

3 +         &

$\left|f\left(x\right)\right|\le 800\Rightarrow 2n^2-n-1\le 800$

$\Rightarrow 2n^2-n-801\le 0$

${\displaystyle \sum _{x\in S}^{}f\left(x\right)={\displaystyle \sum _{}^{}\left(2x^2-x-1\right)}}$

$=2\left(19^2+18^2+........1^2+0^2+1^2+.....+20^2\right)$

= 10620

$f\left(x\right)=\left|5x-7\right|+\left[x^2+2x\right]$

$=\left|5x-7\right|+\left[{\left(x+1\right)}^2\right]-1$

f $\left(\frac{7}{4}\right)=0+4=4$

as both |5x – 7| and x² + 2x are increasing in nature after x = 7/5

f (2) = 3 + 8 = 11

$f{\left(\frac{7}{5}\right)}_{min}-4andf{\left(2\right)}_{max}=11$

Sum is 4 + 11 = 15

$li{m}_{n\to \infty }{\left(\frac{n+1}{n}\right)}^{k-1}\frac{1}{n}{\displaystyle {\sum }_{r=1}^n\left(k+\frac{r}{n}\right)}$

= $33{\displaystyle {\int }_0^1{x}^{k}dx}$