The value of the integral ${\displaystyle \underset{1}{\overset{3}{\int }}\left[x^2-2x-2\right]dx,}$  where [x] denotes the greatest integer less than or equal to x, is

${\displaystyle \int \frac{\left(sinx-cosx\right)si{n}^{2}x}{tanx\left(si{n}^{3}x+co{s}^{3}x\right)}dx}$

${\displaystyle \int \frac{\left(sinx-cosx\right)sinxcosx}{si{n}^{3}x+co{s}^{3}x}dx}$ , put sin³x + cos³x = t(3 sin²x×cosx – 3cos²xsinx) dx = dt

-> $\frac{1}{3}{\displaystyle \int \frac{dt}{t}}$

$=\frac{lnt}{3}+c$

$=\frac{ln\left|si{n}^{3}x+co{s}^{3}x\right|}{3}+c$

    $l={\displaystyle \underset{0}{\overset{2}{\int }}f\left(x\right)dx={\left[xf\left(x\right)\right]}_0^2-{\displaystyle \underset{0}{\overset{2}{\int }}xf\text{'}\left(x\right)dx=2e^2-{\displaystyle \underset{0}{\overset{2}{\int }}xf\text{'}\left(x\right)dx}}}$    ........(A)

Put $l_1={\displaystyle \underset{0}{\overset{2}{\int }}xf\text{'}\left(x\right)dx}$               .........(i)

Using properties ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx={\displaystyle \underset{a}{\overset{b}{\int }}f\left(a+b-x\right)dx}}$

$l_1={\displaystyle \underset{0}{\overset{2}{\int }}\left(2-x\right)f\text{'}\left(2-x\right)dx={\displaystyle \underset{0}{\overset{2}{\int }}\left(2-x\right)f\text{'}\left(x\right)dx}}$ ...........(ii)

Adding (i) and (ii) we get

$2l_1=2{\displaystyle \underset{0}{\overset{2}{\int }}f\text{'}\left(x\right)dx\Rightarrow l_1={\left[f\left(x\right)\right]}_0^2}$

f(2) – f(0) = e² – 1

From (A) l = 2e² – e² + 1 = e² + 1

Given $l_{m,n}={\displaystyle {\int }_0^1{x}^{m-1}{\left(1-x\right)}^{n-1}dx............\left(i\right)}$  

put 1 - x = $t\{\begin{array}{l}x=0,t=1\\ x=1,t=0\end{array}$  

dx = -dt

From (i) $l_{m,n}={\displaystyle {\int }_1^0{\left(1-t\right)}^{m-1}.t^{n-1}\left(-dt\right)}$  

$l_{m,n}={\displaystyle {\int }_0^1{t}^{n-1}{\left(1-t\right)}^{m-1}dt={\displaystyle {\int }_0^1{x}^{n-1}{\left(1-x\right)}^{m-1}dx..........\left(ii\right)}}$                              

(i)   $l_{m,n}={\displaystyle {\int }_{\infty }^0\frac{1}{{\left(1+y\right)}^{m-1}}}{\left(1-\frac{1}{1+y}\right)}^{n-1}\left(\frac{-dy}{{\left(1+y\right)}^2}\right)={\displaystyle {\int }_0^{\infty }\frac{y^{n-1}}{{\left(1+y\right)}^{m+n}}dy}..........\left(iii\right)$

Similarly by (ii) $l_{m,n}={\displaystyle {\int }_0^{\infty }\frac{y^{m-1}}{{\left(y+1\right)}^{m+n}}dy..........\left(iv\right)}$  

Adding (iii) & (iv) $2l_{m,n}={\displaystyle {\int }_0^{\infty }\frac{y^{n-1}+y^{m+1}}{{\left(y+1\right)}^{m+n}}}$  

Putting   $\frac{1}{z}\{\begin{array}{l}y=1,z=1\\ y=\infty ,z=0\end{array}$ $\Rightarrow dy=\frac{-1}{z^2}dz$  

Hence  $l_{m,n}={\displaystyle {\int }_0^1\frac{x^{m-1}+x^{n-1}}{{\left(1+x\right)}^{m+n}}}$ d

$l=\frac{48}{{\pi }^4}{\displaystyle {\int }_0^{\pi }\left[{\left(\frac{\pi }{2}-x\right)}^3-\frac{3{\pi }^2}{4}\left(\frac{\pi }{2}-x\right)+\frac{{\pi }^3}{4}\right]\frac{sinxdx}{1+co{s}^{2}x}}$

Using ${\displaystyle {\int }_a^{b}f\left(x\right)dx={\displaystyle {\int }_a^{b}f\left(a+b-x\right)dx}}$

we get $l=\frac{48}{{\pi }^4}{\displaystyle {\int }_0^{\pi }\left[-{\left(\frac{\pi }{2}-x\right)}^3+\frac{3{\pi }^2}{4}\left(\frac{\pi }{2}-x\right)+\frac{{\pi }^3}{4}\right]\frac{sinxdx}{1+co{s}^{2}x}}$

Adding these two equations, we get

$\Rightarrow l=\frac{12}{\pi }{\left[-ta{n}^{-1}\left(cosx\right)\right]}_0^{\pi }=\frac{12}{\pi }.\frac{\pi }{2}=6$

$\therefore {\displaystyle \underset{-6}{\overset{0}{\int }}f\left(x\right)dx=2\times \frac{1}{2}\left(2+5\right)\times 3=21}$