Question Number 85020 by M±th+et£s last updated on 18/Mar/20
$${solve}\:{integration} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} {x}\:{d}\lfloor{x}^{\mathrm{2}} \rfloor \\ $$
Commented by mr W last updated on 18/Mar/20
$$\mathrm{1}\leqslant{x}<\sqrt{\mathrm{2}}:\:\lfloor{x}^{\mathrm{2}} \rfloor=\mathrm{1},\:{d}\left(\lfloor{x}^{\mathrm{2}} \rfloor\right)=\mathrm{0}\:{dx} \\ $$$$\sqrt{\mathrm{2}}\leqslant{x}<\sqrt{\mathrm{3}}:\:\lfloor{x}^{\mathrm{2}} \rfloor=\mathrm{2},\:{d}\left(\lfloor{x}^{\mathrm{2}} \rfloor\right)=\mathrm{0}\:{dx} \\ $$$$\sqrt{\mathrm{3}}\leqslant{x}<\mathrm{2}:\:\lfloor{x}^{\mathrm{2}} \rfloor=\mathrm{3},\:{d}\left(\lfloor{x}^{\mathrm{2}} \rfloor\right)=\mathrm{0}\:{dx} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} {x}\:{d}\lfloor{x}^{\mathrm{2}} \rfloor=\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} {x}\:{d}\lfloor{x}^{\mathrm{2}} \rfloor+\int_{\sqrt{\mathrm{2}}} ^{\sqrt{\mathrm{3}}} {x}\:{d}\lfloor{x}^{\mathrm{2}} \rfloor+\int_{\sqrt{\mathrm{3}}} ^{\mathrm{2}} {x}\:{d}\lfloor{x}^{\mathrm{2}} \rfloor \\ $$$$=\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} {x}×\mathrm{0}\:{dx}+\int_{\sqrt{\mathrm{2}}} ^{\sqrt{\mathrm{3}}} {x}×\mathrm{0}\:{dx}+\int_{\sqrt{\mathrm{3}}} ^{\mathrm{2}} {x}×\mathrm{0}\:{dx} \\ $$$$=\mathrm{0} \\ $$
Commented by M±th+et£s last updated on 18/Mar/20
![i think the solution is 2+(√3) +(√2) u=x →→ du=dx v=d⌊x^2 ⌋→→ v=⌊x^2 ⌋ I=[⌊x^2 ⌋]_1 ^2 −∫_1 ^2 ⌊x^2 ⌋ dx (8−1)−[(∫_1 ^(√2) ⌊x^2 ⌋dx +∫_(√2) ^(√3) ⌊x^2 ⌋dx+∫_(√3) ^2 ⌊x^2 ⌋dx)] 7−[[1]_1 ^(√2) +[2]_(√2) ^(√3) +[3]_(√3) ^(√2) ] 7−[(√2) −1+2((√3) −(√2))+3(2−(√3))] 7−[−(√2) −(√3) +5] 2+(√3)+(√2)](Q85032.png)
$${i}\:{think}\:{the}\:{solution}\:{is}\:\mathrm{2}+\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{2}} \\ $$$${u}={x}\:\rightarrow\rightarrow\:{du}={dx} \\ $$$${v}={d}\lfloor{x}^{\mathrm{2}} \rfloor\rightarrow\rightarrow\:{v}=\lfloor{x}^{\mathrm{2}} \rfloor \\ $$$${I}=\left[\lfloor{x}^{\mathrm{2}} \rfloor\right]_{\mathrm{1}} ^{\mathrm{2}} \:−\int_{\mathrm{1}} ^{\mathrm{2}} \lfloor{x}^{\mathrm{2}} \rfloor\:{dx} \\ $$$$\left(\mathrm{8}−\mathrm{1}\right)−\left[\left(\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \lfloor{x}^{\mathrm{2}} \rfloor{dx}\:+\int_{\sqrt{\mathrm{2}}} ^{\sqrt{\mathrm{3}}} \lfloor{x}^{\mathrm{2}} \rfloor{dx}+\int_{\sqrt{\mathrm{3}}} ^{\mathrm{2}} \lfloor{x}^{\mathrm{2}} \rfloor{dx}\right)\right] \\ $$$$\mathrm{7}−\left[\left[\mathrm{1}\right]_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} +\left[\mathrm{2}\right]_{\sqrt{\mathrm{2}}} ^{\sqrt{\mathrm{3}}} +\left[\mathrm{3}\right]_{\sqrt{\mathrm{3}}} ^{\sqrt{\mathrm{2}}} \right] \\ $$$$\mathrm{7}−\left[\sqrt{\mathrm{2}}\:−\mathrm{1}+\mathrm{2}\left(\sqrt{\mathrm{3}}\:−\sqrt{\mathrm{2}}\right)+\mathrm{3}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\right] \\ $$$$\mathrm{7}−\left[−\sqrt{\mathrm{2}}\:−\sqrt{\mathrm{3}}\:+\mathrm{5}\right] \\ $$$$\mathrm{2}+\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}} \\ $$$$ \\ $$
Commented by mr W last updated on 18/Mar/20
$${if}\:{f}\left({x}\right)=\lfloor{x}^{\mathrm{2}} \rfloor,\:{say}\:\mathrm{1}\leqslant{x}<\sqrt{\mathrm{2}}, \\ $$$${what}\:{is}\:\frac{{df}\left({x}\right)}{{dx}}=? \\ $$
Commented by M±th+et£s last updated on 18/Mar/20
$${i}\:{dont}\:{think}\:{i}\:{can}\:{derivative}\:{this}\: \\ $$$${but}\:{becuse}\:{dv}={d}\lfloor{x}^{\mathrm{2}} \rfloor\:{so}\:{v}=\lfloor{x}^{\mathrm{2}} \rfloor \\ $$