Question Number 227346 by Spillover last updated on 17/Jan/26
Answered by Kassista last updated on 18/Jan/26
![Note that: ⌊x⌋= n, if n≤x<n+1 (n∈Z) similarly: ⌈x⌉=n+1, n≤x<n+1 (n∈Z) ∴ I = ∫_0 ^( 26) ⌊x⌋x⌈x⌉ dx = Σ_(n=0) ^(25) ∫_n ^( n+1) (n+1)(n)(x)dx = Σ_(n=0) ^(25) (n^2 +n)[(x^2 /2)]_n ^(n+1) = (1/2)Σ_(n=0) ^(25) (n^2 +n)(n^2 +2n+1−n^2 ) (1/2)Σ_(n=0) ^(25) (n^2 +n)(2n+1)=(1/2)Σ_(n=0) ^(25) 2n^3 +3n^2 +n (1/2)[2(((25.26)/2))^2 +3.((25.26.51)/6)+((25.26)/2)] ⇒I= 114,075](https://www.tinkutara.com/question/Q227363.png)
$$ \\ $$$${Note}\:{that}: \\ $$$$\lfloor{x}\rfloor=\:{n},\:{if}\:{n}\leqslant{x}<{n}+\mathrm{1}\:\left({n}\in\mathbb{Z}\right) \\ $$$${similarly}: \\ $$$$\lceil{x}\rceil={n}+\mathrm{1},\:{n}\leqslant{x}<{n}+\mathrm{1}\:\left({n}\in\mathbb{Z}\right) \\ $$$$ \\ $$$$\therefore\:{I}\:=\:\int_{\mathrm{0}} ^{\:\:\mathrm{26}} \lfloor{x}\rfloor{x}\lceil{x}\rceil\:{dx}\:=\:\underset{{n}=\mathrm{0}} {\overset{\mathrm{25}} {\sum}}\int_{{n}} ^{\:{n}+\mathrm{1}} \left({n}+\mathrm{1}\right)\left({n}\right)\left({x}\right){dx} \\ $$$$ \\ $$$$=\:\underset{{n}=\mathrm{0}} {\overset{\mathrm{25}} {\sum}}\left({n}^{\mathrm{2}} +{n}\right)\left[\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right]_{{n}} ^{{n}+\mathrm{1}} =\:\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\mathrm{25}} {\sum}}\left({n}^{\mathrm{2}} +{n}\right)\left({n}^{\mathrm{2}} +\mathrm{2}{n}+\mathrm{1}−{n}^{\mathrm{2}} \right) \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\mathrm{25}} {\sum}}\left({n}^{\mathrm{2}} +{n}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\mathrm{25}} {\sum}}\mathrm{2}{n}^{\mathrm{3}} +\mathrm{3}{n}^{\mathrm{2}} +{n} \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{2}\left(\frac{\mathrm{25}.\mathrm{26}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{3}.\frac{\mathrm{25}.\mathrm{26}.\mathrm{51}}{\mathrm{6}}+\frac{\mathrm{25}.\mathrm{26}}{\mathrm{2}}\right]\:\Rightarrow{I}=\:\mathrm{114},\mathrm{075} \\ $$
Answered by mathmax last updated on 19/Jan/26
![I=∫_0 ^(26) x[x]^2 dx=Σ_(n=0) ^(25) ∫_n ^(n+1) x[x]^2 dx =Σ_(n=0) ^(25) ∫_n ^(n+1) n^2 xdx =Σ_(n=0) ^(25) n^2 [(x^2 /2)]_n ^(n+1) dx =(1/2)Σ_(n=0) ^(25) n^2 ((n+1)^2 −n^2 ) =(1/2)Σ_(n=0) ^(25) n^2 (2n+1) =Σ_(n=0) ^(25) n^3 +(1/2)Σ_(n=0) ^(25) n^2 =(((25.26)^2 )/4)+(1/2).25(25+1)(2.25+1)/6 =(((25.26)^2 )/4)+((25.26.51)/(12))](https://www.tinkutara.com/question/Q227381.png)
$${I}=\int_{\mathrm{0}} ^{\mathrm{26}} {x}\left[{x}\right]^{\mathrm{2}} {dx}=\sum_{{n}=\mathrm{0}} ^{\mathrm{25}} \:\int_{{n}} ^{{n}+\mathrm{1}} {x}\left[{x}\right]^{\mathrm{2}} {dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\mathrm{25}} \int_{{n}} ^{{n}+\mathrm{1}} {n}^{\mathrm{2}} \:{xdx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\mathrm{25}} {n}^{\mathrm{2}} \:\left[\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right]_{{n}} ^{{n}+\mathrm{1}} {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\sum_{{n}=\mathrm{0}} ^{\mathrm{25}} {n}^{\mathrm{2}} \left(\left({n}+\mathrm{1}\right)^{\mathrm{2}} −{n}^{\mathrm{2}} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\sum_{{n}=\mathrm{0}} ^{\mathrm{25}} {n}^{\mathrm{2}} \left(\mathrm{2}{n}+\mathrm{1}\right) \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\mathrm{25}} {n}^{\mathrm{3}} \:+\frac{\mathrm{1}}{\mathrm{2}}\sum_{{n}=\mathrm{0}} ^{\mathrm{25}} {n}^{\mathrm{2}} \\ $$$$=\frac{\left(\mathrm{25}.\mathrm{26}\right)^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{25}\left(\mathrm{25}+\mathrm{1}\right)\left(\mathrm{2}.\mathrm{25}+\mathrm{1}\right)/\mathrm{6} \\ $$$$=\frac{\left(\mathrm{25}.\mathrm{26}\right)^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{25}.\mathrm{26}.\mathrm{51}}{\mathrm{12}} \\ $$