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Question Number 181020 by mr W last updated on 20/Nov/22

solve for x>0 ∫_0 ^x ⌊t⌋^2 dt=2(x−1) (Q180780 reposted)

$${solve}\:{for}\:{x}>\mathrm{0} \\ $$ $$\int_{\mathrm{0}} ^{{x}} \lfloor{t}\rfloor^{\mathrm{2}} {dt}=\mathrm{2}\left({x}−\mathrm{1}\right) \\ $$ $$ \\ $$ $$\left({Q}\mathrm{180780}\:{reposted}\right) \\ $$

Answered by mr W last updated on 20/Nov/22

say x=n+f with 0≤f<1 ⌊x⌋=n ∫_0 ^x ⌊t⌋^2 dt=Σ_(k=0) ^(n−1) ∫_k ^(k+1) ⌊t⌋^2 dt+∫_n ^x ⌊t⌋^2 dt =Σ_(k=0) ^(n−1) ∫_k ^(k+1) k^2 dt+∫_n ^x n^2 dt =Σ_(k=0) ^(n−1) k^2 +n^2 (x−n) =(((n−1)n(2n−1))/6)+n^2 (x−n) (((n−1)n(2n−1))/6)+n^2 (x−n)=2(x−1) (((n−1)n(2n−1))/6)+n^2 f=2(n+f−1) (2n^2 −n−12)(n−1)=6(2−n^2 )f n=1: f=0 ⇒x=1 n=2: (8−2−12)(2−1)=6(2−4)f f=(1/2) ⇒x=2+(1/2)=(5/2) i.e. there are two roots: x=1, (5/2).

$${say}\:{x}={n}+{f}\:{with}\:\mathrm{0}\leqslant{f}<\mathrm{1} \\ $$ $$\lfloor{x}\rfloor={n} \\ $$ $$\int_{\mathrm{0}} ^{{x}} \lfloor{t}\rfloor^{\mathrm{2}} {dt}=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\int_{{k}} ^{{k}+\mathrm{1}} \lfloor{t}\rfloor^{\mathrm{2}} {dt}+\int_{{n}} ^{{x}} \lfloor{t}\rfloor^{\mathrm{2}} {dt} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\int_{{k}} ^{{k}+\mathrm{1}} {k}^{\mathrm{2}} {dt}+\int_{{n}} ^{{x}} {n}^{\mathrm{2}} {dt} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{k}^{\mathrm{2}} +{n}^{\mathrm{2}} \left({x}−{n}\right) \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left({n}−\mathrm{1}\right){n}\left(\mathrm{2}{n}−\mathrm{1}\right)}{\mathrm{6}}+{n}^{\mathrm{2}} \left({x}−{n}\right) \\ $$ $$\frac{\left({n}−\mathrm{1}\right){n}\left(\mathrm{2}{n}−\mathrm{1}\right)}{\mathrm{6}}+{n}^{\mathrm{2}} \left({x}−{n}\right)=\mathrm{2}\left({x}−\mathrm{1}\right) \\ $$ $$\frac{\left({n}−\mathrm{1}\right){n}\left(\mathrm{2}{n}−\mathrm{1}\right)}{\mathrm{6}}+{n}^{\mathrm{2}} {f}=\mathrm{2}\left({n}+{f}−\mathrm{1}\right) \\ $$ $$\left(\mathrm{2}{n}^{\mathrm{2}} −{n}−\mathrm{12}\right)\left({n}−\mathrm{1}\right)=\mathrm{6}\left(\mathrm{2}−{n}^{\mathrm{2}} \right){f} \\ $$ $${n}=\mathrm{1}:\: \\ $$ $${f}=\mathrm{0}\:\Rightarrow{x}=\mathrm{1} \\ $$ $${n}=\mathrm{2}: \\ $$ $$\left(\mathrm{8}−\mathrm{2}−\mathrm{12}\right)\left(\mathrm{2}−\mathrm{1}\right)=\mathrm{6}\left(\mathrm{2}−\mathrm{4}\right){f} \\ $$ $${f}=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{x}=\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$ $${i}.{e}.\:{there}\:{are}\:{two}\:{roots}:\:{x}=\mathrm{1},\:\frac{\mathrm{5}}{\mathrm{2}}. \\ $$

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