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solve-for-x-gt-0-0-x-t-2-dt-2-x-1-Q180780-reposted-

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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)
solveforx>0∫0x⌊t⌋2dt=2(x−1)(Q180780reposted)
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).
sayx=n+fwith0⩽f<1⌊x⌋=n∫0x⌊t⌋2dt=∑n−1k=0∫kk+1⌊t⌋2dt+∫nx⌊t⌋2dt=∑n−1k=0∫kk+1k2dt+∫nxn2dt=∑n−1k=0k2+n2(x−n)=(n−1)n(2n−1)6+n2(x−n)(n−1)n(2n−1)6+n2(x−n)=2(x−1)(n−1)n(2n−1)6+n2f=2(n+f−1)(2n2−n−12)(n−1)=6(2−n2)fn=1:f=0⇒x=1n=2:(8−2−12)(2−1)=6(2−4)ff=12⇒x=2+12=52i.e.therearetworoots:x=1,52.

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