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Question Number 88955 by M±th+et£s last updated on 14/Apr/20
hello floor function ∫_a ^b ⌊x⌋ dx a,b∈z and b>a =∫_0 ^b ⌊x⌋ dx −∫_0 ^a ⌊x⌋ dx =((b^2 −b)/2)−((a^2 −a)/2) ....(1) now ∫_m ^k ⌊x⌋ dx when m,k∉z when m<a<b<k b=[k] and a=[m] ∫_m ^a ⌊x⌋dx +∫_a ^b ⌊x⌋dx +∫_b ^k ⌊x⌋dx =(a−m)⌊m⌋+((b^2 −b)/2)−((a^2 −a)/2)+(k−b)⌊k⌋ =(⌊m⌋−m)⌊m⌋+((⌊k⌋^2 −⌊k⌋)/2)−((⌊m⌋^2 −⌊m⌋)/2)+(k−⌊k⌋)⌊k⌋ ⌊m⌋^2 −m⌊m⌋ +(1/2)⌊k⌋^2 −(1/2)⌊k⌋−(1/2)⌊m⌋^2 −(1/2)⌊m⌋+k⌊k⌋−⌊k⌋^2 =k⌊k⌋−m⌊m⌋+(1/2)⌊m⌋^2 −(1/2)⌊k⌋^2 +(1/2)⌊m⌋−(1/2)⌊k⌋ ∴∫_m ^k ⌊x⌋dx=k⌊k⌋−m⌊m⌋+(1/2)(⌊m⌋−⌊k⌋)(⌊m⌋+⌊k⌋)+1 example ∫_(−1.5) ^(3.7) ⌊x⌋dx=(3.7)3−((−1.5)(−2))+(1/2)(−2−3)(−2+3+1) =11.1−3−5=3.1
hellofloorfunction∫ab⌊x⌋dxa,b∈zandb>a=∫0b⌊x⌋dx−∫0a⌊x⌋dx=b2−b2−a2−a2….(1)now∫mk⌊x⌋dxwhenm,k∉zwhenm<a<b<kb=[k]anda=[m]∫ma⌊x⌋dx+∫ab⌊x⌋dx+∫bk⌊x⌋dx=(a−m)⌊m⌋+b2−b2−a2−a2+(k−b)⌊k⌋=(⌊m⌋−m)⌊m⌋+⌊k⌋2−⌊k⌋2−⌊m⌋2−⌊m⌋2+(k−⌊k⌋)⌊k⌋⌊m⌋2−m⌊m⌋+12⌊k⌋2−12⌊k⌋−12⌊m⌋2−12⌊m⌋+k⌊k⌋−⌊k⌋2=k⌊k⌋−m⌊m⌋+12⌊m⌋2−12⌊k⌋2+12⌊m⌋−12⌊k⌋∴∫mk⌊x⌋dx=k⌊k⌋−m⌊m⌋+12(⌊m⌋−⌊k⌋)(⌊m⌋+⌊k⌋)+1example∫−1.53.7⌊x⌋dx=(3.7)3−((−1.5)(−2))+12(−2−3)(−2+3+1)=11.1−3−5=3.1
Commented by M±th+et£s last updated on 14/Apr/20
nice work sir
niceworksir
Commented by mathmax by abdo last updated on 14/Apr/20
let A =∫_n ^m [x]dx with m≥n ⇒∃q∈N /m=n+q ⇒ A =∫_n ^(n+q) [x]dx =_(x=n+t) ∫_0 ^q [n+t]dt =∫_0 ^q ndt +∫_0 ^q [t]dt =nq +Σ_(p=0) ^(q−1) ∫_p ^(p+1) p dt =nq +Σ_(p=0) ^(q−1) p =nq +(((q−1)q)/2) but q =m−n ⇒ A =n(m−n)+(((m−n−1)(m−n))/2) ⇒ A =((2n(m−n))/2) +((m−n)/2)(m−n−1) =((m−n)/2){ 2n+m−n−1} =((m−n)/2)(n+m−1) example ∫_2 ^7 [x]dx =((7−2)/2)(2+7−1) =(5/2)×8 =20 let verify ∫_2 ^7 [x]dx =∫_2 ^3 [x]dx +∫_3 ^4 [x]dx +∫_4 ^5 [x]dx +∫_5 ^6 [x]dx +∫_6 ^7 [x]dx =2+3 +4 +5 +6 =20 (the relation is correct) ∫_(−2) ^5 [x]dx =((5+2)/2)(−2+5−1) =(7/2)×2 =7 let verify ∫_(−2) ^5 [x]dx =∫_(−2) ^(−1) [x]dx +∫_(−1) ^0 [x]dx +∫_0 ^1 [x]dx +∫_1 ^2 [x]dx +∫_2 ^3 [x]dx +∫_3 ^4 x]dx +∫_4 ^5 [x]dx =−2 +(−1)+0 +1 +2+3+4 =−3+3+3+4=7 (correct!)
letA=∫nm[x]dxwithm⩾n⇒∃q∈N/m=n+q⇒A=∫nn+q[x]dx=x=n+t∫0q[n+t]dt=∫0qndt+∫0q[t]dt=nq+∑p=0q−1∫pp+1pdt=nq+∑p=0q−1p=nq+(q−1)q2butq=m−n⇒A=n(m−n)+(m−n−1)(m−n)2⇒A=2n(m−n)2+m−n2(m−n−1)=m−n2{2n+m−n−1}=m−n2(n+m−1)example∫27[x]dx=7−22(2+7−1)=52×8=20letverify∫27[x]dx=∫23[x]dx+∫34[x]dx+∫45[x]dx+∫56[x]dx+∫67[x]dx=2+3+4+5+6=20(therelationiscorrect)∫−25[x]dx=5+22(−2+5−1)=72×2=7letverify∫−25[x]dx=∫−2−1[x]dx+∫−10[x]dx+∫01[x]dx+∫12[x]dx+∫23[x]dx+∫34x]dx+∫45[x]dx=−2+(−1)+0+1+2+3+4=−3+3+3+4=7(correct!)
Commented by mr W last updated on 14/Apr/20
please check what you get: ∫_(−2) ^4 ⌊x⌋dx=? you get 7, but it should be 3.
pleasecheckwhatyouget:∫−24⌊x⌋dx=?youget7,butitshouldbe3.
Commented by M±th+et£s last updated on 14/Apr/20
where you see a mistake in the solution
whereyouseeamistakeinthesolution
Commented by M±th+et£s last updated on 14/Apr/20
if a,b∈z ((b^2 −b)/2)−((a^2 −a)/2)=((16−4)/2)−((4+2)/2)=6−3=3?? its right
ifa,b∈zb2−b2−a2−a2=16−42−4+22=6−3=3??itsright
Commented by mr W last updated on 14/Apr/20
with k=4, m=−2 you get from ∫_m ^k ⌊x⌋dx=k⌊k⌋−m⌊m⌋+(1/2)(⌊m⌋−⌊k⌋)(⌊m⌋+⌊k⌋)+1 the result 7.
withk=4,m=−2yougetfrom∫mk⌊x⌋dx=k⌊k⌋−m⌊m⌋+12(⌊m⌋−⌊k⌋)(⌊m⌋+⌊k⌋)+1theresult7.
Commented by M±th+et£s last updated on 14/Apr/20
this is for a,b∉z
thisisfora,b∉z
Commented by mr W last updated on 14/Apr/20
i see. but we should find a formula for any values of a and b. my formula works for any values of a and b.
isee.butweshouldfindaformulaforanyvaluesofaandb.myformulaworksforanyvaluesofaandb.
Commented by M±th+et£s last updated on 14/Apr/20
you are right sir thank you
youarerightsirthankyou
Commented by mathmax by abdo last updated on 15/Apr/20
thankx
thankx

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