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](https://www.tinkutara.com/question/Q88955.png)
$${hello}\: \\ $$$${floor}\:{function} \\ $$$$\int_{{a}} ^{{b}} \lfloor{x}\rfloor\:\:{dx}\:\:\:\:\:\:\:\:\:\:{a},{b}\in{z}\:\:\:{and}\:{b}>{a} \\ $$$$=\int_{\mathrm{0}} ^{{b}} \lfloor{x}\rfloor\:{dx}\:−\int_{\mathrm{0}} ^{{a}} \lfloor{x}\rfloor\:{dx}\:=\frac{{b}^{\mathrm{2}} −{b}}{\mathrm{2}}−\frac{{a}^{\mathrm{2}} −{a}}{\mathrm{2}}\:….\left(\mathrm{1}\right) \\ $$$${now} \\ $$$$\int_{{m}} ^{{k}} \lfloor{x}\rfloor\:{dx}\:\:\:{when}\:{m},{k}\notin{z}\:\:\:\:{when}\:{m}<{a}<{b}<{k} \\ $$$${b}=\left[{k}\right]\:\:\:\:{and}\:{a}=\left[{m}\right] \\ $$$$\int_{{m}} ^{{a}} \lfloor{x}\rfloor{dx}\:+\int_{{a}} ^{{b}} \lfloor{x}\rfloor{dx}\:+\int_{{b}} ^{{k}} \lfloor{x}\rfloor{dx} \\ $$$$=\left({a}−{m}\right)\lfloor{m}\rfloor+\frac{{b}^{\mathrm{2}} −{b}}{\mathrm{2}}−\frac{{a}^{\mathrm{2}} −{a}}{\mathrm{2}}+\left({k}−{b}\right)\lfloor{k}\rfloor \\ $$$$=\left(\lfloor{m}\rfloor−{m}\right)\lfloor{m}\rfloor+\frac{\lfloor{k}\rfloor^{\mathrm{2}} −\lfloor{k}\rfloor}{\mathrm{2}}−\frac{\lfloor{m}\rfloor^{\mathrm{2}} −\lfloor{m}\rfloor}{\mathrm{2}}+\left({k}−\lfloor{k}\rfloor\right)\lfloor{k}\rfloor \\ $$$$\lfloor{m}\rfloor^{\mathrm{2}} −{m}\lfloor{m}\rfloor\:+\frac{\mathrm{1}}{\mathrm{2}}\lfloor{k}\rfloor^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\lfloor{k}\rfloor−\frac{\mathrm{1}}{\mathrm{2}}\lfloor{m}\rfloor^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\lfloor{m}\rfloor+{k}\lfloor{k}\rfloor−\lfloor{k}\rfloor^{\mathrm{2}} \\ $$$$={k}\lfloor{k}\rfloor−{m}\lfloor{m}\rfloor+\frac{\mathrm{1}}{\mathrm{2}}\lfloor{m}\rfloor^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\lfloor{k}\rfloor^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\lfloor{m}\rfloor−\frac{\mathrm{1}}{\mathrm{2}}\lfloor{k}\rfloor \\ $$$$\therefore\int_{{m}} ^{{k}} \lfloor{x}\rfloor{dx}={k}\lfloor{k}\rfloor−{m}\lfloor{m}\rfloor+\frac{\mathrm{1}}{\mathrm{2}}\left(\lfloor{m}\rfloor−\lfloor{k}\rfloor\right)\left(\lfloor{m}\rfloor+\lfloor{k}\rfloor\right)+\mathrm{1} \\ $$$${example} \\ $$$$\int_{−\mathrm{1}.\mathrm{5}} ^{\mathrm{3}.\mathrm{7}} \lfloor{x}\rfloor{dx}=\left(\mathrm{3}.\mathrm{7}\right)\mathrm{3}−\left(\left(−\mathrm{1}.\mathrm{5}\right)\left(−\mathrm{2}\right)\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(−\mathrm{2}−\mathrm{3}\right)\left(−\mathrm{2}+\mathrm{3}+\mathrm{1}\right) \\ $$$$=\mathrm{11}.\mathrm{1}−\mathrm{3}−\mathrm{5}=\mathrm{3}.\mathrm{1} \\ $$$$ \\ $$$$ \\ $$
Commented by M±th+et£s last updated on 14/Apr/20
$${nice}\:{work}\:{sir} \\ $$
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!)](https://www.tinkutara.com/question/Q89036.png)
$${let}\:{A}\:=\int_{{n}} ^{{m}} \left[{x}\right]{dx}\:\:{with}\:{m}\geqslant{n}\:\Rightarrow\exists{q}\in{N}\:/{m}={n}+{q}\:\Rightarrow \\ $$$${A}\:=\int_{{n}} ^{{n}+{q}} \left[{x}\right]{dx}\:=_{{x}={n}+{t}} \:\:\int_{\mathrm{0}} ^{{q}} \left[{n}+{t}\right]{dt}\:=\int_{\mathrm{0}} ^{{q}} {ndt}\:+\int_{\mathrm{0}} ^{{q}} \:\left[{t}\right]{dt} \\ $$$$={nq}\:+\sum_{{p}=\mathrm{0}} ^{{q}−\mathrm{1}} \:\int_{{p}} ^{{p}+\mathrm{1}} {p}\:{dt}\:={nq}\:+\sum_{{p}=\mathrm{0}} ^{{q}−\mathrm{1}} {p}\:\:={nq}\:+\frac{\left({q}−\mathrm{1}\right){q}}{\mathrm{2}} \\ $$$${but}\:{q}\:={m}−{n}\:\Rightarrow\:{A}\:={n}\left({m}−{n}\right)+\frac{\left({m}−{n}−\mathrm{1}\right)\left({m}−{n}\right)}{\mathrm{2}}\:\Rightarrow \\ $$$${A}\:=\frac{\mathrm{2}{n}\left({m}−{n}\right)}{\mathrm{2}}\:+\frac{{m}−{n}}{\mathrm{2}}\left({m}−{n}−\mathrm{1}\right) \\ $$$$=\frac{{m}−{n}}{\mathrm{2}}\left\{\:\mathrm{2}{n}+{m}−{n}−\mathrm{1}\right\}\:=\frac{{m}−{n}}{\mathrm{2}}\left({n}+{m}−\mathrm{1}\right) \\ $$$${example}\:\:\int_{\mathrm{2}} ^{\mathrm{7}} \left[{x}\right]{dx}\:=\frac{\mathrm{7}−\mathrm{2}}{\mathrm{2}}\left(\mathrm{2}+\mathrm{7}−\mathrm{1}\right)\:=\frac{\mathrm{5}}{\mathrm{2}}×\mathrm{8}\:=\mathrm{20}\:{let}\:{verify} \\ $$$$\int_{\mathrm{2}} ^{\mathrm{7}} \left[{x}\right]{dx}\:=\int_{\mathrm{2}} ^{\mathrm{3}} \left[{x}\right]{dx}\:+\int_{\mathrm{3}} ^{\mathrm{4}} \left[{x}\right]{dx}\:+\int_{\mathrm{4}} ^{\mathrm{5}} \left[{x}\right]{dx}\:+\int_{\mathrm{5}} ^{\mathrm{6}} \left[{x}\right]{dx}\:+\int_{\mathrm{6}} ^{\mathrm{7}} \left[{x}\right]{dx} \\ $$$$=\mathrm{2}+\mathrm{3}\:+\mathrm{4}\:+\mathrm{5}\:+\mathrm{6}\:=\mathrm{20}\:\:\left({the}\:{relation}\:{is}\:{correct}\right) \\ $$$$\int_{−\mathrm{2}} ^{\mathrm{5}} \left[{x}\right]{dx}\:=\frac{\mathrm{5}+\mathrm{2}}{\mathrm{2}}\left(−\mathrm{2}+\mathrm{5}−\mathrm{1}\right)\:=\frac{\mathrm{7}}{\mathrm{2}}×\mathrm{2}\:=\mathrm{7}\:{let}\:{verify} \\ $$$$\int_{−\mathrm{2}} ^{\mathrm{5}} \left[{x}\right]{dx}\:=\int_{−\mathrm{2}} ^{−\mathrm{1}} \left[{x}\right]{dx}\:+\int_{−\mathrm{1}} ^{\mathrm{0}} \left[{x}\right]{dx}\:+\int_{\mathrm{0}} ^{\mathrm{1}} \left[{x}\right]{dx}\:+\int_{\mathrm{1}} ^{\mathrm{2}} \left[{x}\right]{dx}\:+\int_{\mathrm{2}} ^{\mathrm{3}} \left[{x}\right]{dx} \\ $$$$\left.+\int_{\mathrm{3}} ^{\mathrm{4}} \:{x}\right]{dx}\:+\int_{\mathrm{4}} ^{\mathrm{5}} \left[{x}\right]{dx}\:=−\mathrm{2}\:+\left(−\mathrm{1}\right)+\mathrm{0}\:+\mathrm{1}\:+\mathrm{2}+\mathrm{3}+\mathrm{4} \\ $$$$=−\mathrm{3}+\mathrm{3}+\mathrm{3}+\mathrm{4}=\mathrm{7}\:\left({correct}!\right) \\ $$
Commented by mr W last updated on 14/Apr/20
$${please}\:{check}\:{what}\:{you}\:{get}: \\ $$$$\int_{−\mathrm{2}} ^{\mathrm{4}} \lfloor{x}\rfloor{dx}=? \\ $$$${you}\:{get}\:\mathrm{7},\:{but}\:{it}\:{should}\:{be}\:\mathrm{3}. \\ $$
Commented by M±th+et£s last updated on 14/Apr/20
$${where}\:{you}\:{see}\:{a}\:{mistake}\:{in}\:{the}\:{solution} \\ $$
Commented by M±th+et£s last updated on 14/Apr/20
$${if}\:{a},{b}\in{z} \\ $$$$\frac{{b}^{\mathrm{2}} −{b}}{\mathrm{2}}−\frac{{a}^{\mathrm{2}} −{a}}{\mathrm{2}}=\frac{\mathrm{16}−\mathrm{4}}{\mathrm{2}}−\frac{\mathrm{4}+\mathrm{2}}{\mathrm{2}}=\mathrm{6}−\mathrm{3}=\mathrm{3}?? \\ $$$${its}\:{right} \\ $$
Commented by mr W last updated on 14/Apr/20
$${with}\:{k}=\mathrm{4},\:{m}=−\mathrm{2}\:{you}\:{get}\:{from} \\ $$$$\int_{{m}} ^{{k}} \lfloor{x}\rfloor{dx}={k}\lfloor{k}\rfloor−{m}\lfloor{m}\rfloor+\frac{\mathrm{1}}{\mathrm{2}}\left(\lfloor{m}\rfloor−\lfloor{k}\rfloor\right)\left(\lfloor{m}\rfloor+\lfloor{k}\rfloor\right)+\mathrm{1} \\ $$$${the}\:{result}\:\mathrm{7}. \\ $$
Commented by M±th+et£s last updated on 14/Apr/20
$${this}\:{is}\:{for}\:{a},{b}\notin{z} \\ $$
Commented by mr W last updated on 14/Apr/20
$${i}\:{see}.\:{but}\:{we}\:{should}\:{find}\:{a}\:{formula} \\ $$$${for}\:{any}\:{values}\:{of}\:{a}\:{and}\:{b}.\:{my}\:{formula} \\ $$$${works}\:{for}\:{any}\:{values}\:{of}\:{a}\:{and}\:{b}. \\ $$
Commented by M±th+et£s last updated on 14/Apr/20
$${you}\:{are}\:{right}\:{sir}\:{thank}\:{you} \\ $$
Commented by mathmax by abdo last updated on 15/Apr/20
$${thankx} \\ $$