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Question Number 2645 by Filup last updated on 24/Nov/15

A=∫_N_1 ^N_2 ⌊x⌋dx (N_1 , N_2 )∈Z, N_1 <N_2 Solve for A

$${A}=\int_{{N}_{\mathrm{1}} } ^{{N}_{\mathrm{2}} } \lfloor{x}\rfloor{dx} \\ $$ $$\left({N}_{\mathrm{1}} ,\:{N}_{\mathrm{2}} \right)\in\mathbb{Z},\:\:\:{N}_{\mathrm{1}} <{N}_{\mathrm{2}} \\ $$ $$ \\ $$ $$\mathrm{Solve}\:\mathrm{for}\:{A} \\ $$

Answered by Filup last updated on 24/Nov/15

let N_1 =N, N_2 =N+k (N_1 , N_2 )∈Z, ∴(N, k)∈Z N_1 <N_2 , ∴k>0 N≤x<N+1⇒⌊x⌋=N N+1≤x<N+2⇒⌊x⌋=N+1 ⋮ N+k−1≤x<N+k⇒⌊x⌋=N+k−1 The area between a→a+1 for ⌊x⌋ =∫_a ^( a+1) ⌊x⌋=a ∵It is a rectangle with l=1, h=a For the total area of A, we need to sum all the rectangular areas ∫_N ^(N+k) ⌊x⌋dx=N+(N+1)+...+(N+k−1) =N(k)+(1+2+...+(k−1)) =Nk+Σ_(i=1) ^(k−1) i =Nk+(1/2)(k−1)(1+k−1) =Nk+(1/2)k(k−1) ∴A=∫_N ^( N+k) ⌊x⌋dx=Nk+(1/2)k(k−1) Test: ∫_2 ^( 5) ⌊x⌋dx=9 (wolfram) N=2, N+k=5⇒k=3 A=2×3+(1/2)3(3−1) =6+(1/2)(2)3 =9 =True

$$\mathrm{let}\:\:\:\:{N}_{\mathrm{1}} ={N},\:\:\:\:{N}_{\mathrm{2}} ={N}+{k} \\ $$ $$\left({N}_{\mathrm{1}} ,\:{N}_{\mathrm{2}} \right)\in\mathbb{Z},\:\:\:\therefore\left({N},\:{k}\right)\in\mathbb{Z} \\ $$ $${N}_{\mathrm{1}} <{N}_{\mathrm{2}} ,\:\:\:\therefore{k}>\mathrm{0} \\ $$ $$ \\ $$ $${N}\leqslant{x}<{N}+\mathrm{1}\Rightarrow\lfloor{x}\rfloor={N} \\ $$ $${N}+\mathrm{1}\leqslant{x}<{N}+\mathrm{2}\Rightarrow\lfloor{x}\rfloor={N}+\mathrm{1} \\ $$ $$\vdots \\ $$ $${N}+{k}−\mathrm{1}\leqslant{x}<{N}+{k}\Rightarrow\lfloor{x}\rfloor={N}+{k}−\mathrm{1} \\ $$ $$ \\ $$ $$\mathrm{The}\:\mathrm{area}\:\mathrm{between}\:{a}\rightarrow{a}+\mathrm{1}\:\mathrm{for}\:\lfloor{x}\rfloor \\ $$ $$=\int_{{a}} ^{\:{a}+\mathrm{1}} \lfloor{x}\rfloor={a} \\ $$ $$\because\mathrm{It}\:\mathrm{is}\:\mathrm{a}\:\mathrm{rectangle}\:\mathrm{with}\:\:{l}=\mathrm{1},\:\:{h}={a} \\ $$ $$ \\ $$ $$\mathrm{For}\:\mathrm{the}\:\mathrm{total}\:\mathrm{area}\:\mathrm{of}\:{A},\:\mathrm{we}\:\mathrm{need}\:\mathrm{to}\:\mathrm{sum} \\ $$ $$\mathrm{all}\:\mathrm{the}\:\mathrm{rectangular}\:\mathrm{areas} \\ $$ $$ \\ $$ $$\int_{{N}} ^{{N}+{k}} \lfloor{x}\rfloor{dx}={N}+\left({N}+\mathrm{1}\right)+...+\left({N}+{k}−\mathrm{1}\right) \\ $$ $$={N}\left({k}\right)+\left(\mathrm{1}+\mathrm{2}+...+\left({k}−\mathrm{1}\right)\right) \\ $$ $$={Nk}+\underset{{i}=\mathrm{1}} {\overset{{k}−\mathrm{1}} {\sum}}{i} \\ $$ $$={Nk}+\frac{\mathrm{1}}{\mathrm{2}}\left({k}−\mathrm{1}\right)\left(\mathrm{1}+{k}−\mathrm{1}\right) \\ $$ $$={Nk}+\frac{\mathrm{1}}{\mathrm{2}}{k}\left({k}−\mathrm{1}\right) \\ $$ $$ \\ $$ $$\therefore{A}=\int_{{N}} ^{\:{N}+{k}} \lfloor{x}\rfloor{dx}={Nk}+\frac{\mathrm{1}}{\mathrm{2}}{k}\left({k}−\mathrm{1}\right) \\ $$ $$ \\ $$ $${Test}: \\ $$ $$\int_{\mathrm{2}} ^{\:\mathrm{5}} \lfloor{x}\rfloor{dx}=\mathrm{9}\:\:\:\left({wolfram}\right) \\ $$ $${N}=\mathrm{2},\:\:\:{N}+{k}=\mathrm{5}\Rightarrow{k}=\mathrm{3} \\ $$ $${A}=\mathrm{2}×\mathrm{3}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{3}\left(\mathrm{3}−\mathrm{1}\right) \\ $$ $$=\mathrm{6}+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}\right)\mathrm{3} \\ $$ $$=\mathrm{9}\:\: \\ $$ $$={True} \\ $$

Commented byYozzi last updated on 24/Nov/15

Σ_(i=0) ^(k−1) i=0+Σ_(i=1) ^(k−1) i=Σ_(i=1) ^(k−1) i=((k−1)/2)(k−1+1)=((k(k−1))/2)≠(1/2)(k−1)^2

$$\underset{{i}=\mathrm{0}} {\overset{{k}−\mathrm{1}} {\sum}}{i}=\mathrm{0}+\underset{{i}=\mathrm{1}} {\overset{{k}−\mathrm{1}} {\sum}}{i}=\underset{{i}=\mathrm{1}} {\overset{{k}−\mathrm{1}} {\sum}}{i}=\frac{{k}−\mathrm{1}}{\mathrm{2}}\left({k}−\mathrm{1}+\mathrm{1}\right)=\frac{{k}\left({k}−\mathrm{1}\right)}{\mathrm{2}}\neq\frac{\mathrm{1}}{\mathrm{2}}\left({k}−\mathrm{1}\right)^{\mathrm{2}} \\ $$ $$ \\ $$

Commented byFilup last updated on 24/Nov/15

Thanks, i just noticed the mistake myself haha

$${Thanks},\:\mathrm{i}\:\mathrm{just}\:\mathrm{noticed}\:\mathrm{the}\:\mathrm{mistake}\:\mathrm{myself} \\ $$ $$\mathrm{haha} \\ $$

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