Question Number 22843 by gourav~ last updated on 22/Oct/17
![calculate the value of ∫_0 ^(1000) e^(x−[x]) dx =? where.... [x] is greatest integer functon](Q22843.png)
$${calculate}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\mathrm{1000}} {e}^{{x}−\left[{x}\right]} {dx}\:=? \\ $$$${where}....\:\left[{x}\right]\:{is}\:{greatest}\:{integer}\:{functon} \\ $$
Answered by ajfour last updated on 23/Oct/17
![let I=∫_0 ^( 1000) e^(x−[x]) dx =Σ_0 ^(999) ∫_m ^( m+1) e^(x−[x]) dx let x−[x]={x} ⇒ dx=d{x} for each interval so I=Σ_0 ^(999) ∫_0 ^( 1) e^({x}) d{x}=1000(e−1) .](Q22845.png)
$${let}\:{I}=\int_{\mathrm{0}} ^{\:\:\mathrm{1000}} {e}^{{x}−\left[{x}\right]} {dx} \\ $$$$\:\:=\underset{\mathrm{0}} {\overset{\mathrm{999}} {\sum}}\int_{{m}} ^{\:\:{m}+\mathrm{1}} {e}^{{x}−\left[{x}\right]} {dx} \\ $$$${let}\:{x}−\left[{x}\right]=\left\{{x}\right\} \\ $$$$\:\:\:\:\:\Rightarrow\:\:{dx}={d}\left\{{x}\right\}\:{for}\:{each}\:{interval} \\ $$$${so}\:\:{I}=\underset{\mathrm{0}} {\overset{\mathrm{999}} {\sum}}\int_{\mathrm{0}} ^{\:\:\mathrm{1}} {e}^{\left\{{x}\right\}} {d}\left\{{x}\right\}=\mathrm{1000}\left({e}−\mathrm{1}\right)\:. \\ $$
Commented by gourav~ last updated on 23/Oct/17
$${yes}\:{sir}...\:{but}\:{how} \\ $$
Commented by jota+ last updated on 03/Dec/17
![m≤x<m+1 ⇒ x−[x]=x−m=u then x=m+u ∫_m ^(m+1) dx → ∫_0 ^1 du](Q25106.png)
$${m}\leqslant{x}<{m}+\mathrm{1}\:\Rightarrow\:{x}−\left[{x}\right]={x}−{m}={u} \\ $$$${then}\:\:{x}={m}+{u} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\int_{{m}} ^{{m}+\mathrm{1}} {dx}\:\rightarrow\:\int_{\mathrm{0}} ^{\mathrm{1}} {du} \\ $$