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Question Number 16088 by hamil last updated on 17/Jun/17

 ∫_( 0) ^(1000) e^(x−[x]) dx =

$$\:\underset{\:\mathrm{0}} {\overset{\mathrm{1000}} {\int}}{e}^{{x}−\left[{x}\right]} {dx}\:= \\ $$

Commented by prakash jain last updated on 17/Jun/17

x=[x]+{x} where 0≤{x}<1  ∫_0 ^( 1000) e^(x−[x]) dx=1000∫_0 ^1 e^x dx  =1000(e^x −1)

$${x}=\left[{x}\right]+\left\{{x}\right\}\:\mathrm{where}\:\mathrm{0}\leqslant\left\{{x}\right\}<\mathrm{1} \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{1000}} {e}^{{x}−\left[{x}\right]} {dx}=\mathrm{1000}\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{{x}} {dx} \\ $$$$=\mathrm{1000}\left({e}^{{x}} −\mathrm{1}\right) \\ $$

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