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Question Number 49898 by Raj Singh last updated on 12/Dec/18

Commented by Abdo msup. last updated on 12/Dec/18

I =∫_1 ^(400[x]) e^([t]) dt =Σ_(k=1) ^(400[x]−1) ∫_k ^(k+1) e^k dt =Σ_(k=1) ^(400[x]−1) e^k = ((1−e^(400[x]) )/(1−e)) −1 =((1−e^(400[x]) −1+e)/(1−e)) =((e−e^(400[x]) )/(1−e)) =

I=∫1400[x]e[t]dt=∑k=1400[x]−1∫kk+1ekdt=∑k=1400[x]−1ek=1−e400[x]1−e−1=1−e400[x]−1+e1−e=e−e400[x]1−e=

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