Question-49898

Tinku Tara June 4, 2023 Integration 0 Comments

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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 = \int_{1}^{400 [x]} e^{[t]} d t = \sum_{k = 1}^{400 [x] - 1} \int_{k}^{k + 1} e^{k} d t

= \sum_{k = 1}^{400 [x] - 1} e^{k} = \frac{1 - e^{400 [x]}}{1 - e} - 1 = \frac{1 - e^{400 [x]} - 1 + e}{1 - e}

= \frac{e - e^{400 [x]}}{1 - e}

=

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