What is the integral from 0 to 40 of 1-e^{-x/(22)} ?

The integral from 0 to 40 of 1-e^{-x/(22)} is (22)/(e^{20/11)}+18

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integral from 0 to 40 of 1-e^{-x/(22)}

\int_{0}^{40} 1 - e^{- \frac{x}{22}} d x

\frac{22}{e ^{\frac{20}{11}}} + 18

Decimal Notation

21.57105 \dots

Solution steps

\int_{0}^{40} 1 - e^{- \frac{x}{22}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{40} 1 d x - \int_{0}^{40} e^{- \frac{x}{22}} d x

\int_{0}^{40} 1 d x = 40

\int_{0}^{40} e^{- \frac{x}{22}} d x = 22 (1 - \frac{1}{e ^{\frac{20}{11}}})

= 40 - 22 (1 - \frac{1}{e ^{\frac{20}{11}}})

Simplify

40 - 22 (1 - \frac{1}{e ^{\frac{20}{11}}}) : \frac{22}{e ^{\frac{20}{11}}} + 18

= \frac{22}{e ^{\frac{20}{11}}} + 18

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What is the integral from 0 to 40 of 1-e^{-x/(22)} ?
The integral from 0 to 40 of 1-e^{-x/(22)} is (22)/(e^{20/11)}+18