What is the integral from 0 to 60 of 3-3e^{-0.15x} ?

The integral from 0 to 60 of 3-3e^{-0.15x} is (20)/(e^9)+160

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integral from 0 to 60 of 3-3e^{-0.15x}

\int_{0}^{60} 3 - 3 e^{- 0.15 x} d x

\frac{20}{e ^{9}} + 160

Solution steps

\int_{0}^{60} 3 - 3 e^{- 0.15 x} 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}^{60} 3 d x - \int_{0}^{60} 3 e^{- 0.15 x} d x

\int_{0}^{60} 3 d x = 180

\int_{0}^{60} 3 e^{- 0.15 x} d x = 20 (1 - \frac{1}{e ^{9}})

= 180 - 20 (1 - \frac{1}{e ^{9}})

Simplify

180 - 20 (1 - \frac{1}{e ^{9}}) : \frac{20}{e ^{9}} + 160

= \frac{20}{e ^{9}} + 160

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What is the integral from 0 to 60 of 3-3e^{-0.15x} ?
The integral from 0 to 60 of 3-3e^{-0.15x} is (20)/(e^9)+160