What is the integral from 0 to 120 of 3-3e^{-0.15t} ?

The integral from 0 to 120 of 3-3e^{-0.15t} is (20)/(e^{18)}+340

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integral from 0 to 120 of 3-3e^{-0.15t}

\int_{0}^{120} 3 - 3 e^{- 0.15 t} d t

\frac{20}{e ^{18}} + 340

Solution steps

\int_{0}^{120} 3 - 3 e^{- 0.15 t} d t

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}^{120} 3 d t - \int_{0}^{120} 3 e^{- 0.15 t} d t

\int_{0}^{120} 3 d t = 360

\int_{0}^{120} 3 e^{- 0.15 t} d t = 20 (1 - \frac{1}{e ^{18}})

= 360 - 20 (1 - \frac{1}{e ^{18}})

Simplify

360 - 20 (1 - \frac{1}{e ^{18}}) : \frac{20}{e ^{18}} + 340

= \frac{20}{e ^{18}} + 340

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What is the integral from 0 to 120 of 3-3e^{-0.15t} ?
The integral from 0 to 120 of 3-3e^{-0.15t} is (20)/(e^{18)}+340