What is the integral from 0 to 0.2 of 5(1-e^{-2t}) ?

The integral from 0 to 0.2 of 5(1-e^{-2t}) is 0.17580…

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integral from 0 to 0.2 of 5(1-e^{-2t})

\int_{0}^{0.2} 5 (1 - e^{- 2 t}) d t

0.17580 \dots

Solution steps

\int_{0}^{0.2} 5 (1 - e^{- 2 t}) d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 5 \cdot \int_{0}^{0.2} 1 - e^{- 2 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

= 5 (\int_{0}^{0.2} 1 d t - \int_{0}^{0.2} e^{- 2 t} d t)

\int_{0}^{0.2} 1 d t = 0.2

\int_{0}^{0.2} e^{- 2 t} d t = 0.16483 \dots

= 5 (0.2 - 0.16483 \dots)

Simplify

= 0.17580 \dots

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What is the integral from 0 to 0.2 of 5(1-e^{-2t}) ?
The integral from 0 to 0.2 of 5(1-e^{-2t}) is 0.17580…