What is the integral from 1 to 2 of 1-4e^{-3x} ?

The integral from 1 to 2 of 1-4e^{-3x} is 1-(4(e^3-1))/(3e^6)

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integral from 1 to 2 of 1-4e^{-3x}

\int_{1}^{2} 1 - 4 e^{- 3 x} d x

1 - \frac{4 ( e ^{3} - 1 )}{3 e ^{6}}

Decimal Notation

0.93692 \dots

Solution steps

\int_{1}^{2} 1 - 4 e^{- 3 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_{1}^{2} 1 d x - \int_{1}^{2} 4 e^{- 3 x} d x

\int_{1}^{2} 1 d x = 1

\int_{1}^{2} 4 e^{- 3 x} d x = \frac{4 ( e ^{3} - 1 )}{3 e ^{6}}

= 1 - \frac{4 ( e ^{3} - 1 )}{3 e ^{6}}

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What is the integral from 1 to 2 of 1-4e^{-3x} ?
The integral from 1 to 2 of 1-4e^{-3x} is 1-(4(e^3-1))/(3e^6)