What is the integral of 4-10e^{-0.2x} ?

The integral of 4-10e^{-0.2x} is 4x+50e^{-0.2x}+C

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integral of 4-10e^{-0.2x}

\int 4 - 10 e^{- 0.2 x} d x

4 x + 50 e^{- 0.2 x} + C

Solution steps

\int 4 - 10 e^{- 0.2 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 4 d x - \int 10 e^{- 0.2 x} d x

\int 4 d x = 4 x

\int 10 e^{- 0.2 x} d x = - 50 e^{- 0.2 x}

= 4 x - (- 50 e^{- 0.2 x})

Simplify

= 4 x + 50 e^{- 0.2 x}

Add a constant to the solution

= 4 x + 50 e^{- 0.2 x} + C

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What is the integral of 4-10e^{-0.2x} ?
The integral of 4-10e^{-0.2x} is 4x+50e^{-0.2x}+C