What is the integral of 400e^{-0.1x}+8 ?

The integral of 400e^{-0.1x}+8 is -4000e^{-0.1x}+8x+C

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integral of 400e^{-0.1x}+8

\int 400 e^{- 0.1 x} + 8 d x

- 4000 e^{- 0.1 x} + 8 x + C

Solution steps

\int 400 e^{- 0.1 x} + 8 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 400 e^{- 0.1 x} d x + \int 8 d x

\int 400 e^{- 0.1 x} d x = - 4000 e^{- 0.1 x}

\int 8 d x = 8 x

= - 4000 e^{- 0.1 x} + 8 x

Add a constant to the solution

= - 4000 e^{- 0.1 x} + 8 x + C

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What is the integral of 400e^{-0.1x}+8 ?
The integral of 400e^{-0.1x}+8 is -4000e^{-0.1x}+8x+C