What is the integral of 250(e^{-0.2x}-1) ?

The integral of 250(e^{-0.2x}-1) is 250(-5e^{-0.2x}-x)+C

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integral of 250(e^{-0.2x}-1)

\int 250 (e^{- 0.2 x} - 1) d x

250 (- 5 e^{- 0.2 x} - x) + C

Solution steps

\int 250 (e^{- 0.2 x} - 1) d x

Take the constant out:

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

= 250 \cdot \int e^{- 0.2 x} - 1 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

= 250 (\int e^{- 0.2 x} d x - \int 1 d x)

\int e^{- 0.2 x} d x = - 5 e^{- 0.2 x}

\int 1 d x = x

= 250 (- 5 e^{- 0.2 x} - x)

Add a constant to the solution

= 250 (- 5 e^{- 0.2 x} - x) + C

What is the integral of 250(e^{-0.2x}-1) ?
The integral of 250(e^{-0.2x}-1) is 250(-5e^{-0.2x}-x)+C