What is the integral of 200e^{0.1t}+120e^{-0.3t} ?

The integral of 200e^{0.1t}+120e^{-0.3t} is 2000e^{0.1t}-399.99999…e^{-0.3t}+C

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integral of 200e^{0.1t}+120e^{-0.3t}

\int 200 e^{0.1 t} + 120 e^{- 0.3 t} d t

2000 e^{0.1 t} - 399.99999 \dots e^{- 0.3 t} + C

Solution steps

\int 200 e^{0.1 t} + 120 e^{- 0.3 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

= \int 200 e^{0.1 t} d t + \int 120 e^{- 0.3 t} d t

\int 200 e^{0.1 t} d t = 2000 e^{0.1 t}

\int 120 e^{- 0.3 t} d t = - 399.99999 \dots e^{- 0.3 t}

= 2000 e^{0.1 t} - 399.99999 \dots e^{- 0.3 t}

Add a constant to the solution

= 2000 e^{0.1 t} - 399.99999 \dots e^{- 0.3 t} + C

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What is the integral of 200e^{0.1t}+120e^{-0.3t} ?
The integral of 200e^{0.1t}+120e^{-0.3t} is 2000e^{0.1t}-399.99999…e^{-0.3t}+C