What is the integral from 0 to t of 12e^{2x}-12e^x ?

The integral from 0 to t of 12e^{2x}-12e^x is -12e^t+6e^{2t}+6

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integral from 0 to t of 12e^{2x}-12e^x

\int_{0}^{t} 12 e^{2 x} - 12 e^{x} d x

- 12 e^{t} + 6 e^{2 t} + 6

Solution steps

\int_{0}^{t} 12 e^{2 x} - 12 e^{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_{0}^{t} 12 e^{2 x} d x - \int_{0}^{t} 12 e^{x} d x

\int_{0}^{t} 12 e^{2 x} d x = 6 (e^{2 t} - 1)

\int_{0}^{t} 12 e^{x} d x = 12 (e^{t} - 1)

= 6 (e^{2 t} - 1) - 12 (e^{t} - 1)

Simplify

6 (e^{2 t} - 1) - 12 (e^{t} - 1) : - 12 e^{t} + 6 e^{2 t} + 6

= - 12 e^{t} + 6 e^{2 t} + 6

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What is the integral from 0 to t of 12e^{2x}-12e^x ?
The integral from 0 to t of 12e^{2x}-12e^x is -12e^t+6e^{2t}+6