What is the integral from 0 to 10 of (2200e^{0.024t}-1460e^{0.018t}) ?

The integral from 0 to 10 of (2200e^{0.024t}-1460e^{0.018t}) is 8867.98599…

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integral from 0 to 10 of (2200e^{0.024t}-1460e^{0.018t})

\int_{0}^{10} (2200 e^{0.024 t} - 1460 e^{0.018 t}) d t

8867.98599 \dots

Solution steps

\int_{0}^{10} 2200 e^{0.024 t} - 1460 e^{0.018 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_{0}^{10} 2200 e^{0.024 t} d t - \int_{0}^{10} 1460 e^{0.018 t} d t

\int_{0}^{10} 2200 e^{0.024 t} d t = 24864.50544 \dots

\int_{0}^{10} 1460 e^{0.018 t} d t = 15996.51945 \dots

= 24864.50544 \dots - 15996.51945 \dots

Simplify

= 8867.98599 \dots

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What is the integral from 0 to 10 of (2200e^{0.024t}-1460e^{0.018t}) ?
The integral from 0 to 10 of (2200e^{0.024t}-1460e^{0.018t}) is 8867.98599…