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…

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

\frac{\partial}{\partial x} (4 e^{x} ln (y))

\int 11 x ln (x) d x

f (x) = - \frac{2}{x ^{3}}

\int (1 - x^{2})^{- 0.5} d x

e x p an d - \frac{z}{e ^{z + 1} ( z - 1 ) ^{2}}

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…