## Calculus: Early Transcendentals 8th Edition

$\int_{0}^{30}u(t)~dt=C_0~(1-e^{-30r/V})$ This expression tells us how many $mg$ of urea have been removed from the blood after 30 minutes.
We can evaluate the integral: $\int_{0}^{30}u(t)~dt = \int_{0}^{30}\frac{r}{V}C_0~e^{-rt/V}~dt$ Let $u = -\frac{rt}{V}$ $\frac{du}{dt} = -\frac{r}{V}$ $dt = -\frac{V~du}{r}$ When $t = 0$, then $u = 0$ When $t= 30$, then $u = -\frac{30~r}{V}$ $\int_{0}^{-30r/V}(\frac{r}{V}C_0~e^u)~(-\frac{V~du}{r})$ $=~-\int_{0}^{-30r/V}C_0~e^u~du$ $=\int_{-30r/V}^{0}C_0~e^u~du$ $=C_0~e^u~\vert_{-30r/V}^{0}$ $=C_0~(e^0-e^{-30r/V})$ $=C_0~(1-e^{-30r/V})$ This expression tells us how many $mg$ of urea have been removed from the blood after 30 minutes.