Answer
$\int_{0}^{30}u(t)~dt=C_0~(1-e^{-30r/V})$
This expression tells us how many $mg$ of urea have been removed from the blood after 30 minutes.
Work Step by Step
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.
