Substitute u = t ³/2592 and du = t ²/864 dt. Then
[tex]P(T)=1-\displaystyle\int_0^T f(t)\,\mathrm dt=1-\displaystyle864\left(1.1574\times10^{-3}\right)\int_0^{\frac{T^3}{2592}}e^{-u}\,\mathrm du[/tex]
The probability that a 12 hour surgery is successful is P (12), so letting T = 12 gives
[tex]P(12)\approx1-\displaystyle\int_0^{\frac23}e^{-u}\,\mathrm du\approx e^{-\frac23} \approx \boxed{0.5134}[/tex]
