I have this question:

1. The amount a fish grows follows the differential equation

$\displaystyle dL/dt = 6.48e^-0.09t$,

with initial condition L(0) = 5, where t is measured in years and L is measured in centimetres. How much does the fish grow between ages t = 0.5 and t = 1.5?

(Give your answer to two decimal places.)

I integrated it to $\displaystyle L(t) = -72e^-0.09t$ and did the definite integral calculation getting ~6cm. However, when I sub in L(0), I don't get 5. Is that relevant?