Consider again the deposition of sediment on the bottom of a lake.
Here's the picture again to illustrate
the physical problem.
If the rate of deposition is constant, say 4mm a year, then we can see immediately that after 5 years the sediment will be 20mm thick.
Now consider a more realisitic scenario, in which the rate of deposition is changing in time. Let's call the rate of sediment deposition ds/dt, where s(t) is the thickness of the sediment.
Suppose the sediment is being deposited at a rate ds/dt=e-2t.
That means it's slowing down in time, for larger values of t the rate is tending to zero.
To find the thickness at any time t, we can simply integrate this rate of deposition.
The constant of integration is determined by how thick the sediment is at t=0. (Actually any fixed value of t that we choose, but it's most straightforward if we use t=0.)
Let's say that we start our "time" when the thickness is 10mm.
What will the thickness be at time t=1?
Well, we need to integrate the rate of deposition, ds/dt=e-2t.
This gives us s(t)=-.5e-2t + c. (Remember that we need that -.5 in the front to cancel with the -2 that comes down from the power when we differentiate)
The "c" is the constant of integration. We now fix this by saying that
we want s(t) to be 10 when t=0.
Substituting these values in to our equation for s(t), we find
10=-.5e0+c.
This tells us that c must be 10.5 (since e0=1).
So now we've got an equation that we can use to find s(t), for all values of t: s(t)=-.5e-2t + 10.5
In particular we were asked what is s(t) when t=1. So we substitute t=1 into our equation to find s(1)=-.5e-2 + 10.5, which is about 10.43.
We can also see what will happen as time goes on. At large values of t, the first term, the exponential one, becomes smaller and smaller, approaching (but never reaching) zero. So the thickness becomes closer to 10.5 as time goes on.