Water is poured into a bucket according to the rate F(t)=(t+7)/(t+2) , and at the same time empties out through a hole in the bottom at the rate E(t)=(ln(t+4))/t+2 , with both F(t) and E(t) measured in pints per minute. How much water, to the nearest pint, is in the bucket at time t = 5 minutes. You must show your setup but can use your calculator for all evaluations.
I have no idea how to do this problem the only thing I know is that it will equal the definite integral from [0,5] of F(t)-E(t)dt
correct. So, just do it...
∫[0,5] F(t)-E(t) dt
= ∫[0,5] (t+7)/(t+2) - ln(t+4)/(t+2) dt
∫(t+7)/(t+2) dt = ∫1 + 5/(t+2) dt
= t + 5ln(t+2)
For ∫ln(t+4)/(t+2) dt, you're in trouble. It is not doable with elementary functions. You sure there's no typo?
∫ln(u)/u du = 1/2 ln^2(u)
but you don't quite have that.
I can ask my teacher! how would we do it if it was actually just (t-4)/(t+2)?
oops I meant (t+4)/(t+2)
then it's just like the one I showed first. divide and hen integrate.
To find the amount of water in the bucket at t = 5 minutes, we need to calculate the net flow of water into the bucket during the time interval [0, 5] minutes.
The rate at which water flows into the bucket is given by F(t) = (t + 7)/(t + 2) pints per minute.
The rate at which water flows out of the bucket is given by E(t) = (ln(t + 4))/(t + 2) pints per minute.
To find the net flow, we subtract the rate at which water flows out of the bucket from the rate at which water flows into the bucket:
Net Rate = F(t) - E(t)
Net Rate = (t + 7)/(t + 2) - (ln(t + 4))/(t + 2)
To find the amount of water in the bucket at t = 5 minutes, we integrate the net rate of flow over the interval [0, 5]:
Amount of Water at t = 5 minutes = ∫[0,5] (F(t) - E(t)) dt
Amount of Water at t = 5 minutes = ∫[0,5] ((t + 7)/(t + 2) - (ln(t + 4))/(t + 2)) dt
Now we can evaluate this integral using a calculator or software.