## anonymous 5 years ago HELPP?? This problem is driving me nuts. Its Definite Integral Problem. Thanks a ton in advance!! If t is in years since 1990 , the population, P , of the world in billions can be modelled by P=5*3e^(0.014t) What does this model give for the world population in 1990 ? In 2000 ? Round answers to one decimal place. Use the Fundamental Theorem to find the average population of the world during the 1990's. Round the answer to one decimal place.

1. anonymous

hmmm, looks like we should find the derivative and then do an integral. Is that the idea?

2. anonymous

the population P is already given by the equation, so i'm assuming you have no trouble in finding the first two answers, am I right?

3. anonymous

ill brb

4. anonymous

Okay, for the population of the world in 1990 and 2000, that's just plugging in 0 and 10 for t in the equation. I need these values for the next part, so I'll go ahead and answer those. P(0)=15, and P(10)=17.254 approximately. The average value of a function is $[(1/(b-a)]\int\limits_{a}^{b}f(x)dx$ $[(1/(10-0)]\int\limits_{0}^{10}(5*3e^{.014t})dt$ $(1/10)*161.0076$ 16.101 billion is the average population during the 1990's.

5. anonymous

Although rounded to one decimal place it's 16.1 billion. Oops :p

6. anonymous

cant have done it better myself. how do you get the equations to work?

7. anonymous

What do you mean? How do I get them to show up correctly? If that's what you're asking, I just use the Equation button when I'm typing and make sure I use all the correct [] () and {} to make it look right.

8. anonymous

it's not showing up for me somehow, when i press the eqn button. guess im using an unsupported browser

9. anonymous

I guess, or your computer's old or something haha. Sorry!