Question

Seems simple..but idk why I am getting it wrong :/ I did 3000e^(0.04*1) and I got =3122.43

If you invest $3,000 at 4% interest compounded continuously, what is the average amount in your account over one year? (Round your answer to the nearest cent.)

Answer 1

what does "average amount" mean? are you supposed to integrate?

Answer 2

I have no idea that is all it says in the problem, but that would suck if I had to integrate :/

Answer 3

not really since it is pretty easy to integrate \(e^x\)

Answer 4

oh then that would just be e^x

Answer 5

and since the time interval is one year
maybe it is \[\int _0^1e^{,04t}dt\]

Answer 6

times 3000 of course

Answer 7

ok let me see

Answer 8

i will try it and see what i get
i think that is what it means by "average value' maybe i am wrong

Answer 9

we put in the 1 and 0 for e?

Answer 10

or nvm we raise it to those powers right?

Answer 11

i get 3060.81

Answer 12

the anti derivative of \(e^{.04t}\) is \(25e^{.04t}\) by a simple u sub

Answer 13

evaluate at 1 and 0, and subtract

Answer 14

what did you substitute to get 25e^.04t? sorry I am confused :/

Answer 15

you get
\[3000\times 25\left(e^{.04}-e^0\right)\] or \[75000\left(e^{.04}-1\right)\]

Answer 16

oh ok you have \(e^{.04t}\) and you want the anti derivative

Answer 17

the u sub is \(u-.04t, du =.04dt, 25du = dt\) since 25 is the reciprocal of \(.04\)

Answer 18

i meant \(u=.04t\)

Answer 19

aka \(u=\frac{1}{25}t\)

Answer 20

clear or no?

Answer 21

yes I think I got it!!

Answer 22

ok good luck
if it is an on line class you can check the number i got above and see if it works

Answer 23

yup it did! I appreciate the help! thanks! :)

Answer 24

if not, then i am lost
oh whew!

Answer 25

yw

Answer 26

:)