
Using Integrals :/
I need somewhere to start this problem. Were working on integrals in class and I need help with this Application.
Question: Water flows into an empty storage tank at a rate of V'(t)= 40e^(0.01t) liters per minute. How much water is in the storage tank after 20 minutes?
What I did to set it up was I setup the integral from 0 to 20 with the integrand being 40e^(0.01t). Now from here do I just take the antiderivative and then plug in my times and subract them from eachother?

Re: Using Integrals :/
Yes, you would have:
$\displaystyle V(20)V(0)=\int_0^{20}V'(t)\,dt$
Since the tank is initially empty, i.e. $\displaystyle V(0)=0$ we then have:
$\displaystyle V(20)=40\int_0^{20}e^{0.01t}\,dt$

Re: Using Integrals :/
Okay I got everything you said. The only thing is I try to do U substitution but I'm not sure if that would work. Im getting U= .01t, DU= .01dt. So I dont know if i multiply the 40 by the reciprocal of .01 or what?

Re: Using Integrals :/
Okay I got it to be 40(e^.01(20)) and got it to be 792.03986. To the nearest thousandth would be 792.039 liters

Re: Using Integrals :/
I did something wrong. Damn it.

Re: Using Integrals :/
Your $\displaystyle u$ substitution is what I would use:
$\displaystyle u=0.01t\,\therefore\,du=0.01\,dt$
and so we now have:
$\displaystyle V(20)=40(100)\int_{u(0)}^{u(20)}e^{u}\,(0.01\,dt)=4000\int_{\frac{1}{5}}^0 e^u\,du$
Note: For the last step I used the rule $\displaystyle \int_a^b f(x)\,dx=\int_b^a f(x)\,dx$

Re: Using Integrals :/
For the answer I got 8.008 liters? Did I do something wrong?
4000*[e^0  e^(.01)*(1/5)] and got 8.008 liters.

Re: Using Integrals :/
In liters, we have:
$\displaystyle V(20)=4000\int_{\frac{1}{5}}^0 e^u\,du=4000\left[e^u \right]_{\frac{1}{5}}^0=4000\left(e^0e^{\frac{1}{5}} \right)=$
$\displaystyle 4000\left(1e^{\frac{1}{5}} \right)\approx725.0769876880727$

Re: Using Integrals :/
What happens to the .01 inside the e function?

Re: Using Integrals :/
We did away with that when we substituted. We rewrote the definite integral in terms of u.