# Using Integrals :/

• Nov 26th 2012, 08:30 PM
illicitkush
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 anti-derivative and then plug in my times and subract them from eachother?
• Nov 26th 2012, 08:37 PM
MarkFL
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:

$\displaystyle V(20)=40\int_0^{20}e^{-0.01t}\,dt$
• Nov 26th 2012, 08:45 PM
illicitkush
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?
• Nov 26th 2012, 08:56 PM
illicitkush
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
• Nov 26th 2012, 09:03 PM
illicitkush
Re: Using Integrals :/
I did something wrong. Damn it.
• Nov 26th 2012, 09:03 PM
MarkFL
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 $\displaystyle -\int_a^b f(x)\,dx=\int_b^a f(x)\,dx$
• Nov 26th 2012, 09:13 PM
illicitkush
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.
• Nov 26th 2012, 09:47 PM
MarkFL
Re: Using Integrals :/
In liters, we have:

$\displaystyle V(20)=4000\int_{-\frac{1}{5}}^0 e^u\,du=4000\left[e^u \right]_{-\frac{1}{5}}^0=4000\left(e^0-e^{-\frac{1}{5}} \right)=$

$\displaystyle 4000\left(1-e^{-\frac{1}{5}} \right)\approx725.0769876880727$
• Nov 26th 2012, 09:49 PM
illicitkush
Re: Using Integrals :/
What happens to the -.01 inside the e function?
• Nov 26th 2012, 10:01 PM
MarkFL
Re: Using Integrals :/
We did away with that when we substituted. We rewrote the definite integral in terms of u.