Rate of Change and Integrals

[olicoh]

I already posted about this but I redid the problem and got another answer:

1. Homework Statement
The volume of an air mattress is 15 cubic feet. Air escapes at a rate of r(t)=0.25e^(−0.05t), where r is in cubic feet per second. Assuming the mattress is still completely inflated when the valve is opened, how much air is released in the first 30 seconds?

Homework Equations

Here is what I have so far:
V(30) = integral[0,30] (0.25e^(-0.05t))
= [-(1/2) (e^(-1/2t))]³⁰₀

The Attempt at a Solution

= 0.499999847 cubic feet/sec


I think I did this correctly but the answer seems too small...

 

[LCKurtz]

I already posted about this but I redid the problem and got another answer:

1. Homework Statement
The volume of an air mattress is 15 cubic feet. Air escapes at a rate of r(t)=0.25e^(−0.05t), where r is in cubic feet per second. Assuming the mattress is still completely inflated when the valve is opened, how much air is released in the first 30 seconds?

Homework Equations

Here is what I have so far:
V(30) = integral[0,30] (0.25e^(-0.05t))
= [-(1/2) (e^(-1/2t))]³⁰₀

The Attempt at a Solution

= 0.499999847 cubic feet/sec


I think I did this correctly but the answer seems too small...

You have written e^-.05t and worked the problem as though it was e^-.5t. And when you are asked how much air has been released in 30 seconds, would you really give an answer in units of cubic ft / second?

 

[olicoh]

You have written e-.05t and worked the problem as though it was e-.5t

What do you mean by that?

The antiderivative of (0.25e^(-0.05t)) = [-(1/2) (e^(-1/2t))] doesn't it?

 

[LCKurtz]

You have written e^-.05t and worked the problem as though it was e^-.5t. And when you are asked how much air has been released in 30 seconds, would you really give an answer in units of cubic ft / second?

What do you mean by that?

The antiderivative of (0.25e^(-0.05t)) = [-(1/2) (e^(-1/2t))] doesn't it?

-.05 is not equal to -1/2.