# Homework Help: Why take the integral of this instead of just plugging in the number?

1. Dec 25, 2012

### Xetman

1. The problem statement, all variables and given/known data

An oil storage tank ruptures at time t = 0 and oil leaks from the tank at a rate of r(t) = 100 e^(-0.01t) liters per minute. How much oil leaks out during the first hour?

2. Relevant equations

r(t) = 100 e^(-0.01t)

3. The attempt at a solution

Eh, I already got the answer, I just did a definite integral from 0 to 60 for the equation.
My question is: Why would you find the integral instead of just plugging in 60 for t into that equation? Is it because that equation is a rate, so it's a derivative and you have to take the integral to get the original equation?

2. Dec 25, 2012

### symbolipoint

Rate multiplied by Time give the amount of oil leaked, but the rate is not constant. As you guessed, you integrate because the rate changes with time. NOT constant.

Another explanation: What if you just plug in the number?
If you just plug in, what... 60? Then this give you r(60)=100e^(-0.01*60)=54.88 liters per minute. But from time 0 to time 60, the rate was NOT 54.88, so you would be calculating the liters for 60 minutes at the mostly wrong rate of 54.88 liters per minute. This is a very large underestimation. The actual rate is different at every point in time from 0 to 60, and this rate VARIES with time. So,... integrate.

Last edited: Dec 25, 2012
3. Dec 26, 2012

### Ray Vickson

Why did you integrate? That was the correct thing to do, but you claim you did it without knowing why. Did somebody tell you to do it, or what?

4. Dec 30, 2012

### Xetman

Nah lol the question was in the U-Substitution section of my book so I figured it wouldn't be that easy just to plug in a number.

5. Dec 30, 2012

### Curious3141

You've kind of answered your own question here. Yes, the expression you're given is a rate (an instantaneous volume per unit time), so you need to integrate with respect to time to get the total volume that leaks out over a time interval. The definite integral just sets the bounds (i.e. answers the question: what's the total volume leaking out from the start [t=0] till t = 60?).