# Accumulation: Need an Amount?

Accumulation 1: If you need an amount, look around for a rate to integrate.

While most textbooks barely mention the word, the concept of accumulation is extremely useful in a variety of situations. This is the first of several posts will be on the topic of accumulation. We will discuss what accumulation is all about, look at some typical rate problems, discuss functions defined by integrals, see how the concept can be applied to graphing and differential equations.

The Fundamental Theorem of Calculus tells us that $\displaystyle \int_{a}^{b}{{f}'\left( x \right)x}=f\left( b \right)-f\left( a \right)$

The integral of a rate of change (derivative) gives the net amount of change of the function over the interval. This equation tells us a lot more than just how to evaluate a definite integral. It tell us that if you are looking for the amount that something changes by, you need only integrate the rate at which it is changing: If you need an amount, integrate a rate.

So if water is pumped into a tank at the rate of $\sqrt{t+1}$ gallons per minute then the amount that is pumped in, in 5 minutes is $\displaystyle \int_{0}^{5}{\sqrt{t+1}\,dt}$ gallons. It’s as simple as that!

If a car is traveling with a velocity (rate) given by $v\left( t \right)={{t}^{2}}+6t+12$ miles per hour then the distance it travels (amount of miles) in three hours is $\displaystyle \int_{0}^{3}{{{t}^{2}}+6t+12}\,dt$ miles.

But it gets better. By changing the variables in the FTC equation a little you can write $f\left( x \right)=f\left( a \right)+\int_{a}^{x}{{f}'\left( t \right)dt}$

This defines the function f (x) in terms of an integral whose upper limit of integration is the independent variable x. The integral gives the amount of change from t = a to t = x. This is added to the initial amount, f (a). The amount, f (x), at any time x is the initial amount,  f (a), plus the amount of change between t = a and t = x, given by the integral.

If there was 100 gallons of water in the tank in the first example above, then after 5 hours there is $\displaystyle f(5)=100+\int_{0}^{5}{\sqrt{t+1}\,dt}$ gallons in the tank.

If in the second example above the car is 53 miles from its starting point and traveling in a straight line, then after 3 hours it is $\displaystyle s\left( 3 \right)=53+\int_{0}^{3}{{{t}^{2}}+6t+12}\,dt$ miles from where it started.

If you need an amount, integrate a rate.

This site uses Akismet to reduce spam. Learn how your comment data is processed.