Calculus, Presented at about a Third Grade Level

Integration is really just a really high-class way of ADDING! (That's all, you're done!)

It will help to use some examples. Say that we hear a buzzer sound at exactly 2 pm and then again at exactly 3 pm. And we want to know how much TIME passed between those two events. (Yes, this is a REALLY simple example, but there is a reason!)

You already KNOW how to solve that one, I know! But in Calculus, we could describe this problem as an Integral, and write it as:

The tall wiggly line is the Integral symbol, and the numbers at the bottom and top tell us what numbers to start and to end our calculation. The dt after the Integral sign indicates a " Δtime" which means ANY time interval that you want to use to count by! In this case, you might decide to count by MINUTES, or by SECONDS. But there are a lot of more complicated problems where it is more accurate to count by MILLIONTHS OF A SECOND or BILLIONTHS OF A SECOND! For this simple problem, we would get the exact same answer in any of these cases. We might count off 60 minutes, and then call it one hour! OR, we might count off 3,600 seconds and still call it one hour! OR, we might count off 3,600,000,000,000 billionths of a second, and STILL call it one hour! The answer is the same, for whatever SIZE OF INTERVAL we use. (Again, for a lot of problems in Calculus, there are advantages to using really, really small intervals, and that is actually why Integration is better than normal adding!

In our case, time always passes at exactly the same rate, so we always get the same answer, no matter how big or small we pick our interval.

We now look up in a Table of Integrals to see what the solution is for our equation. For our simple dt, that Table shows that the answer is t, as shown here:

We show a vertical line, with the bottom and top numbers near it again. This indicates that this solution is to be calculated for the top value and then calculated again for the bottom value, with the second subtracted from the first, as:

We can see that this gives us the one hour answer that we already were sure of!

(We were able to move the v outside the Integral process because it was a CONSTANT for this problem.) So we can see that if you walked at a constant 5 miles per hour, for our one hour interval, you would walk a total of 5 miles.

OK. Now, we have already covered ALL of what Integration is, so no more LEARNING right now! But we want to show WHY this is actually a good way to approach many problems. Say that we will NOT have you walking at a constant speed, but that you start out with virtually no speed at all, and every second, you speed up. At the end of the hour, you may be moving really fast!

Now, trying to ADD UP the total distance you will go can still work, but it is sort of a pain, because you need to figure out how far you went in the first minute (or first second) and then the DIFFERENT distance you went in the second, and so on. By ADDING, you might be able to get a reasonable estimate, but Integration gives us an EXACT distance, and actually a lot easier and quicker!

We are replacing v here with the quantity k*t, meaning some constant RATE OF CHANGE of velocity times the amount of time that has passed. We actually have a special word for this, acceleration. You are accelerating (smoothly) from zero velocity up to some final velocity.

So we are dealing with a situation here where your velocity is CONSTANTLY CHANGING, which would make the addition method really time consuming to do!

Note that we had to look up in the Table of Integrals again, this time for t*dt and we found t²/2 as the solution. We chose to count the starting time as being ZERO, to simplify the calculations. We have a number k in there, which we would call an acceleration rate. In this case, it might be 0.1 mile per hour per minute. In that case, we would get an answer of 1/2 * 0.1/hr/min * 60 minutes * 1 hour (units are selected to make the units cancel out properly) or exactly 3.0 miles! Without Calculus, it would be hard to calculate that accurately!

We can use these things we have found here in some more ways. From this last example, we can see that if an acceleration is constant then the velocity is just that constant times the amount of time that has passed. Or v = k * t. In our lives, we are always subject to gravity, and that happens to be a constant acceleration, commonly called g. So a falling object would have a velocity of g * t.

We know that g is about 32 feet/sec² or 9.8 meters/sec². So, one second after we dropped something, if it has not hit anything yet, it should be falling at 32 * 1 or 32 ft/second velocity. After another second, it will be falling at 32 * 2 or 64 ft/sec. After 10 seconds, it should be falling at at 32 * 10 or 320 ft/sec.

Using our other Integral examples above, we can Integrate this equation (in exactly the same way we already did) and then also know the distance the object falls after any amount of time.

After one second, it would have fallen 32/2 *1² or 16 feet. After two seconds, it would have fallen 32/2 * 2² or 64 feet. After ten seconds, it should have fallen 32/2 * 10² or 1600 feet.

(It turns out that there is another effect which influences what is going on here, the fact that the Earth has an atmosphere! As the object gets falling faster and faster, there is more air resistance which exists, which acts to fight the acceleration due to gravity. In fact, there is some specific downward velocity which causes an amount of air resistance force which exactly balances out the acceleration due to gravity, and the object then stops accelerating downward [still falling though!]. This is usually called a Terminal Velocity for that object in that fluid [air] under those conditions.)

GUESS WHAT? It turns out that a very simple application of Calculus can calculate the air resistance effect! We have a different web-page which presents that solution, at Free-Fall and Terminal Velocity

Without Calculus, it would be extremely difficult to calculate these things! Actually, Newton invented Calculus (which he called Fluxions) just so he could solve problems like this. It turns out that Calculus has thousands of other and more complicated uses! But every one of them is really just a ritzy way of adding!

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C Johnson, Theoretical Physicist, Physics Degree from Univ of Chicago