# Relationship between Derivatives and Integrals

1. Apr 3, 2013

### BraneChild

Hi,
I've recently taken a Calculus 1 (Differential Calculus) course and I've been looking ahead to see what sort of material is covered in the Calculus 2 (Integral Calculus) course. I am wondering about the relationship between derivatives and integrals.

From what I understand, an integral is the area under a curve that is bounded on either side by random points and below by the x-axis. Is this right? If so, how is it geometricallyrelated to the derivative, which is the slope of the tangent at a particular point along a curve?

I had this problem a while back with limits. Although I knew how to compute limits, I didn't know until much later on exactly what a limit was. Same with integrals. I feel like if I can't paint a picture of what I'm doing I don't really understand it.

2. Apr 3, 2013

### Vorde

Well, you're learn all this in your next course. But the key here (to be brief) is the difference between definite integrals and indefinite integrals.

Indefinite integrals are obviously related to derivatives because its just the opposite of differentiation (no magic involved here, that's just how they're defined).

The magic part is the fact that the definite integral, which is defined as the area under a curve (or something equivalent to that), is related to the indefinite integral in an astonishingly simple and remarkable way. The way I like to look at it is the following:

Let's say I have a function which gives the position of an object. You know that if I take the derivative of this position function I get a function which gives me the velocity of this object.

Well if you know any physics you know that Velocity = Distance Elapsed/Time Elapsed, and the x-axis represents time. Imagine the velocity function is a constant so that the space under the curve is a rectangle (see note at bottom), multiplying the height (value) of the curve by the length of time beneath it gives you the area under the curve, but it also gives you the distance traveled because V=D/T so V*T=D.

So we know that the area under this velocity curve, whatever it is, equals the distance traveled. But there is another way of finding out that distance; the obvious way! You take the distance (value) on the right side of the position function and subtract it from the value on the left side, giving you the distance elapsed. Remember that the velocity function is the derivative of the position function, which I'm going to instead phrase as the position function is the antiderivative (indefinite integral) of the velocity function.

So what does this tell us? It tells us that the area under a curve is equal to the value of its anti-derivative at the right side of an interval minus the value on this left side! This is one part (and the more important part) of the fundamental theorem of calculus, and is the idea behind pretty much everything you'll do in that course.

NOTE: The function doesn't have to be constant for this to work, because you can approximate any function by a series of rectangles with height equal to the function at given points. As the number of rectangles increases (and the width of each rectangle decreases) the approximation approaches the actual function and so you can do this with any function.

Last edited: Apr 3, 2013
3. Apr 4, 2013

### BraneChild

Thanks a lot!
It actually seems like a simple idea, I guess it just took me to see it from a different perspective before it clicked.
Integral Calculus here I come!