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MATH 1121.81 Calculus I

Chapter 4:  Antiderivatives, Integrals, and the Fundamental Theorem of Calculus

Part I:  Antiderivatives:

Given a particular derivative function, can you tell where it came from?

The answer is, "up to a point," or "yes, if you have some extra information."

Think about the following:

Suppose you know that f ' (x) = cos(x).  What function was f(x)?

How about f(x) = sin(x)?  Sure, since Dx[sin(x)] = cos(x).

But, what about:  f(x) = sin(x) + 2?  Dx[sin(x) + 2] = cos(x), as well.

In fact, since derivatives are linear operators (you can break them up at plus signs, in particular), and the derivative of a constant is 0, you can see that:

Dx[sin(x) + C] = cos(x), for any constant, C.

This also makes sense, when you think of what adding a constant does to a function -- merely displacing it vertically, and otherwise leaving it unaffected.

Because of this, tangent lines for a particular value of x will all be parallel to each other:

Applet

The next question logically becomes, did we find all the functions with derivative = cos(x)?

That is, we know that for all constant values, C, Dx[sin(x)+C] = cos(x), might there be some other function, that is not just the sine function translated by a constant, whose derivative is the cosine function?

The answer is, NO.  (Or, the answer to the first question about finding them all, is, YES.)

Here's the theorem that proves it, and it is our first window into the Fundamental Theorems of Calculus:

Theorem:  If f(x) and g(x) are two functions, differentiable on an interval (a, b), and if f ' (x) = g ' (x) on (a, b), then g(x) = f(x) + C, for some constant, C.

Proof:

Consider the function h(x) = f(x) - g(x).  By linearity, h ' (x) = f ' (x) - g ' (x), so that, for all x on (a, b), h ' (x) = 0.  This theorem then claims that h(x) = C (is a constant function), for some C.

Now, we'll use that very convenient theorem, the Mean Value Theorem.  Suppose h(x) is NOT a constant function on (a, b).  Then, for some x1 < x2 in (a, b), either h(x1) < h(x2), or else h(x1) > h(x2).  In either case, the difference quotient,

(h(x2) - h(x1))/(x2 - x1) is not equal to zero.  But, by the Mean Value Theorem, since h(x) is differentiable throughout (a, b) (why?), there exists a c, x1 < c < x2 , with h ' (c) = (h(x2) - h(x1))/(x2 - x1) ; i.e, h ' (c)  is not equal to zero.  This contradicts our assumption.  Therefore, for all x1, x2 in (a, b), h(x1) = h(x2); i.e., h (x) is a constant function on (a, b), or,

g(x) = f(x) + C.

As a matter of convention, we usually write antiderivatives as capital letter functions (F(x), G(x), etc.)

Putting all of this together, we get the following:

If F(x) is such that F ' (x) = f(x) for all x in (a , b), then G(x) is such that G ' (x) = f(x) for all x in (a , b) if and only if G(x) = F(x) + C, for some constant C, on (a , b).

We call F(x) an antiderivative of f(x) in this case.  Stated another way, the family of antiderivatives of any particular function, f(x), must differ by constants from each other.

In practical terms, if you can find one antiderivative, then you know them all:

Ex:  Find all functions y = F(x) such that dy/dx = cos(x).

Ans:  Since one such function is F(x) = sin(x), then all such functions are the family, {F(x) | F(x) = sin(x) + C, for real C}.  (We abbreviate this as:  F(x) = sin(x) + C.)

Notation:  For reasons that will become clear in the next sections, we use an elongated "S" as the symbol for antidifferentiation.  This symbol is called an "integral" symbol, and the antiderivative is often referred to as the "indefinite integral" (indefinite both because we get a whole family of functions, and because we don't get a numerical answer, as we do with the "definite integral" we'll discuss soon.)

Read this as "The integral of cosine x, with respect to x, is sine x plus a constant."  For now, think of the "dx" as simply identifying the variable in the particular problem.

Notice that the process of indefinite integration, or antidifferentiation, unlike the operation of differentiation, returns a FAMILY of functions, rather than a lone, particular function.  If we want a particular antiderivative, we need some bit of extra information that allows us to solve for the constant, C.

The problem of finding the family of all antiderivatives for a particular function is called a "differential equation," (they actually get quite a bit more complicated than this!), and the problem of finding a particular function is often of the form called an "initial-value problem."

Examples:

This "by observation and brain-power" method just isn't going to cut it.  We'll need some techniques to find antiderivatives

Next:  Basic Antidifferentiation Rules: