On-line Math 21

5.2  The Fundamental Theorem of Calculus

For a function f , defined on [a,b] , define G(x) to be the definite integral of f from a to x . So, now the upper ``limit of integration'', the upper value of the variable that forms the end of the region, becomes the variable for a function.

Note how I kept away from naming the variable of f . That's because of our repeated use of x as the variable all the time. Now, we can't write

G(x): = ó õ x a  f(x)dx,

even though that's what we're thinking, because each letter is supposed to only stand for one thing. Is x the variable for f , or is it the upper limit of integration? Now, a rose is a rose and all that, it doesn't matter what we call the variable of f . But we can't use x , since it's already spoken for as the variable of G , so we write t as the variable for f , and we have:

G(x): = ó õ x a  f(t)dt.

5.2.1  Preliminary result, another MVT

In order to prove the FTC, we need a slightly different version of the MVT (Mean Value Theorem) that applies to integrals.

Proposition 1 [MVT, integral form] If f(x) is continuous on [a,b] , then there is a c Î [a,b] so that

f(c) = ó õ b a  f(x)dx b-a .

Remark 2 If you are used to thinking about the definite integral as an antiderivative, then this is really just the old MVT applied to the antiderivative of f , but we're not there, just yet. This is a slightly different statement from the regular MVT, but is even easier to prove. We are assuming that any continuous function is integrable, which we proved earlier.

Proof:

5.2.2  The Theorem

The way most texts describe it these days, there are two ``Fundamental Theorems'' of calculus. The first one will show that the general function G(x) defined as

G(x): = ó õ x a  f(t)dt

has derivative G¢(x) = f(x) . That is really the most serious part of this theorem. What it shows is not so much that f has an antiderivative, although that is theoretically useful, but it shows that the definite integral is related to that antiderivative.

The second part of the Fundamental Theorem uses that information to show the ``easy way'' to compute a definite integral, by evaluating an antiderivative of the function (which you find by integration techniques).

Theorem 1, [FTC, part I] Let f be a continuous function on an open interval containing [a,b] . Set

G(x): = ó õ x a  f(t)dt.

Then, G(x) is differentiable on [a,b] and its derivative is f , that is, G¢(x) = f(x).

Proof:

Example 1 Find an antiderivative of

f(x) = Ö   2+sin2x   .

Real Stupid Solution

Example 2 Find the derivative of

F(x) = ó õ x2 0  Ö   1+t6   dt.

Solution

Exercise 1 Find

d dx æ è ó õ x2 x  sin(t2)dt ö ø =

Theorem 2, [FTC, part II] Let f be a continuous function on an open interval containing [a,b] . Let F(x) be any antiderivative of f . Then:

ó õ b a  f(x)dx = F(b)-F(a): = F(x)| ba.

The notation on the last line,

F(x)| ba,

is defined by

F(b)-F(a): = F(x)| ba.

It just makes it easier when you are working out an integration problem, to have the chance to write down the indefinite integral F before you evaluate it at the endpoints.

Proof:

Example 3

ó õ p/2 0  cos(x)dx.

Solution

Exercise 2

ó õ 3 1  1 x3 dx =

Example 4

ó õ 2 -1  |x|dx.

Solution

Example 5

ó õ 5 2  f(x)dx.

where

f(x) = ì í î 3x-4, if x ³ 4 x+2, if x < 4

Solution

Example 6

ó õ 2 0  x Ö   4-x2   dx

Solution

Example 7

ó õ 2 0  Ö   4-x2   dx

Solution

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