We now come to what is perhaps the culminating point of the course - the connection between integral and differential calculus!

Since integration of a continuous function on a finite interval is always defined, we can form the function

g(x) = Integral f(t) dt from t=a to t=x, a < x < b (Sorry for the lack of inline mathematics notation!)

We can ask the question - Does g(x) have a derivative? . The answer is yes. Since we do not have a formula, we must do it the hard way - find a limit.

g(x+h) = Integral f(t) dt from t=a to t=x+h, and

g(x) = Integral f(t) dt from t=a to t=x,

therefore, g(x+h)-g(x) = Integral f(t) dt from t=x to t=x+h.

This quantity is approximately f(x^*)h, for some x^* in (x,x+h). In fact we have

m*h <= Integral f(t) dt from t=x to t=x+h <= M*h

where m is the min of f on (x,x+h) and M is the max of f on (x,x+h). Dividing by h, and taking the limit as h -> 0, and noting that m->f(x) and M->f(x), we conclude that

d/dx g(x) = f(x)

This implies that differentiation and integration are inverse to one another.

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If F(x) is any other anti-derivative, then F(x) = g(x) + C. This implies that F(b)-F(c) = (g(b)+C) - (g(a)+C) = g(b) - g(a) = Integral of f(t) dt from t=a to t=b.

This is the ``second fundamental theorem of calculus'', definite integration is the same as anti-differentiation!

In order to compute the definite integral of any function f(x), it is sufficient to know an anti-derivative, F(X), and compute F(b)-F(a).

Last modified Wed Nov 6 23:04:49 CST 1996