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The Fundamental Theorem of Calculus

Module by: Lawrence Baggett. E-mail the author

Summary: This module covers two important theorems, including the fundamental theorem of calculus.

We begin this section with a result that is certainly not a surprise, but we will need it at various places in later proofs, so it's good to state it precisely now.

Theorem 1

Suppose fI([a,b]),fI([a,b]), and suppose a<c<b.a<c<b. Then fI([a,c]),fI([a,c]),fI([c,b]),fI([c,b]), and

a b f = a c f + c b f . a b f = a c f + c b f .
(1)

Proof

Suppose first that hh is a step function on [a,b],[a,b], and let P={x0<x1<...<xn}P={x0<x1<...<xn} be a partition of [a,b][a,b] such that h(x)=aih(x)=ai on the subinterval (xi-1,xi)(xi-1,xi) of P.P. Of course, we may assume without loss of generality that cc is one of the points of P,P, say c=xk.c=xk. Clearly hh is a step function on both intervals [a,c][a,c] and [c,b].[c,b].

Now, let Q1={a=x0<x1<...<c=xk}Q1={a=x0<x1<...<c=xk} be the partition of [a,c][a,c] obtained by intersecting PP with [a,c],[a,c], and let Q2={c=xk<xk+1<...<xn=b}Q2={c=xk<xk+1<...<xn=b} be the partition of [c,b][c,b] obtained by intersecting PP with [c,b].[c,b]. We have that

a b h = S P ( h ) = i = 1 n a i ( x i - x i - 1 ) = i = 1 k a i ( x i - x i - 1 ) + i = k + 1 n a i ( x i - x i - 1 ) = S Q 1 ( h ) + S Q 2 ( h ) = a c h + c b h , a b h = S P ( h ) = i = 1 n a i ( x i - x i - 1 ) = i = 1 k a i ( x i - x i - 1 ) + i = k + 1 n a i ( x i - x i - 1 ) = S Q 1 ( h ) + S Q 2 ( h ) = a c h + c b h ,
(2)

which proves the theorem for step functions.

Now, write f=limhn,f=limhn, where each hnhn is a step function on [a,b].[a,b]. Then clearly f=limhnf=limhn on [a,c],[a,c], which shows that fI([a,c]),fI([a,c]), and

a c f = lim a c h n . a c f = lim a c h n .
(3)

Similarly, f=limhnf=limhn on [c,b],[c,b], showing that fI([c,b]),fI([c,b]), and

c b f = lim c b h n . c b f = lim c b h n .
(4)

Finally,

a b f = lim a b h n = lim ( a c h n + c b h n ) = lim a c h n + lim c b h n = a c f + c b f , a b f = lim a b h n = lim ( a c h n + c b h n ) = lim a c h n + lim c b h n = a c f + c b f ,
(5)

as desired.

I's time for the trumpets again! What we call the Fundamental Theorem of Calculus was discovered by Newton and Leibniz more or less simultaneously in the seventeenth century, and it is without doubt the cornerstone of all we call mathematical analysis today. Perhaps the main theoretical consequence of this theorem is that it provides a procedure for inventing “new” functions. Polynomials are rather natural functions, power series are a simple generalization of polynomials, and then what? It all came down to thinking of a function of a variable xx as being the area beneath a curve between a fixed point aa and the varying point x.x. By now, we have polished and massaged these ideas into a careful, detailed development of the subject, which has substantially obscured the original ingenious insights of Newton and Leibniz. On the other hand, our development and proofs are complete, while theirs were based heavily on their intuition. So, here it is.

Theorem 2: Fundamental Theorem of Calculus

Suppose ff is an arbitrary element of I([a,b]).I([a,b]). Define a function FF on [a,b][a,b] by F(x)=axf.F(x)=axf. Then:

  1.  FF is continuous on [a,b],[a,b], and F(a)=0.F(a)=0.
  2. If ff is continuous at a point c(a,b),c(a,b), then FF is differentiable at cc and F'(c)=f(c).F'(c)=f(c).
  3. Suppose that ff is continuous on [a,b].[a,b]. If GG is any continuous function on [a,b][a,b] that is differentiable on (a,b)(a,b) and satisfies G'(x)=f(x)G'(x)=f(x) for all x(a,b),x(a,b), then
    abf(t)dt=G(b)-G(a).abf(t)dt=G(b)-G(a).
    (6)

REMARK Part (2) of this theorem is the heart of it, the great discovery of Newton and Leibniz, although most beginning calculus students often think of part (3) as the main statement. Of course it is that third part that enables us to actually compute integrals.

Proof

Because fI([a,b]),fI([a,b]), we know that fI([a,x])fI([a,x]) for every x[a,b],x[a,b], so that F(x)F(x) at least is defined.

Also, we know that ff is bounded; i.e., there exists an MM such that |f(t)|M|f(t)|M for all t[a,b].t[a,b]. Then, if x,y[a,b]x,y[a,b] with xy,xy, we have that

| F ( x ) - F ( y ) | = | a x f - a y f | = | a y f + y x f - a y f | = | y x f | y x | f | y x M = M ( x - y ) , | F ( x ) - F ( y ) | = | a x f - a y f | = | a y f + y x f - a y f | = | y x f | y x | f | y x M = M ( x - y ) ,
(7)

so that |F(x)-F(y)|M|x-y|<ϵ|F(x)-F(y)|M|x-y|<ϵ if |x-y|<δ=ϵ/M.|x-y|<δ=ϵ/M. This shows that FF is (uniformly) continuous on [a,b].[a,b]. Obviously, F(a)=aaf=0,F(a)=aaf=0, and part (1) is proved.

Next, suppose that ff is continuous at c(a,b),c(a,b), and write L=f(c).L=f(c). Let ϵ>0ϵ>0 be given. To show that FF is differentiable at cc and that F'(c)=f(c),F'(c)=f(c), we must find a δ>0δ>0 such that if 0<|h|<δ0<|h|<δ then

| F ( c + h ) - F ( c ) h - L | < ϵ . | F ( c + h ) - F ( c ) h - L | < ϵ .
(8)

Since ff is continuous at c,c, choose δ>0δ>0 so that |f(t)-f(c)|<ϵ|f(t)-f(c)|<ϵ if |t-c|<δ.|t-c|<δ. Now, assuming that h>0h>0 for the moment, we have that

F ( c + h ) - F ( c ) = a c + h f - a c f = a c f + c c + h f - a c f = c c + h f , F ( c + h ) - F ( c ) = a c + h f - a c f = a c f + c c + h f - a c f = c c + h f ,
(9)

and

L = c c + h L h . L = c c + h L h .
(10)

So, if 0<h<δ,0<h<δ, then

| F ( c + h ) - F ( c ) h - L | = | c c + h f ( t ) d t h - c c + h L h | = | c c + h ( f ( t ) - L ) d t h | c c + h | f ( t ) - L | d t h = c c + h | f ( t ) - f ( c ) | d t h c c + h ϵ h = ϵ , | F ( c + h ) - F ( c ) h - L | = | c c + h f ( t ) d t h - c c + h L h | = | c c + h ( f ( t ) - L ) d t h | c c + h | f ( t ) - L | d t h = c c + h | f ( t ) - f ( c ) | d t h c c + h ϵ h = ϵ ,
(11)

where the last inequality follows because for t[c,c+h],t[c,c+h], we have that |t-c|h<δ.|t-c|h<δ. A similar argument holds if h<0.h<0. (See the following exercise.) This proves part (2).

Suppose finally that GG is continuous on [a,b],[a,b], differentiable on (a,b),(a,b), and that G'(x)=f(x)G'(x)=f(x) for all x(a,b).x(a,b). Then, F-GF-G is continuous on [a,b],[a,b], differentiable on (a,b),(a,b), and by part (2) (F-G)'(x)=F'(x)-G'(x)=f(x)-f(x)=0(F-G)'(x)=F'(x)-G'(x)=f(x)-f(x)=0 for all x(a,b).x(a,b). It then follows from (Reference) that F-GF-G is a constant function C,C, whence,

G ( b ) - G ( a ) = F ( b ) + C - F ( a ) - C = F ( b ) = a b f ( t ) d t , G ( b ) - G ( a ) = F ( b ) + C - F ( a ) - C = F ( b ) = a b f ( t ) d t ,
(12)

and the theorem is proved.

Exercise 1

  1. Complete the proof of part (2) of the preceding theorem; i.e., take care of the case when h<0.h<0. HINT: In this case, a<c+h<c.a<c+h<c. Then, write acf=ac+hf+c+hcf.acf=ac+hf+c+hcf.
  2. Suppose ff is a continuous function on the closed interval [a,b],[a,b], and that f'f' exists and is continuous on the open interval (a,b).(a,b). Assume further that f'f' is integrable on the closed interval [a,b].[a,b]. Prove that f(x)-f(a)=axf'f(x)-f(a)=axf' for all x[a,b].x[a,b]. Be careful to understand how this is different from the Fundamental Theorem.
  3. Use the Fundamental Theorem to prove that for x1x1 we have
    ln(x)=F(x)1x1tdt,ln(x)=F(x)1x1tdt,
    (13)
    and for 0<x<10<x<1 we have
    ln(x)=F(x)-x11tdt.ln(x)=F(x)-x11tdt.
    (14)
    HINT: Show that these two functions have the same derivative and agree at x=1.x=1.

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