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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.
Suppose $f\in I\left(\right[a,b\left]\right),$ and suppose $a<c<b.$ Then $f\in I\left(\right[a,c\left]\right),$ $f\in I\left(\right[c,b\left]\right),$ and
Suppose first that $h$ is a step function on $[a,b],$ and let $P=\{{x}_{0}<{x}_{1}<...<{x}_{n}\}$ be a partition of $[a,b]$ such that $h\left(x\right)={a}_{i}$ on the subinterval $({x}_{i-1},{x}_{i})$ of $P.$ Of course, we may assume without loss of generality that $c$ is one of the points of $P,$ say $c={x}_{k}.$ Clearly $h$ is a step function on both intervals $[a,c]$ and $[c,b].$
Now, let ${Q}_{1}=\{a={x}_{0}<{x}_{1}<...<c={x}_{k}\}$ be the partition of $[a,c]$ obtained by intersecting $P$ with $[a,c],$ and let ${Q}_{2}=\{c={x}_{k}<{x}_{k+1}<...<{x}_{n}=b\}$ be the partition of $[c,b]$ obtained by intersecting $P$ with $[c,b].$ We have that
which proves the theorem for step functions.
Now, write $f=lim{h}_{n},$ where each ${h}_{n}$ is a step function on $[a,b].$ Then clearly $f=lim{h}_{n}$ on $[a,c],$ which shows that $f\in I\left(\right[a,c\left]\right),$ and
Similarly, $f=lim{h}_{n}$ on $[c,b],$ showing that $f\in I\left(\right[c,b\left]\right),$ and
Finally,
as desired.
I's time for the trumpets again! What we call the Fundamental Theorem of Calculuswas 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 naturalfunctions, power series are a simple generalization of polynomials, and then what? It all came down to thinking of a function of a variable $x$ as being the area beneath a curve between a fixed point $a$ and the varying point $x.$ By now, we have polished and massaged these ideas into a careful, detailed development of the subject, which has substantially obscured the originalingenious 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.
Suppose $f$ is an arbitrary element of $I\left(\right[a,b\left]\right).$ Define a function $F$ on $[a,b]$ by $F\left(x\right)={\int}_{a}^{x}f.$ Then:
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.
Because $f\in I\left(\right[a,b\left]\right),$ we know that $f\in I\left(\right[a,x\left]\right)$ for every $x\in [a,b],$ so that $F\left(x\right)$ at least is defined.
Also, we know that $f$ is bounded; i.e., there exists an $M$ such that $\left|f\right(t\left)\right|\le M$ for all $t\in [a,b].$ Then, if $x,y\in [a,b]$ with $x\ge y,$ we have that
so that $\left|F\right(x)-F(y\left)\right|\le M|x-y|<\u03f5$ if $|x-y|<\delta =\u03f5/M.$ This shows that $F$ is (uniformly) continuous on $[a,b].$ Obviously, $F\left(a\right)={\int}_{a}^{a}f=0,$ and part (1) is proved.
Next, suppose that $f$ is continuous at $c\in (a,b),$ and write $L=f\left(c\right).$ Let $\u03f5>0$ be given. To show that $F$ is differentiable at $c$ and that ${F}^{\text{'}}\left(c\right)=f\left(c\right),$ we must find a $\delta >0$ such that if $0<\left|h\right|<\delta $ then
Since $f$ is continuous at $c,$ choose $\delta >0$ so that $\left|f\right(t)-f(c\left)\right|<\u03f5$ if $|t-c|<\delta .$ Now, assuming that $h>0$ for the moment, we have that
and
So, if $0<h<\delta ,$ then
where the last inequality follows because for $t\in [c,c+h],$ we have that $|t-c|\le h<\delta .$ A similar argument holds if $h<0.$ (See the following exercise.) This proves part (2).
Suppose finally that $G$ is continuous on $[a,b],$ differentiable on $(a,b),$ and that ${G}^{\text{'}}\left(x\right)=f\left(x\right)$ for all $x\in (a,b).$ Then, $F-G$ is continuous on $[a,b],$ differentiable on $(a,b),$ and by part (2) ${(F-G)}^{\text{'}}\left(x\right)={F}^{\text{'}}\left(x\right)-{G}^{\text{'}}\left(x\right)=f\left(x\right)-f\left(x\right)=0$ for all $x\in (a,b).$ It then follows from [link] that $F-G$ is a constant function $C,$ whence,
and the theorem is proved.
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