$$$$

29  The Fundamental Theorem

This chapter brings together the two calculus tools of differentiation and integration, proving the fundamental theorem of calculus. We then use this theorem to derive familiar techniques for computing integrals.

The fundamental theorem of calculus is a beautiful result for many different reasons. One of course, is that it forges a deep connection between the theory of areas and the theory of derivatives - something missed by the ancients and left undiscovered until the modern advent of the calculus. But second, it shows how incredibly constraining our simple axioms are: we will not prove the fundamental theorem of calculus for any particular definition of the integral (Riemann’s, Lebesgue’s, Darboux’s, etc) but rather showed that if continuous functions are integrable then your theory of integration has no choice whatsoever on how to integrate them!

Theorem 29.1 (The Fundamental Theorem of Calculus) Let $f$ be a continuous function and assume that $f$ is integrable on $[a,b]$. Denote its area function by $$F(x)={\int }_{[a,x]}f$$ Then $F$ is differentiable, and for all points $x\in (a,b)$, $$F^{\prime }=f$$

Proof. Because $f$ is continuous on a closed interval, it is bounded (by the Extreme Value theorem), and so the area function $F$ is continuous (Theorem 25.1).

Choose an arbitrary $c\in (a,b)$. We wish to show that $F^{\prime }(c)=f(c)$: that is, we need $$\underset{x\to c}{lim}\frac{F(x)-F(c)}{x-c}=f(c)$$ In terms of $\epsilon$s and $\delta$s, this means for arbitrary $\epsilon$ we need to find a $\delta$ such that if $x$ is within $\delta$ of $c$, this difference quotient is within $\epsilon$ of $f(c)$.

It will be convenient to separate this argument into two cases, depending on if $x<c$ or $c<x$ (both arguments are analogous, all that changes is whether the interval in question is $[c,x]$ or $[x,c]$). Below we proceed under the assumption that $c<x$. In this case, looking at the numerator, we see by subidvision (Axiom III) that

$$\begin{array}{rl}F(x)& ={\int }_{[a,x]}f\\ & ={\int }_{[a,c]}f+{\int }_{[c,x]}f\\ & =F(c)+{\int }_{[c,x]}f\end{array}$$ $$⟹F(x)-F(c)={\int }_{[c,x]}f$$

Thus the real quantity of interest is this integral over $[c,x]$. Choose $\epsilon >0$. Since $f$ is continuous, there is some $\delta >0$ where $|x-c|<\delta$ implies $|f(x)-f(c)|<\epsilon$. Equivalently, for all $x\in [c-\delta ,c+\delta ]$ we have $$f(c)-\epsilon <f(x)<f(c)+\epsilon$$

By subdivison (Axiom III), we know that $f$ is integrable on $[c,x]$, and so by comparison (Axiom II) and the area of rectangles (Axiom I) we have $$(f(c)-\epsilon )(x-c)\le {\int }_{[c,x]}f\le (f(c)+\epsilon )(x-c)$$

Dividing through by $x-c$

$$f(c)-\epsilon \le \frac{{\int }_{[c,x]}f}{x-c}\le f(c)+\epsilon$$

and subtracting $f(c)$ $$-\epsilon \le \frac{{\int }_{[c,x]}f}{x-c}-f(c)\le \epsilon$$

We arrive at the inequality

$$\left|\frac{{\int }_{[c,x]}f}{x-c}\right|<\epsilon$$

But the numerator here is none other than $F(x)-F(c)$! So, we’ve done it: for all $x>c$ with $|x-c|<\delta$, we have the difference quotient within $\epsilon$ of $f(c)$. This implies the limit exists, and that $$F^{\prime }(c)=f(c)$$

Exercise 29.1 Write out the case for $x<c$ following the same logic as above.

This tells us that the area function of $f$ is one of its antiderivatives! The theory of area is the inverse of the theory of rates of change. But which antiderivative? The mean value theorem assures us that the collection of all possible antiderivatives are easy to understand - any two differ by a constant (Corollary 22.2). So to uniquely specify an antiderivative its enough to give its value at one point. And we can do this! But we first need a small lemma.

One technical condition that will be useful to us later is to think about what happens when the interval is of zero size: intuitively the ‘net area’ over a point should be zero, but can we prove that from the axioms? Indeed we can!

Proposition 29.1 (Integrating over a Degenerate Interval) If $\{c\}$ is the degenerate closed interval containing a single point, and $f$ is a function which is integrable on any interval containing $a$, then $${\int }_{\{a\}}f=0$$

Proof. Let $f$ be integrable on the interval $[u,v]$ and $a\in [u,v]$ be a point. Without loss of generality we can in fact take $a$ to be one of the endpoints of the interval, by subdivision: if $a\in (u,v)$ then Axiom III implies that $f$ is integrable on $[u,a]$ and on $[a,v]$ as well.

Thus, we assume $f$ is integrable on $[a,v]$, and further subdivide this interval as $$[a,v]=[a,a]\cup [a,v]=\{a\}\cup [a,v]$$

By subdivision, we see that $f$ is integrable on $\{a\}$ and that

$${\int }_{[a,v]}f={\int }_{\{a\}}f+{\int }_{[a,v]}f$$

Subtracting the common integral over $[a,v]$ from both sides yields the result,

$${\int }_{\{a\}}f=0$$

Now that we know integration over a point is zero, we know $F(a)={\int }_{[a,a]}f=0$, which determines the precise antiderivative that appears.

Corollary 29.1 Let $f$ be a continuous function which is integrable on $[a,b]$. Then the function $F(x)={\int }_{[a,x]}f$ is uniquely determined as the antiderivative of $F$ such that $F(a)=0$.

This connection of integration with antidifferentiation and the classification of antiderivatives has a useful corollary for computation, which is often called the second fundamental theorem

Theorem 29.2 (FTC Part II) Let $f$ be continuous and integrable on $[a,b]$ and let $F$ be any antiderivative of $f$. Then $${\int }_{[a,b]}f=F(b)-F(a)$$

Proof. Denote the area function for $f$ as $A(x)={\int }_{[a,x]}f$. Then the quantity we want to compute is $A(b)$.

Now, let $F$ be any antiderivative of $f$. The first part of the fundamental theorem assures us that $A$ is an antiderivative of $f$, and so Corollary 22.2 implies there is some constant $C$ such that $A(x)-F(x)=C$, or $F(x)=A(x)+C$. Now computing,

$$\begin{array}{rl}F(b)-F(a)& =(A(b)+C)-(A(a)+C)\\ & =A(b)-A(a)+(C-C)\\ & =A(b)-A(a)\\ & =A(b)\end{array}$$

Where the last equality comes from the fact that $A(a)={\int }_{\{a\}}f=0$.

We are going to have a lot of endpoint-subtraction going on, so its nice to have a notation for it.

Definition 29.1 Let $[a,b]$ be an interval and $f$ a function. We write $$f{|}_{[a,b]}=f(b)-f(a)$$ as a shorthand for evaluation at the endpoints.

Remark 29.1. It is often convenient when doing calculations to introduce a slight generalization of the integral, which depends on an oriented interval. A natural notation for this is already in use in calculus, using the top and bottom of the integral sign for the locations of the ‘ending’ and ‘starting’ bound respectively:

$${\int }_a^{b}f=\{\begin{array}{ll}{\int }_{[a,b]}f& a\le b\\ -{\int }_{[b,a]}f& a\ge b\end{array}$$

Show that using this notation, we have a clean generalized subdivision rule: for **all points $a,b,c$ irrespective of their orderings, $${\int }_a^{b}f={\int }_a^{c}f+{\int }_c^{b}f$$

This notation helps shorten the computations in the proof of the fundamental theorem (at the expense of adding one new thing to remember).

29.1 Linearity

We already have general proofs that the integral is linear, over any axiomatically integrable functions (in the chapter Properties, of the previous part). But now with the fundamental theorem in hand we can provide much simpler proofs, at least when restricted to continuous functions. We give these arguments here.

Theorem 29.3 Let $f$ be a continuous function $[a,b]$, and $k\in \mathbb{R}$. Then $${\int }_{[a,b]}kf=k{\int }_{[a,b]}f$$

Proof. Since $f$ is continuous on $[a,b]$ it is integrable. Set $F(x)={\int }_{[a,x]}f$; and note by the fundamental theorem $F$ we have $F^{\prime }=f$. Since $kf$ is a constant multiple of a continuous function it is also continuous, and hence integrable. Using the fundamental theorem, we can compute its integral by finding an antiderivative. But this is easy! Since the derivative is linear, $${(kF(x))}^{\prime }=k{(F(x))}^{\prime }=kf(x)$$ So $kF$ is an antideriavtive of $kf$. Thus we can use it to evaluate our integral, $${\int }_{[a,b]}kf=kF(b)-kF(a)$$ Factoring out the $k$, this is $k(F(b)-F(a))$, and (by our definition of $F$!) $F(b)-F(a)={\int }_{[a,b]}f$. Putting this all togehter gives the claimed identity, $${\int }_{[a,b]}kf=k{\int }_{[a,b]}f$$

Theorem 29.4 Let $f,g$ be continuous on $[a,b]$. Then $${\int }_{[a,b]}(f+g)={\int }_{[a,b]}f+{\int }_{[a,b]}g$$

Proof. Since $f,g$ are both continuous so is $f+g$, and hence all three are integrable. Let $F,G$ be antiderivatives of $f,g$ respectively, and note by the linearity of the derivative that $$(F+G{)}^{\prime }=F^{\prime }+G^{\prime }=f+g$$ Thus $F+G$ is an antiderivative of $f+g$, so we can use it to evaluate the integral: $${\int }_{[a,b]}(f+g)=(F+G){|}_{[a,b]}=(F(b)+G(b))-(F(a)+G(a))$$ Regrouping the right hand side as $(F(b)-F(a))+(G(b)-G(a))$ we recognize each as the result of applying the fundamental theorem to calculate the integrals of $f,g$ respectively. Thus $${\int }_{[a,b]}(f+g)={\int }_{[a,b]}f+{\int }_{[a,b]}g$$

29.2 Integration Techniques

We can use the fundamental theorem to justify the main integration techniques learned in calculus courses - substitution and integration by parts - as simply antiderivatives of the chain rule and product rule.
:::{#thm-integral-substitution} ## Substitution Let $g$ be a continuously differentiable function on $[a,b]$ and $f$ be continuous on the range of $g$, with $F$ an antiderivative of $f$. Then $${\int }_{[a,b]}f(g(x))g^{\prime }(x)=F\circ g{|}_{[a,b]}$$ ::: :::{.proof} Note $g$ is continuous as it is differentiable. As compositions and products of continuous functions are continuous, $f(g(x))g^{\prime }(x)$ is a continuous function, and hence integrable. Thus by the fundamental theorem of calculus we can evaluate its integral by finding an antiderivative. The chain rule readily confirms $F\circ g$ is such a function as $$(F(g(x)){)}^{\prime }=F^{\prime }(g(x))g^{\prime }(x)=f(g(x))g^{\prime }(x)$$ Thus ${\int }_{[a,b]}f(g(x))g^{\prime }(x)=F(g(b))-F(g(a))$. :::

This justifies the familiar use of u-substitution in Calculus

Example 29.1 To integrate $(2x+1{)}^5$ on the interval $[a,b]$, note that we may write $$(2x+1{)}^5=f(g(x))$$ for $f(x)=x^5$ and $g(x)=2x+1$. Then $2(2x+1{)}^5$ is the derivative of $\frac{1}{6}(2x+1{)}^6$, so $${\int }_{[a,b]}2(2x+1{)}^5=\frac{1}{6}(2x+1{)}^6{|}_{[a,b]}$$ By the linearity of the integral ${\int }_{[a,b]}2(2x+1{)}^5=2{\int }_{[a,b]}(2x+1{)}^5$ and solving for this yields $${\int }_{[a,b]}(2x+1{)}^5=\frac{(2x+1{)}^6}{12}{|}_{[a,b]}=\frac{(2b+1{)}^6-(2a+1{)}^5}{12}$$

Theorem 29.5 (Integration by Parts) Let $f$ be continuous and $g$ continuously differentiable on $[a,b]$. Then $${\int }_{[a,b]}f(x)g(x)=F(x)g(x){|}_{[a,b]}-{\int }_{[a,b]}F(x)g^{\prime }(x)$$ where $F$ is an antiderivative of $f$.

Proof. Since $f$ is continuous we know it is integrable, so let $F(x)={\int }_{[a,x]}f$. Then $F$ is differentiable (by the fundamental theorem) and so is $g$ (by assumption), so the product $F(x)g(x)$ is a differentiable function. Taking the derivative with the product rule yields $$(F(x)g(x){)}^{\prime }=F^{\prime }(x)g(x)+F(x)g^{\prime }(x)=f(x)g(x)+F(x)g^{\prime }(x)$$ where in the last equality we used that $F^{\prime }=f$ from the fundamental theorem. Thus $F(x)g(x)$ is an antiderivative of the sum on the right hand side, so integrating gives $${\int }_{[a,b]}(f(x)g(x)+F(x)g^{\prime }(x))=F(x)g(x){|}_{[a,b]}$$ Distributing the integral over addition

Corollary 29.2 (Iterated Integration by Parts) Applying twice,

$${\int }_{[a,b]}fg=Fg{|}_{[a,b]}-{\int }_{[a,b]}F{g}^{\prime }$$ $$=Fg{|}_{[a,b]}-(\mathcal{F}{g}^{\prime }{|}_{[a,b]}-{\int }_{[a,b]}\mathcal{F}{g}^{\prime \prime })$$ $$=F_g-\mathcal{F}{g}^{\prime }+{\int }_{[a,b]}\mathcal{F}{g}^{\prime \prime }$$ Where $F$ is an antiderivative of $f$, and $\mathcal{F}$ is an antiderivative of $F$. Continuing in this fashion, we can replace our integral with one containing $n$ derivatives of $g$, at the cost of having to take $n$ antiderivatives of $f$.