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3.4 Fundamental theorem of calculus and applications

We can now connect integration and differentiation in a very elegant way. Let us denote $[b,a]:=[a,b]$ and

\int_{b}^{a}f(x)\operatorname{d}x:=-\int_{a}^{b}f(x)\operatorname{d}x

if $a<b$ and $f:[a,b]\rightarrow\mathbb{R}$ is continuous.

Theorem 3.4.1.

Let $I$ be an interval, $f:I\rightarrow\mathbb{R}$ continuous and $c\in I$ . Then the indefinite integral with base point $c$ ,

F:I\rightarrow\mathbb{R},\quad F(x)=\int_{c}^{x}f(t)\operatorname{d}t,

is differentiable and $F^{\prime}(x)=f(x)$ , for all $x\in I$ .

The theorem states that $F$ becomes “automatically” differentiable if $f$ is continuous!

Comment: As defined, the indefinite integral $F$ of $f$ is a function again, while the (definite) integral of $f$ in Theorem 1.2.12 is a number. However, if $I=[a,b]$ then, taking as base point $c=a$ , this number is exactly the value $F(b)$ .

Proof.

In order to prove differentiability at some fixed $x\in I$ , we consider the difference quotient:

\frac{F(x+h)-F(x)}{h}=\frac{\int_{c}^{x+h}f(t)\operatorname{d}t-\int_{c}^{x}f(% t)\operatorname{d}t}{h}=\frac{1}{h}\int_{x}^{x+h}f(t)\operatorname{d}t,

where $h\not=0$ is such that $[x,x+h]\subset I$ ; here the last equality follows from Proposition 1.3.4. We consider the last integral and remember that

\inf_{s\in[0,h]}f(x+s)\leq f(t)\leq\sup_{s\in[0,h]}f(x+s)\quad t\in[x,x+h].

Thus, by Proposition 1.3.1,

\inf_{s\in[0,h]}f(x+s)\leq\frac{1}{h}\int_{x}^{x+h}f(t)\operatorname{d}t\leq% \sup_{s\in[0,h]}f(x+s).

Remember that $h$ may be negative, as explained above.

We claim that

\lim_{h\rightarrow 0}\inf_{s\in[0,h]}f(x+s)=f(x)=\lim_{h\rightarrow 0}\sup_{s% \in[0,h]}f(x+s),{}

which will prove that

f(x)\leq\lim_{h\rightarrow 0}\frac{1}{h}\int_{x}^{x+h}f(t)\operatorname{d}t% \leq f(x),

thus

\lim_{h\rightarrow 0}\frac{F(x+h)-F(x)}{h}=\lim_{h\rightarrow 0}\frac{1}{h}% \int_{x}^{x+h}f(t)\operatorname{d}t=f(x).

To prove our claim $(*)$ , let us recall that $f$ is continuous at $x$ . Thus, for every $\varepsilon>0$ , there is $\delta>0$ such that if $|s|<\delta$ then $|f(x+s)-f(x)|<\varepsilon$ . Take $h$ such that $0<|h|<\delta$ and $[x,x+h]\subset I$ , which is legitimate since we are interested in the limit $h\rightarrow 0$ in $(*)$ . Then

|\inf_{s\in[0,h]}f(x+s)-f(x)|<\varepsilon,\quad|\sup_{s\in[0,h]}f(x+s)-f(x)|<\varepsilon,

which proves $(*)$ . ∎

Key idea of proof.

bound $\frac{1}{h}\int_{x}^{x+h}f(t)\operatorname{d}t$ between $\inf_{s\in[0,h]}f(x+s)$ and $\sup_{s\in[0,h]}f(x+s)$ , use continuity of $f$ , and study the limit $h\rightarrow 0$ .

Example 3.4.2.

Consider the function

f:[0,\infty)\rightarrow\mathbb{R},\quad f(x)=x.

We would like to find

F(x)=\int_{0}^{x}f(t)\operatorname{d}t=\int_{0}^{x}t\operatorname{d}t,

for fixed $x$ , the indefinite integral with base point $0$ , from scratch (without looking at any integral tables). We compute the lower and upper approximating step functions:

\ell(f,[0,x],n,w)=\sum_{i=1}^{2^{n}}\frac{(i-1)x}{2^{n}}\mathbf{1}_{[(i-1)x/2^% {n},ix/2^{n})}(w),\quad w\in[0,x]

and

u(f,[0,x],n,w)=\sum_{i=1}^{2^{n}}\frac{ix}{2^{n}}\mathbf{1}_{[(i-1)x/2^{n},ix/% 2^{n})}(w),\quad w\in[0,x].

For $x=1.5$ , we get the following picture (with full blue the upper and dashed red the lower approximating step functions):

The integral of each of these is

	$\displaystyle\ell_{n}^{(f)}=$	$\displaystyle\int_{0}^{x}\ell(f,[0,x],n,w)\operatorname{d}x=\sum_{i=1}^{2^{n}}% \frac{(i-1)x}{2^{n}}\cdot\frac{x}{2^{n}}=\frac{x^{2}}{4^{n}}\sum_{i=1}^{2^{n}}% (i-1)$
	$\displaystyle=$	$\displaystyle\frac{x^{2}}{4^{n}}\frac{2^{n}(2^{n}-1)}{2}=\frac{x^{2}}{2}\frac{% 2^{n}-1}{2^{n}}$

and

	$\displaystyle u_{n}^{(f)}=$	$\displaystyle\int_{0}^{x}u(f,[0,x],n,w)\operatorname{d}x=\sum_{i=1}^{2^{n}}% \frac{ix}{2^{n}}\cdot\frac{x}{2^{n}}=\frac{x^{2}}{4^{n}}\sum_{i=1}^{2^{n}}i$
	$\displaystyle=$	$\displaystyle\frac{x^{2}}{4^{n}}\frac{(2^{n}+1)2^{n}}{2}=\frac{x^{2}}{2}\frac{% 2^{n}+1}{2^{n}}.$

Thus $\ell_{n}^{(f)}\leq\frac{x^{2}}{2}\leq u_{n}^{(f)}$ and

\ell_{n}^{(f)},u_{n}^{(f)}\rightarrow\frac{x^{2}}{2},\quad n\rightarrow\infty,

so our indefinite integral with base point $0$ is

F(x)=\frac{x^{2}}{2},\quad x\in[0,\infty).

We know from Example 3.1.2 that $F$ is differentiable with $F^{\prime}(x)=2\frac{x}{2}=x$ , for all $x\in[0,\infty)$ , so $F^{\prime}=f$ , confirming the theorem.

Exercise W5.1.

In analogy to the preceding example, compute the indefinite integral of

f:\mathbb{R}\rightarrow\mathbb{R},\quad f(x)=x^{2},

with base point $0$ . You may use the formula

\sum_{k=1}^{n}k^{2}=\frac{n(n+1)(2n+1)}{6}.

For other functions, this procedure of calculating the indefinite integral by hand would be very inefficient, so we would like to build up more machinery first.

Theorem 3.4.3.

Let $G:I\rightarrow\mathbb{R}$ be differentiable on $I$ with continuous derivative, and let $c\in I$ . Then

G(x)-G(c)=\int_{c}^{x}G^{\prime}(t)\operatorname{d}t,\quad x\in I.

Proof.

Let $F:I\rightarrow\mathbb{R}$ be the indefinite integral for $G^{\prime}$ with base point $c$ . From Theorem 3.4.1 we know that $F$ is differentiable with $F^{\prime}(x)=G^{\prime}(x)$ , for all $x\in I$ , so $(F-G)^{\prime}(x)=0$ , for all $x\in I$ . Corollary 3.3.2(iii) then shows that there must be $\lambda\in\mathbb{R}$ such that $(F-G)(x)=\lambda$ , for all $x\in I$ . Since $F(c)=0$ , we get $\lambda=-G(c)$ , so $F(x)=G(x)-G(c)$ , which proves the theorem. ∎

Remark 3.4.4.

Theorems 3.4.1 and 3.4.3 together say that “composing integration and differentiation gives you the identity operation up to a constant”, symbolically (but not mathematically rigorously)

``\quad\frac{\operatorname{d}}{\operatorname{d}x}\int f=f=\int\frac{% \operatorname{d}}{\operatorname{d}x}f+C\quad"

if the function $f$ is sufficiently nice.

Exercise W5.3.

Elaborate Remark 3.4.4: what precise conditions does $f$ have to fulfil? How exactly are indefinite integral and derivative related? How should the formula read? Can we iterate this, i.e., integrate 10 times and then differentiate 10 times in order to get the original function back?

Corollary 3.4.5.

Given a continuous function $f:I\rightarrow\mathbb{R}$ and a differentiable function $G:I\rightarrow\mathbb{R}$ such that $G^{\prime}=f$ . Such a $G$ is called an anti-derivative of $f$ . Then the indefinite integral of $f$ with base point $c$ is given by

F(x)=G(x)-G(c),\quad x\in I.

Hence, all anti-derivatives of $f$ differ only by a constant, i.e., they are of the form $G+C$ , with some constant $C\in\mathbb{R}$ .

Proof.

Exercise W5.2 ∎

Comment: So in practice, how do we compute the indefinite integral or an anti-derivative efficiently? – Well, there is no standard “protocol” or “recipe”. You have to somehow guess $G$ and then check whether its derivative $G^{\prime}$ equals $f$ . You find the most important ones collected in tables and you will also memorise some of them over time, or you might remember tables and integration techniques like substitution etc. from MATH101/102.

Example 3.4.6.

(1)

Consider the function $f:\mathbb{R}\rightarrow\mathbb{R}$ , defined by $f(x)=x^{2}+\frac{2}{3}x^{5}$ , which is continuous and so its indefinite integral

$F:\mathbb{R}\rightarrow\mathbb{R},\quad F(x)=\int_{1}^{x}f(x)\operatorname{d}x$

with base point $1$ can be defined. We know that $G:\mathbb{R}\rightarrow\mathbb{R}$ defined by $G(x)=\frac{1}{3}x^{3}+\frac{1}{9}x^{6}$ is differentiable with $G^{\prime}(x)=x^{2}+\frac{2}{3}x^{5}=f(x)$ , so

$F(x)=G(x)-G(1)=\frac{1}{3}x^{3}+\frac{1}{9}x^{6}-\frac{4}{9}.$
(2)

Consider the function $f:\mathbb{R}\rightarrow\mathbb{R}$ , defined by $f(x)=\sin(2x)$ , which is continuous, and so its indefinite integral $F$ with base point $c$ can be defined. In order to compute it explicitly we proceed as follows. We know that the function $G:\mathbb{R}\rightarrow\mathbb{R}$ defined by $G(x)=\frac{1}{2}\cos(2x)$ is differentiable with $G^{\prime}(x)=\sin(2x)=f(x)$ . Thus $F(x)=G(x)-G(c)=\frac{1}{2}\cos(2x)-\frac{1}{2}\cos(2c)$ .