# 8.2. Integrals

Unfortunately primitive functions are not uniquely determined by the presented integrable function. Due to [8.1.2] (which is in fact an implication of the mean value theorem) however any two primitives of f on an interval differ only in constant addend, which will vanish if we compute the difference of two values of f. This is a crucial feature for the integrals to be introduced now.

For the subsequent text let I be an arbitrary interval.

Definition and Proposition:  Let f be an integrable function on I, i.e. $f\in \mathcal{I}\left(I\right)$ and g be any primitive of f. For any two points $a,b\in I$ the number

 $\underset{a}{\overset{b}{\int }}f≔g{\phantom{\rule{0.1em}{12pt}}|}_{a}^{b}≔g\left(b\right)-g\left(a\right)$ [8.2.1]

does not depend on the choice of g.

It is called the (definite) integral of f from a to b. The symbol $g\phantom{\rule{0.5pt}{12pt}}{|}_{a}^{b}$ is read as "g at the boundaries a b". a and b are thus the boundaries (or lower and upper limit) of the integral [8.2.1], whereas f is called the integrand.

Proof:  If ${g}_{1}$ and ${g}_{2}$ are two primitives of f we find some c according to [8.1.2] such that ${g}_{1}={g}_{2}+c$. Thus we may compute as follows:

$\begin{array}{ll}{g}_{1}{\phantom{\rule{0.1em}{12pt}}|}_{a}^{b}\hfill & ={g}_{1}\left(b\right)-{g}_{1}\left(a\right)\hfill \\ \hfill & =\left({g}_{2}+c\right)\left(b\right)-\left({g}_{2}+c\right)\left(a\right)\hfill \\ \hfill & ={g}_{2}\left(b\right)+c-{g}_{2}\left(a\right)-c\hfill \\ \hfill & ={g}_{2}\left(b\right)-{g}_{2}\left(a\right)\hfill \\ \hfill & ={g}_{2}{\phantom{\rule{0.1em}{12pt}}|}_{a}^{b}\hfill \end{array}$

Consider:

• Continuous functions on intervals are integrable ([8.1.5]). Hence the integral $\underset{a}{\overset{b}{\int }}f$  exists for each $f\in {\mathcal{C}}^{0}\left(I\right)$.

• The term "definite" integral indicates that for its calculation an appropriate primitive function has to be evaluated at the fixed boundaries a and b. With an indefinite integral we would omit this calculation so that the symbol $\underset{}{\overset{}{\int }}f$, or sometimes $\underset{}{\int }f+c$, denotes just some primitive of f. The fundamental theorem of calculus [8.2.13] however demonstrates that this notation conveys a content meaning as well.

• To save parentheses we stipulate that the operators $\underset{a}{\overset{b}{\int }}$ and ${\phantom{\rule{0.5pt}{12pt}}|}_{a}^{b}$ should obtain a higher priority than the basic arithmetics. For example we write

$\underset{2}{\overset{5}{\int }}3{\mathrm{X}}^{2}+2\mathrm{X}={\mathrm{X}}^{3}+{\mathrm{X}}^{2}{\phantom{\rule{0.1em}{12pt}}|}_{2}^{5}$  instead of  $\underset{2}{\overset{5}{\int }}\left(3{\mathrm{X}}^{2}+2\mathrm{X}\right)=\left({\mathrm{X}}^{3}+{\mathrm{X}}^{2}\right){\phantom{\rule{0.1em}{12pt}}|}_{2}^{5}$

• The integral [8.2.1] is often denoted by $\underset{a}{\overset{b}{\int }}f\left(x\right)\phantom{\rule{0.1em}{0ex}}dx$ or $\underset{a}{\overset{b}{\int }}f\phantom{\rule{0.05em}{0ex}}d\mathrm{X}$. This notation originates in an alternative approach to integrals where area measurement is the starting point and has just a symbolic meaning. We will dwell on this only in part 8.4.

Beyond that however $f\left(x\right)\phantom{\rule{0.1em}{0ex}}dx$ and $f\phantom{\rule{0.05em}{0ex}}d\mathrm{X}$ resp. are introduced as a new objects to calculus, so called differential forms i We consider only differential forms of degree 1 in this case which are introduced as follows: Let ${\omega }_{x}:ℝ\to ℝ$ be a linear function for each $x\in I$, that means ${\omega }_{x}\left(\alpha r+\beta s\right)=\alpha {\omega }_{x}\left(r\right)+\beta {\omega }_{x}\left(s\right)$  for all $r,s,\alpha ,\beta \in ℝ$ Linear functions are often called homomorphisms and are understood as elements of $Hom\left(ℝ,ℝ\right)$. The function $\omega :I\to I×Hom\left(ℝ,ℝ\right)$ defined by $\omega \left(x\right)=\left(x,{\omega }_{x}\right)$ is now called a differential form of degree 1 on I (a 1-form on I for short). We naturally assign a 1-form $dg:x↦\left(x,{d}_{x}g\right)$ to each differentiable function $g\in {\mathcal{D}}^{1}\left(I\right)$ by setting ${d}_{x}g\left(r\right)≔{g}^{\prime }\left(x\right)\cdot r$ for $x\in I$. The function ${d}_{x}g$, obviously a homomorphism, is called the differential of g at x whereas $d$ is the differential of g. As an example we have ${d}_{x}\mathrm{X}=\mathrm{X}$ because: ${d}_{x}\mathrm{X}\left(r\right)={\mathrm{X}}^{\prime }\left(x\right)\cdot r=1\cdot r=r$. With ${\omega }_{x}$ being linear all its multiples are homomorphisms as well, thus for any function $f:I\to ℝ$ a differential form $f\cdot \omega$ is given by setting  $f\cdot \omega \left(x\right)≔\left(x,f\left(x\right)\cdot {\omega }_{x}\right)$. If f is integrable for example then $f\phantom{\rule{0.05em}{0ex}}d\mathrm{X}$ is an integrable differential form such that $f\phantom{\rule{0.05em}{0ex}}d\mathrm{X}\left(x\right)=\left(x,f\left(x\right)\cdot \mathrm{X}\right)$.
. [8.2.1] would then define the integral of the differential form $f\phantom{\rule{0.05em}{0ex}}d\mathrm{X}$ and the notation $\underset{a}{\overset{b}{\int }}f\phantom{\rule{0.05em}{0ex}}d\mathrm{X}$  now has a content meaning.

•

Example:

• $\underset{a}{\overset{b}{\int }}1=\mathrm{X}{\phantom{\rule{0.1em}{12pt}}|}_{a}^{b}=\mathrm{X}\left(b\right)-\mathrm{X}\left(a\right)=b-a$

If $a\mathrm{<}b$ the integral of 1 thus calculates the length of the interval given by the integral's bounderies!

• $\underset{-1}{\overset{2}{\int }}{\mathrm{X}}^{2}-3=\frac{1}{3}{\mathrm{X}}^{3}-3\mathrm{X}\phantom{\rule{0.1em}{12pt}}{|}_{-1}^{2}=\frac{8}{3}-6-\left(-\frac{1}{3}+3\right)=-6$

• $\underset{0}{\overset{-\pi }{\int }}\mathrm{sin}=-\mathrm{cos}{\phantom{\rule{0.1em}{12pt}}|}_{0}^{-\pi }=-\mathrm{cos}\left(-\pi \right)-\left(-\mathrm{cos}\left(0\right)\right)=1+1=2$

• $\underset{-\pi }{\overset{\pi }{\int }}\mathrm{sin}\cdot \mathrm{cos}=\frac{1}{2}{\mathrm{sin}}^{2}{\phantom{\rule{0.1em}{12pt}}|}_{-\pi }^{\pi }=0-0=0$    The integrand is of the $f\cdot {f}^{\prime }$ type, see [8.1.11].

Exercise:

• $\underset{0}{\overset{2}{\int }}3{\mathrm{X}}^{2}+4={\text{?}}{\mathrm{X}}^{3}+4\mathrm{X}{\phantom{\rule{0.1em}{12pt}}|}_{0}^{2}=16-0=16$

• $\underset{-\pi }{\overset{\pi }{\int }}\mathrm{X}-\mathrm{cos}={\text{?}}\frac{1}{2}{\mathrm{X}}^{2}-\mathrm{sin}{\phantom{\rule{0.1em}{12pt}}|}_{-\pi }^{\pi }=\frac{1}{2}{\pi }^{2}-\frac{1}{2}{\left(-\pi \right)}^{2}=0$

• $\underset{2}{\overset{1}{\int }}\frac{1}{{\mathrm{X}}^{2}}={\text{?}}-\frac{1}{\mathrm{X}}{\phantom{\rule{0.1em}{12pt}}|}_{2}^{1}=-1-\left(-\frac{1}{2}\right)=-\frac{1}{2}$ i Note that the similar integral $\underset{-2}{\overset{1}{\int }}\frac{1}{{\mathrm{X}}^{2}}$ does not exist as there is no subinterval of the domain ${ℝ}^{\ne 0}$ that contains −2 and 1.You must not integrate over gaps in the domain!

• $\underset{\frac{\pi }{6}}{\overset{\frac{5\pi }{6}}{\int }}\frac{\mathrm{cos}}{{\mathrm{sin}}^{2}}={\text{?}}-\frac{1}{\mathrm{sin}}{\phantom{\rule{0.1em}{12pt}}|}_{\frac{\pi }{6}}^{\frac{5\pi }{6}}=-2-\left(-2\right)=0$   Consider [8.1.12] as the integrand is of the $\frac{{f}^{\prime }}{{f}^{2}}$ type.

We start investigating integrals by looking at the behaviour at their boundaries. One important result is: integrals are additive at their boundaries.

Proposition:  If f is integrable on I then the following holds for all $a,b,c\in I$:

 $\underset{a}{\overset{b}{\int }}f=\underset{a}{\overset{\phantom{\phantom{\rule{0pt}{0ex}}b}c}{\int }}f+\underset{c}{\overset{b}{\int }}f$ [8.2.2]
 $\underset{a}{\overset{a}{\int }}f=0$ [8.2.3]
 $\underset{a}{\overset{b}{\int }}f=-\underset{b}{\overset{a}{\int }}f$ [8.2.4]

Proof:

1.  Let g be a primitive of f. The following identity is a straight forward calculation:

$\underset{a}{\overset{\phantom{\phantom{\rule{0pt}{0ex}}b}c}{\int }}f+\underset{c}{\overset{b}{\int }}f=g{\phantom{\rule{0.1em}{12pt}}|}_{a}^{c}+g{\phantom{\rule{0.1em}{12pt}}|}_{c}^{b}=g\left(c\right)-g\left(a\right)+g\left(b\right)-g\left(c\right)=g{\phantom{\rule{0.1em}{12pt}}|}_{a}^{b}=\underset{a}{\overset{b}{\int }}f$

2.  The assertion comes directly from $\underset{a}{\overset{a}{\int }}f+\underset{a}{\overset{a}{\int }}f=\underset{a}{\overset{a}{\int }}f$  which is valid because of 1.

3.  Combining 1. and 2. yields $\underset{a}{\overset{b}{\int }}f+\underset{b}{\overset{a}{\int }}f=\underset{a}{\overset{a}{\int }}f=0$, from which 3. is immediate.

1. allows to interrupt an integration at an arbitrary point c (c not even needs to be in between a and b!). Occasionally there are advantages from that. We could for instance calculate the integral $\underset{-1}{\overset{1}{\int }}|\mathrm{X}|$ without knowing any primitive of $|\mathrm{X}|$.We just split up the integration at 0 and thus have:

$\underset{-1}{\overset{1}{\int }}|\mathrm{X}|=\underset{-1}{\overset{0}{\int }}|\mathrm{X}|+\underset{0}{\overset{1}{\int }}|\mathrm{X}|=\underset{-1}{\overset{0}{\int }}-\mathrm{X}+\underset{0}{\overset{1}{\int }}\mathrm{X}=-\frac{1}{2}{\mathrm{X}}^{2}\phantom{\rule{0.2em}{12pt}}{|}_{-1}^{0}+\frac{1}{2}{\mathrm{X}}^{2}\phantom{\rule{0.2em}{12pt}}{|}_{0}^{1}=1$.

The next rules (integrals are linear in their integrands) are just integral versions of some derivation rules, namely sum, difference and factor rule.

Proposition:  Let $f,{f}_{1},{f}_{2}$ be integrable on I and $c\in ℝ$. For all $a,b\in I$ the following identities hold:

 $\underset{a}{\overset{b}{\int }}{f}_{1}+{f}_{2}=\underset{a}{\overset{b}{\int }}{f}_{1}+\underset{a}{\overset{b}{\int }}{f}_{2}$ [8.2.5] $\underset{a}{\overset{b}{\int }}{f}_{1}-{f}_{2}=\underset{a}{\overset{b}{\int }}{f}_{1}-\underset{a}{\overset{b}{\int }}{f}_{2}$ [8.2.6] $\underset{a}{\overset{b}{\int }}c\phantom{\rule{0.1em}{0ex}}f=c\underset{a}{\overset{b}{\int }}f$ [8.2.7]

Proof:

1.   If ${g}_{1}$ and ${g}_{2}$ are primitives of ${f}_{1}$ and ${f}_{2}$ resp. then ${g}_{1}+{g}_{2}$ is a primitive of ${f}_{1}+{f}_{2}$ due to [8.1.6] and thus we have:

$\underset{a}{\overset{b}{\int }}{f}_{1}+{f}_{2}={g}_{1}+{g}_{2}{\phantom{\rule{0.5pt}{12pt}}|}_{a}^{b}={g}_{1}\left(b\right)+{g}_{2}\left(b\right)-{g}_{1}\left(a\right)-{g}_{2}\left(a\right)={g}_{1}{\phantom{\rule{0.5pt}{12pt}}|}_{a}^{b}+{g}_{2}{\phantom{\rule{0.5pt}{12pt}}|}_{a}^{b}=\underset{a}{\overset{b}{\int }}{f}_{1}+\underset{a}{\overset{b}{\int }}{f}_{2}$

2. and 3.  follow in a similar way with [8.1.7] and [8.1.8].

Consider:

• We may now integrate polynomials addendwise:

$\underset{a}{\overset{b}{\int }}\sum _{i=0}^{n}{a}_{i}\phantom{\rule{0.1em}{0ex}}{\mathrm{X}}^{i}=\sum _{i=0}^{n}{a}_{i}\underset{a}{\overset{b}{\int }}{\mathrm{X}}^{i}=\sum _{i=0}^{n}\frac{{a}_{i}}{i+1}{\mathrm{X}}^{i+1}{\phantom{\rule{0.1em}{12pt}}|}_{a}^{b}$

for example $\underset{0}{\overset{1}{\int }}8{\mathrm{X}}^{3}+2{\mathrm{X}}^{2}-3=\frac{8}{4}{\mathrm{X}}^{4}{\phantom{\rule{0.1em}{12pt}}|}_{0}^{1}+\frac{2}{3}{\mathrm{X}}^{3}{\phantom{\rule{0.1em}{12pt}}|}_{0}^{1}-3\mathrm{X}{\phantom{\rule{0.1em}{12pt}}|}_{0}^{1}=2+\frac{2}{3}-3=-\frac{1}{3}$.

Product and chain rule are transferable as well. The resulting intergration rules are dealt with in the next part.

We proved several non-trivial properties of differentiable function using the mean value theorem, a basic theorem of calculus and should expect similar strong results from its integral version.

Theorem (mean value theorem, integral version):  Let f be an integrable function on I, i.e. $f\in \mathcal{I}\left(I\right)$. If $a,b\in I$ are any two different points of I then there is an $\stackrel{˜}{x}$ in between a and b such that

 $\underset{a}{\overset{b}{\int }}f=\left(b-a\right)\cdot f\left(\stackrel{˜}{x}\right)$ [8.2.8]

Proof:  Take a primitive g of f. g is differentiable on I with $g={f}^{\prime }$. Due to the mean value theorem [7.9.5] there is thus an $\stackrel{˜}{x}$ in between a and b such that

$\underset{a}{\overset{b}{\int }}f=g{\phantom{\rule{0.1em}{12pt}}|}_{a}^{b}=g\left(b\right)-g\left(a\right)=\left(b-a\right)\cdot {g}^{\prime }\left(\stackrel{˜}{x}\right)=\left(b-a\right)\cdot f\left(\stackrel{˜}{x}\right)$

Consider:

• The proof of [8.2.8] is actually the proof of the implication

mean value theorem, differential version$\text{ }⇒\text{ }$mean value theorem, integral version

In fact however both versions are equivalent as the reverse

mean value theorem, integral version$\text{ }⇒\text{ }$mean value theorem, differential version

is valid as well.

Proof:  ? $f\in {\mathcal{D}}^{1}\left(\left[a,b\right]\right)$

$f\left(b\right)-f\left(a\right)=f\phantom{\phantom{\rule{0pt}{12pt}}}{\phantom{\rule{0.1em}{12pt}}|}_{a}^{b}=\underset{a}{\overset{b}{\int }}{f}^{\prime }=\left(b-a\right)\cdot {f}^{\prime }\left(\stackrel{˜}{x}\right)$

With its integral version [8.2.8] the mean value theorem controls how integrals are affected by properties of their integrands. One example describes integration as a monotone process.

Proposition:  The following implications are valid for all $f,g\in \mathcal{I}\left(\left[a,b\right]\right)$:

 [8.2.9] [8.2.10]

Proof:

1.   According to [8.2.8] there is an $\stackrel{˜}{x}\in \right]a,b\left[$ such that

$\underset{a}{\overset{b}{\int }}f=\underset{>0}{\underbrace{\left(b-a\right)}}\cdot \underset{\ge 0}{\underbrace{f\left(\stackrel{˜}{x}\right)}}\ge 0$

2.   From the premise we know that . We apply 1. to the integrable ([8.1.7]) function $g-f$ and get

$0\le \underset{a}{\overset{b}{\int }}g-f=\underset{a}{\overset{b}{\int }}g-\underset{a}{\overset{b}{\int }}f$

which in fact is the assertion [8.2.10].

As the above proof obviously remains valid if we replace $\le$ by $<$ we see that integration is even strictly monotone.

The monotony itself will yield an important estimate that is needed if we are going to interchange integral and absolute value.

Proposition:  For all $f\in {\mathcal{C}}^{0}\left(\left[a,b\right]\right)$ we have

 $|\underset{a}{\overset{b}{\int }}f|\le \underset{a}{\overset{b}{\int }}|f|$ [8.2.11]

Proof:  First we note that f and $|f|$ are continuous functions on an interval and are thus integrable. And as  $-|f\left(x\right)|\le f\left(x\right)\le |f\left(x\right)|$ for all $x\in \right]a,b\left[$  [8.2.10] guarantees that

$-\underset{a}{\overset{b}{\int }}|f|=\underset{a}{\overset{b}{\int }}-|f|\le \underset{a}{\overset{b}{\int }}f\le \underset{a}{\overset{b}{\int }}|f|$

which is the assertion. i This is in fact a property of the absolute value: The zero distance of a number x does not exceed r if and only if x is in between −r and r, i.e. $|x|\le r\text{ }⇔\text{ }-r\le x\le r$

Consider:

• In general the estimate [8.2.11] could not be tightened to =. For example, as shown above, $\underset{-1}{\overset{1}{\int }}|\mathrm{X}|=1$, but  $|\underset{-1}{\overset{1}{\int }}\mathrm{X}|=0$.

If however the algebraic sign of f is homogeneous on $\left[a,b\right]$, that means $f\left(x\right)\ge 0$ or $f\left(x\right)\le 0$ resp. for all $x\in \left[a,b\right]$, [8.2.11] may be replaced by the identity

 $|\underset{a}{\overset{b}{\int }}f|=\underset{a}{\overset{b}{\int }}|f|$ [8.2.12]

Proof:  According to [8.2.10] we have:

1.   $f\left(x\right)\ge 0$ for all $x\in \left[a,b\right]\text{ }⇒\text{ }\underset{a}{\overset{b}{\int }}f\ge 0$, thus: $|\underset{a}{\overset{b}{\int }}f|=\underset{a}{\overset{b}{\int }}f=\underset{a}{\overset{b}{\int }}|f|$

2.   $f\left(x\right)\le 0$ for all $x\in \left[a,b\right]\text{ }⇒\text{ }\underset{a}{\overset{b}{\int }}f\le 0$, thus: $|\underset{a}{\overset{b}{\int }}f|=-\underset{a}{\overset{b}{\int }}f=\underset{a}{\overset{b}{\int }}-f=\underset{a}{\overset{b}{\int }}|f|$

•

The next statement is an important joining element between differential and integral calculus. Moreover with suitable techniques at hand it will allow us to calculate primitives, which puts the triviality "Calculating integrals needs a good grasp of finding primitives" upside down: "Finding primitives needs a good grasp of calculating integrals".

Theorem (Fundamental theorem of calculus):  Let $f\in \mathcal{I}\left(I\right)$ be integrable on I. For each $c\in I$ the function $g:I\to ℝ$ defined by

 $g\left(x\right)≔\underset{c}{\overset{x}{\int }}f$ [8.2.13]

is a primitive function of f.

Proof:  We evaluate g at x with an arbitrary primitive h of f

$g\left(x\right)=\underset{c}{\overset{x}{\int }}f=h{\phantom{\rule{0.1em}{12pt}}|}_{c}^{x}=h\left(x\right)-h\left(c\right)$

which proves the identity $g=h-h\left(c\right)$, so that along with h g is a primitive of f as well (g and h only differ in the constant addend $h\left(c\right)$!).

Consider:

• [8.2.13] is sometimes called the second fundamental theorem of calculus. The first one would have been our definition [8.2.1] if integrals had been introduced in the alternative way mentioned before.

• As a primitive the function g from [8.2.13] is of course differentiable. Occasionally this is put into the phrase The integral is a differentiable function of its upper limit.

• Sometimes we simply use the symbol $\underset{c}{\overset{}{\int }}f$  for the function g which makes the relationship to the the indefinite integrals, as indicated initially, more transparent. Also the notation $\left(\underset{c}{\overset{}{\int }}f{\right)}^{\prime }=f$  [8.2.13] (combined with $\underset{c}{\overset{}{\int }}\left({f}^{\prime }\right)=f-f\left(c\right)$ ) supports the phrasing "integration and differentiation (nearly) cancel each other out".

• $g\left(c\right)=0$ due to [8.2.3]. Thus for any integrable f  [8.2.13] produces just the one primitive of f that has a zero at c.

• With that property we see that not all primitives could be produced by the fundamental theorem. The primitive ${\mathrm{X}}^{2}+1$ of $2\mathrm{X}$ for instance has no zero and thus would never come out of [8.2.13].

We proved in [8.1.15] that integrability tolerates uniform convergence. Now we will see that even the integral itself is compatible with that kind of convergence..

Proposition:  Let $\left({f}_{n}\right)$ be a sequence of integrable function on an interval I. If $f:I\to ℝ$ is the uniform limit of $\left({f}_{n}\right)$, i.e. ${f}_{n}\underset{uf}{\to }f$, the convergence

 $\underset{a}{\overset{b}{\int }}{f}_{n}\to \underset{a}{\overset{b}{\int }}f$ [8.2.14]

holds for all $a,b\in I$.

Proof:  First we note that f is integrable due to [8.1.15]. Furthermore [8.2.14] is obviously valid if $a=b$ so that we may assume $a\ne b$. Now let $\epsilon >0$ be arbitrary. According to the premise there is an ${n}_{0}\in {ℕ}^{\ast }$ such that

$|{f}_{n}\left(x\right)-f\left(x\right)|<\frac{\epsilon }{|b-a|}$  for all  $n\ge {n}_{0}$ and all $x\in I$.

Now we find an ${\stackrel{˜}{x}}_{n}$ in between a and b for each n (see the mean value theorem [8.2.8]) such that

$\underset{a}{\overset{b}{\int }}{f}_{n}-f=\left(b-a\right)\cdot \left({f}_{n}-f\right)\left({\stackrel{˜}{x}}_{n}\right)$

and therefore we have for all $n\ge {n}_{0}$:

$|\underset{a}{\overset{b}{\int }}{f}_{n}-\underset{a}{\overset{b}{\int }}f|=|\underset{a}{\overset{b}{\int }}{f}_{n}-f|=|b-a|\cdot |{f}_{n}\left({\stackrel{˜}{x}}_{n}\right)-f\left({\stackrel{˜}{x}}_{n}\right)|<|b-a|\cdot \frac{\epsilon }{|b-a|}=\epsilon$. 