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1.3 Basic properties and estimates

Proposition 1.3.1.

If $f:[a,b]\rightarrow\mathbb{R}$ is a continuous function and $\lambda\in\mathbb{R}$ such that $0\leq f(x)\leq\lambda$ , for all $x\in[a,b]$ , i.e., $\lambda$ is an upper bound for $f$ , then

0\leq\int_{a}^{b}f(x)\operatorname{d}x\leq\lambda(b-a).

Proof.

Let $u(f,[a,b],n,\cdot)$ be the upper approximating step function. Since $\max_{x\in[a,b]}f(x)\leq\lambda$ , we have $u(f,[a,b],n,x)\leq\lambda\mathbf{1}_{[a,b]}(x)$ , for all $x\in[a,b]$ and $n\in\mathbb{N}$ , and $\lambda\mathbf{1}_{[a,b]}$ is a step function. Now $\int_{a}^{b}f(x)\operatorname{d}x\leq u_{n}^{(f)}$ , for all $n\in\mathbb{N}$ , and by Lemma 1.2.8, $u_{n}^{(f)}\leq\lambda(b-a)$ . The lower bound $0$ is proved analogously, which concludes the proof. ∎

This holds also for piecewise continuous $f$ .

Let us now extend Lemmas 1.2.10 and 1.2.9 to continuous functions.

Proposition 1.3.2.

If $f,g:[a,b]\rightarrow\mathbb{R}$ are continuous functions such that $f(x)\leq g(x)$ , for all $x\in[a,b]$ . Then

\int_{a}^{b}f(x)\operatorname{d}x\leq\int_{a}^{b}g(x)\operatorname{d}x.

Proof.

W2.1. ∎

This holds also for piecewise continuous $f$ .

Proposition 1.3.3.

If $f,g:[a,b]\rightarrow\mathbb{R}$ are piecewise continuous functions then

\int_{a}^{b}(f(x)+g(x))\operatorname{d}x=\int_{a}^{b}f(x)\operatorname{d}x+% \int_{a}^{b}g(x)\operatorname{d}x.

Proof.

Let us suppose first that $f$ and $g$ are actually continuous. From MATH113 we remember that $f+g$ is continuous again and that the following relation holds: for every compact interval $I$ ,

\max_{x\in I}(f+g)(x)\leq\max_{x\in I}f(x)+\max_{x\in I}g(x).

This means that

u(f+g,[a,b],n,x)\leq u(f,[a,b],n,x)+u(g,[a,b],n,x),\quad x\in[a,b],

which in turn implies

u^{(f+g)}_{n}\leq u^{(f)}_{n}+u^{(g)}_{n},\quad n\in\mathbb{N},

owing to Lemma 1.2.8. So

\int_{a}^{b}(f(x)+g(x))=\lim_{n\rightarrow\infty}u^{(f+g)}_{n}\leq\lim_{n% \rightarrow\infty}u^{(f)}_{n}+\lim_{n\rightarrow\infty}u^{(g)}_{n}=\int_{a}^{b% }f(x)\operatorname{d}x+\int_{a}^{b}g(x)\operatorname{d}x,

as follows from elementary properties of the limit of sums of convergent sequences, see MATH113.

The analogous procedure for lower approximating step functions yields

\ell(f+g,[a,b],n,x)\geq\ell(f,[a,b],n,x)+\ell(g,[a,b],n,x),\quad x\in[a,b],

and

\int_{a}^{b}(f(x)+g(x))=\lim_{n\rightarrow\infty}\ell^{(f+g)}_{n}\geq\lim_{n% \rightarrow\infty}\ell^{(f)}_{n}+\lim_{n\rightarrow\infty}\ell^{(g)}_{n}=\int_% {a}^{b}f(x)\operatorname{d}x+\int_{a}^{b}g(x)\operatorname{d}x,

so combining the two inequalities we get

\int_{a}^{b}(f(x)+g(x))=\int_{a}^{b}f(x)\operatorname{d}x+\int_{a}^{b}g(x)% \operatorname{d}x.

Comment: We say “analogous” when a case is a very simple modification of the previous case, such as a sign switch, or inversion of inequality signs, etc., and every step translates in a straight-forward way to an “analogous” step in the new setting.

Now let us drop our initial assumption of continuity and let $f, g$ be piecewise continuous, so there are $a_{i}$ such that $f\restriction_{[a_{i-1},a_{i})}$ and $g\restriction_{[a_{i-1},a_{i})}$ are continuous, for all $i=1,\ldots,n$ . Notice that the $a_{i}$ are chosen such that they fit both $f$ and $g$ , and so not both $f$ and $g$ need to have a “jump” at each single $a_{i}$ :

Then the previous reasoning applies to each such interval, i.e.,

\int_{a_{i-1}}^{a_{i}}(f(x)+g(x))=\int_{a_{i-1}}^{a_{i}}f(x)\operatorname{d}x+% \int_{a_{i-1}}^{a_{i}}g(x)\operatorname{d}x,

Comment: It is a common little trick or strategy to first reduce a new situation to an easier or previously treated one and then to prove the statement in that easier situation.

for each $i$ . Summing over $i$ on both sides yields

\int_{a}^{b}(f(x)+g(x))=\int_{a}^{b}f(x)\operatorname{d}x+\int_{a}^{b}g(x)% \operatorname{d}x.

∎

Key idea of proof.

split into continuous pieces, there apply the theorem for continuous functions

Proposition 1.3.4.

If $f:[a,b]\rightarrow\mathbb{R}$ is a continuous function and $a\leq c<d\leq b$ . Then $f\restriction_{[c,d]}:[c,d]\rightarrow\mathbb{R}$ is also continuous and one defines

\int_{c}^{d}f(x)\operatorname{d}x:=\int_{c}^{d}(f\restriction_{[c,d]})(x)% \operatorname{d}x.

We have

\int_{a}^{b}f(x)\operatorname{d}x=\int_{a}^{c}f(x)\operatorname{d}x+\int_{c}^{% d}f(x)\operatorname{d}x+\int_{d}^{b}f(x)\operatorname{d}x.

Proof.

We first notice that

\int_{c}^{d}f(x)\operatorname{d}x=\int_{a}^{b}(f\mathbf{1}_{[c,d)})(x)% \operatorname{d}x.{}

Since $f=f\mathbf{1}_{[a,c)}+f\mathbf{1}_{[c,d)}+f\mathbf{1}_{[d,b]}$ , we can apply Proposition 1.3.3 and $(*)$ to obtain

	$\displaystyle\int_{a}^{b}f(x)\operatorname{d}x=$	$\displaystyle\int_{a}^{b}(f\mathbf{1}_{[a,c)})(x)\operatorname{d}x+\int_{a}^{b% }(f\mathbf{1}_{[c,d)})(x)\operatorname{d}x+\int_{a}^{b}(f\mathbf{1}_{[d,b]})(x% )\operatorname{d}x$
	$\displaystyle=$	$\displaystyle\int_{a}^{c}f(x)\operatorname{d}x+\int_{c}^{d}f(x)\operatorname{d% }x+\int_{d}^{b}f(x)\operatorname{d}x.$

∎

Again this holds also for piecewise continuous $f$ .

Example 1.3.5.

As an easy application, consider the integral

\int_{a}^{a+2\pi}\sin(x)\operatorname{d}x.

We can write

\int_{a}^{a+2\pi}\sin(x)\operatorname{d}x=\int_{a}^{a+\pi}\sin(x)\operatorname% {d}x+\int_{a+\pi}^{a+2\pi}\sin(x)\operatorname{d}x.

Now we know that $\sin(x+\pi)=-\sin(x)$ , so

\int_{a+\pi}^{a+2\pi}\sin(x)\operatorname{d}x=\int_{a}^{a+\pi}\sin(x+\pi)% \operatorname{d}x=\int_{a}^{a+\pi}-\sin(x)\operatorname{d}x=-\int_{a}^{a+\pi}% \sin(x)\operatorname{d}x,

which proves that

\int_{a}^{a+2\pi}\sin(x)\operatorname{d}x=\int_{a}^{a+\pi}\sin(x)\operatorname% {d}x+\int_{a+\pi}^{a+2\pi}\sin(x)\operatorname{d}x=0.

We end the section with a brief definition concerning the case of functions defined on all of $\mathbb{R}$ (rather than compact intervals):

Definition 1.3.6.

Given a continuous function

f:\mathbb{R}\rightarrow\mathbb{R},

we define the improper integral of $f$ as

\int_{-\infty}^{\infty}f(x)\operatorname{d}x:=\lim_{a\rightarrow-\infty}\int_{% a}^{0}f(x)\operatorname{d}x+\lim_{b\rightarrow\infty}\int_{0}^{b}f(x)% \operatorname{d}x

if those limits exist; otherwise $\int_{-\infty}^{\infty}f(x)\operatorname{d}x$ remains undefined.

This is treated in more detail in other modules, e.g. MATH102 and MATH210.

Remark 1.3.7.

The idea of upper and lower approximating step functions can be applied to a bigger class of functions which need not be (piecewise) continuous functions or step functions, so-called integrable functions. This will be done in MATH210.