9.4 Indefinite Integrals

9.57   Theorem. Let $a,b,c$ be real numbers. If $f$ is a function that is integrable on each interval with endpoints in $\{ a,b,c\}$ then

$$\int_a^c f=\int_a^b f+\int_b^c f.$$

Proof: The case where $a\leq b\leq c$ is proved in theorem 8.18. The rest of the proof is exactly like the proof of exercise 5.69. $\diamondsuit$

9.58   Exercise. Prove theorem 9.57.

We have proved the following formulas:

$$\int_a^b x^r dx = {{b^{r+1}-a^{r+1}}\over {r+1}} \mbox{ for } 0<a<b \quad r\in \mbox{{\bf Q}}\setminus\{-1\},$$ (9.59)

$$\int_a^b {1\over t}dt = \ln (b)-\ln (a) \mbox{ for } 0< a < b,$$

$$\int_a^b \sin (ct)dt = {{-\cos (cb)+\cos (ca)}\over c} \mbox{ for } a<b, \mbox{ and }c > 0,$$

$$\int_a^b \cos (ct) dt = {{\sin (cb)-\sin (ca)}\over c} \mbox{ for } a<b, \mbox{ and }c > 0.$$ (9.60)

In each case we have a formula of the form

$$\int_a^b f(t)dt= F(b)-F(a).$$

This is a general sort of situation, as is shown by the following theorem.

9.61   Theorem (Existence of indefinite integrals.) Let $J$ be an interval in $\mbox{{\bf R}}$, and let $f\colon J\to\mbox{{\bf R}}$ be a function such that $f$ is integrable on every subinterval $[p,q]$ of $J$. Then there is a function $F\colon J\to\mbox{{\bf R}}$ such that for all $a,b\in J$

$$\int_a^b f(t)dt= F(b)-F(a).$$

Proof: Choose a point $c\in J$ and define

$$F(x)=\int_c^x f(t)dt \mbox{ for all } x\in J.$$

Then for any points $a,b$ in $J$ we have

$$F(b)-F(a)=\int_c^b f(t) dt-\int_c^a f(t) dt=\int_a^b f(t) dt.$$

We've used the fact that

$$\displaystyle { \int_c^b f(t)dt =\int_c^a f(t) dt+\int_a^b f(t)dt} \mbox{ for all } a,b,c\in J.\mbox{ $\diamondsuit$}$$

9.62   Definition (Indefinite integral.) Let $f$ be a function that is integrable on every subinterval of an interval $J$. An indefinite integral for $f$ on $J$ is any function $F\colon J\to\mbox{{\bf R}}$ such that $${\int_a^b f(t)dt=F(b)-F(a)}$$ for all $a,b\in J$.

A function that has an indefinite integral always has infinitely many indefinite integrals, since if $F$ is an indefinite integral for $f$ then so is $F+ c$ for any number $c$:

$$(F+ c)(b)-(F+ c)(a)=(F(b)+c)-(F(a)+c)=F(b)-F(a).$$

The following notation is used for indefinite integrals. One writes $\int f(t)dt$ to denote an indefinite integral for $f$. The $t$ here is a dummy variable and can be replaced by any available symbol. Thus, based on formulas (9.59) - (9.60), we write

$$\begin{eqnarray*} \int x^r dx &=& {{x^{r+1}}\over {r+1}} \mbox{ if } r\in\mbox{{... ... \\ \int\cos (ct) dt &=& {{\sin (ct)}\over c} \mbox{ if } c> 0. \end{eqnarray*}$$

We might also write

$$\int x^r dr= {{x^{r+1}}\over {r+1}}+3.$$

Some books always include an arbitrary constant with indefinite integrals, e.g.,

$$\int x^r dr={{x^{r+1}}\over {r+1}} +C \mbox{ if } r\in\mbox{{\bf Q}}\setminus\{-1\}.$$

The notation for indefinite integrals is treacherous. If you see the two equations

$$\int x^3 dx = {1\over 4} x^4$$

and

$$\int x^3 dx = {1 \over 4}(x^4+1),$$

then you want to conclude

$${1\over 4} x^4 = {1 \over 4}(x^4+1),$$ (9.63)

which is wrong. It would be more logical to let the symbol $\int f(x) dx$ denote the set of all indefinite integrals for $f$. If you see the statements

$${1\over 4}x^4 \in \int x^3 dx$$

and

$${1\over 4}(x^4+1) \in \int x^3 dx,$$

you are not tempted to make the conclusion in (9.63).

9.64   Theorem (Sum theorem for indefinite integrals) Let $f$ and $g$ be functions each of which is integrable on every subinterval of an interval $J$, and let $c,k\in\mbox{{\bf R}}$. Then

$$\int \Big(cf(x)+kg(x)\Big)dx=c\int f(x)dx +k\int g(x)dx.$$ (9.65)

Proof: The statement (9.65) means that if $F$ is an indefinite integral for $f$ and $G$ is an indefinite integral for $G$, then $cF+kG$ is an indefinite integral for $cf+kg$.

Let $F$ be an indefinite integral for $f$ and let $G$ be an indefinite integral for $g$. Then for all $a,b\in J$

$$\begin{eqnarray*} \int_a^b \Big(cf(x)+kg(x)\Big)dx &=& \int_a^b cf(x)dx+\int_a^b... ...kG(b)\Big) -\Big( cF(a)+kG(a)\Big)\\ &=& (cF+kG)(b)-(cF+kG)(a). \end{eqnarray*}$$

It follows that $cF+kG$ is an indefinite integral for $cf+kg$. $\diamondsuit$

9.66   Notation ($F(t)\mid_a^b$.) If $F$ is a function defined on an interval $J$, and if $a,b$ are points in $J$ we write $F(t)\mid_a^b$ for $F(b)-F(a)$. The $t$ here is a dummy variable, and sometimes the notation is ambiguous, e.g. $x^2-t^2\mid_0^1$. In such cases we may write $F(t)\mid_{t=a}^{t=b}$. Thus

$$(x^2-t^2)\mid_{x=0}^{x=1} = (1-t^2)-(0-t^2)=1$$

while

$$(x^2-t^2)\mid_{t=0}^{t=1}=(x^2-1)-(x^2-0)=-1.$$

Sometimes we write $F\mid_a^b$ instead of $F(t)\mid_a^b$.

9.67   Example. It follows from our notation that if $F$ is an indefinite integral for $f$ on an interval $J$ then

$$\int_a^b f(t)dt=F(t)\mid_a^b$$

and this notation is used as follows:

$$\begin{eqnarray*} \int_a^b 3x^2 dx &=& x^3\Big\vert _a^b=b^3-a^3.\\ \int_0^\pi ... ...t _0^2\\ &=& 4\cdot{8\over 3}+3\cdot{4\over 2}+2={{56}\over 3}. \end{eqnarray*}$$

In the last example I have implicitly used

$$\int (4x^2+3x+1) dx=4\int x^2 dx+3\int x\;dx +\int 1\; dx.$$

9.68   Example. By using the trigonometric identities from theorem 9.21 we can calculate integrals of the form $\int_a^b\sin^n (cx)\cos^m(kx)dx$ where $m,n$ are non-negative integers and $c,k\in\mbox{{\bf R}}$. We will find

$$\int_0^{ {\pi\over 2}} \sin^3 (x)\cdot \cos (3x) dx.$$

We have

$$\sin^2 (x)={{1-\cos (2x)}\over 2},$$

so

$$\begin{eqnarray*} \sin^3(x) &=& \sin^2(x)\sin (x)={1\over 2}\sin (x)-{1\over 2}\... ...x)-\sin (x)\Big)\\ &=& {3\over 4}\sin (x) -{1\over 4}\sin (3x). \end{eqnarray*}$$

Thus

$$\begin{eqnarray*} \sin^3 (x)\cdot\cos (3x) &=& {3\over 4}\cos (3x)\sin (x)-{1\ov... ...3x)\\ &=& {3\over 8}[\sin (4x)-\sin (2x) ]-{1\over 8}\sin (6x). \end{eqnarray*}$$

Hence

$${\int_0^{\pi /2}\sin^3 (x)\cdot \cos (3x)\;dx}$$

$$\begin{eqnarray*} &=&\left. {3\over 8} {{(-\cos (4x))}\over 4}\right\vert _0^{{... ...1-1)=-{3\over 8}-{1\over {24}}={{-10}\over {24}}=-{5\over {12}}. \end{eqnarray*}$$

The method here is clear, but a lot of writing is involved, and there are many opportunities to make errors. In practice I wouldn't do a calculation of this sort by hand. The Maple command

> int((sin(x))^3*cos(3*x),x=0..Pi/2);

responds with the value

                      - 5/12

9.69   Exercise. A Calculate the integrals

$$\int_0^{{\pi\over 2}}\sin x\;dx,\; \int_0^{{\pi\over 2}}\sin^2x\;dx \mbox{ and } \int_0^{{\pi\over 2}}\sin^4 x\; dx.$$

Then determine the values of

$$\int_0^{{\pi\over 2}}\cos x\; dx,\; \int_0^{{\pi\over 2}}\cos^2 x\; dx \mbox{ and } \int_0^{{\pi\over 2}}\cos^4 x\; dx$$

without doing any calculations. (But include an explanation of where your answer comes from.)

9.70   Exercise. Find the values of the following integrals. If the answer is geometrically clear then don't do any calculations, but explain why the answer is geometrically clear.

a)

$${\int_1^2 {1\over {x^3}}dx}$$

b)

$${\int_{-1}^1 x^{11}(1+x^2)^3\;dx}$$

c)

$${\int_0^2\sqrt{4-x^2}\;dx}$$

d)

$${\int_0^\pi (x+\sin(2x))dx}$$

e)

$${\int_{-1}^1 {1\over {x^2}}dx}$$

f)

$${\int_1^4 {{4+x}\over x} dx}$$

g)

$${\int_0^1\sqrt x dx}$$

h)

$${\int_1^2 {4\over x}dx}$$

i)

$${\int_0^1 (1-2x)^2 dx}$$

j)

$${\int_0^1 (1-2x)dx}$$

k)

$${\int_0^\pi \sin (7x)\; dx}$$

l)

$${\int_0^\pi\sin (8x)\; dx}$$

9.71   Exercise.

$$\begin{eqnarray*} \mbox{Let } A &=& \int_0^{\pi /2}(\sin (4x))^5 dx\\ B &=& \in... ...\sin (3x))^5 dx \\ C &=& \int_0^{{\pi\over 2}}(\cos (3x))^5 dx. \end{eqnarray*}$$

Arrange the numbers $A,B,C$ in increasing order. Try to do the problem without making any explicit calculations. By making rough sketches of the graphs you should be able to come up with the answers. Sketch the graphs, and explain how you arrived at your conclusion. No `` proof" is needed.