Bear H.S. Understanding Calculus

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The

Indefinite

Integral

In many applications, the information about the dependent variable is given in terms of its

derivative or derivatives. For example, if an object falls to the earth under the influence of

gravity, Newton's law says that the acceleration is a constant. If s is the distance the object

falls in time

t, and v and a are its velocity and acceleration of time t, then

V=-,

a=-=-.

(11.1)

dt dt dt?

Hence, Newton's law in mathematical terms is

= g, (11.2)

where g is a constant that depends on the units used. (If s is measured in feet and t in seconds,

then

g is approximately 32 ft/sec'")

If we are given an expression like (11.2) which describes the derivatives of

s, then

naturally we want to be able to convert it into an explicit formula for s. This process is called

(indefinite) integration. We start with

= dt = g.

Of course,

-it

(gt) = g, and we know that any other function, v, which has this same derivative,

g, must differ from gt by a constant. Therefore,

v = gt + Cl

for some constant Ct. Now use

= v, so

-=gt+Ct.

Clearly, 4gt

+ Cl t has derivative gt +

CI,

1 2

s="2gt

+Clt+C2

64 Understanding Calculus

for someconstantC2. The constantC2 is the value of s when t = 0 (the initial valueof s), and

the constant

CI is the valueof v =

when t = 0 (the initial velocity).

Now we will systematizethis process of going from the derivative to the function. If

F'(x)

ftx),

then

F(x)

is called an antiderivative or indefinite integral of

I(x),

and we

indicatethis with the notation

F(x)

= f

f(x)

dx +c,

where C is understoodto be an arbitrary constant. The rules for integrationare equivalentto

the rules for differentiation, since

f(x)dx

F(x)

means

~F(x)

f(x).

Herearesomeof thedifferentiation formulaswehave,withthecorresponding integration

formulas. In writingintegrationformulas, we will suppressthe arbitraryadditiveconstantand

understand that a formula like

Jx dx =

means that

is one of the infinitely many

integralsof

x, the rest all having the form !x

+c.

~xn

=nx

n-';

fxndX

_1_

(n:/=

-1);

dx n + 1

d . f

- Sinx = cosx;

cosx

dx = sinr:

d . f· d

- cosx = -

SIn

x; sInx x = - cosx;

~ex=ex;

feXdx=e

logx =

dx =logx;

dx x x

d .

1 f dx .

sln-

= sin" x;

dx Jf=X2 Jf=X2

-1

1 f dx

-1

dx tan x = 1 + x

;

1 + x

= tan x.

Noticethat integrationis a linear operationbecausedifferentiation is linear. That is,

(f(x)

g(x))dx

= f

f(x)dx

+ f

g(x)dx,

af(x)dx

=a f

f(x)dx.

For example,

f (3 cosx +5x) dx = 3 f cos x dx + 5 f x dx

. 5 2

=3s1nx+

+c.

EXAMPLE 11.1

Supposethat at everypoint (x, y) of a curvethe slopeof the curveis e' - x, and thecurvepassesthrough

(0,2). Find

y in termsof x.

Chapter

• The Indefinite Integral

Solution

Weare giventhe differential equation

dy x

-x,

with initial condition y(0) = 2. Hence

y = f

(eX

-x)dx

=e" -

-x

+ C.

Since

y (0) = 2, we must have

o 1

2 = e - -

·0

+c,

= 1.

Finally,

The integration process frequently involves some juggling of constants. Consider

!sin2xdx.

We notice that

~(-cos2x)

= 2sin2x,

so - cos 2x is almost the answer, with an extra factor of 2 resulting from the chain rule. We

write

sin2xdx

!(Sin2X)(2)dX,

and notice that the last integral has the form

!sin u du = -

cos u

where u = 2x. Thus, with u = 2x

!sin 2x dx =

!(Sin2X)2dX

sinudu

cosu

-2

cos2x.

This technique is called making a u-substitution. Here are some more examples:

(i) !e

dx .

Make the substitution u = 3x, du

=.3dx,

= 1du, so

Understanding

Calculus

dx = f

eU~du

feUdu

1 u

-e

(ii) fx cos(4x

)

dx.

Let u = 4x

du = 8x dx, x dx = k

duo

Then

fx coe

dx =f

cosu

(~)

fcosudu

1 .

sm s

sin-ix .

(iii) fj1=-:;2 x dx.

In all these examples, the key is to recognize the chain rule in reverse. Here that

means noticing that except for a constant,

x is the derivative

1 - x

•

Hence, we

letu =

l-x

2,du

-2xdx,dx

-!du,

j1=-:;2xdx

= f

.;u(

-~)

. f 3x

(IV)

--2

dx.

4+x

If u =4 + x

then du = 2x

dx,

and 3x dx =

duo

Therefore,

~du

4+x

= 2

10gu

log(4 +x

) .

In the last example we used the formula J

= log u, which is not quite complete.

Chapter 11 • The Indefinite Integral

Recall that logx is

defined

only for x > 0, and consequently

logx =

only

makessensefor x >

However,

log IxImakessensefor all x

-:f=.

0, and it is easy

to see (Problem

11.34)

that

d 1

log [z] =

(11.3)

for all

-:f=.

Consequently, the properintegration versionof (11.3)is

=log [x].

For example, the curve y such that

and y(

-e)

=3 is determined by:

y =f

~dX

= log lx] +c.

Since y(

-e)

=3,

= log I- e] + c

= loge +c

= 1

+c,

so c = 2, and y = log [xI+ 2.

PROBLEMS

Find the indefinite integrals.

11.1 fx

11.2 f

11.3 f

11.4 f

~dX

11.5 f

5x-~

11.6 f

11.7 f

(2JX

+3x

) dx

11.8 f x

11.9 fsin3x dx

11.10 fcos5x dx

11.11 f(Sin

- 3cos2x) dx

11.12 fx

cos r ' dx

11.13 fsintl +x) dx

11.14 fx costl +x

) dx

11.15 f

.j4+x

2dx

Understanding Calculus

11.16 f(1 + X

2)IO

11.17

f(3x +5e

)

11.18 f(1 +e

)2dx

Hint:

Squarethe binomial.

11.19 fxe

x 2

11.20

fc cos

11.21

fx : 3 dx

11.22

f 3

~:3

11.23

f 2x +1

+X +5

11.24 f

- 4x

11.25 f dx = f

J4-x

2Jl-

(I)2

11.26

+9x

11 27 f

.-!!!.-

- f

• 9 +x

9 (1 +

11.28

x+5

Find the function y that satisfies the givenconditions.

11.29 dy = 3x

+x - 7,

y(l)

= 2

dy x 1

11. - = e +

--,

y(O) = 2

dx 1

dy 1

11.31

dx = 1 + x

y(l)

= 0

dy 1

11.32

-d

v'f=X2'

y(O) =5

x 1

-x

11.33 Supposea car's lockedbrakesprovidea constantnegative acceleration

-k)

which

will stop the car going60 mph(88 ft/sec) in 4 seconds. What is k?

11.34 Showthat

log

I =

Hint:

If x > 0, there is nothingto

show,

so assumex < 0 and

log

[x]

= log(

-x).

(12.1)

(12.2)

The

Definite

Integral

Thedefiniteintegralweconsiderinthissectionhasmanyinterpretations andmanyapplications.

Westartwiththesimplegeometricideaoftheareaunderacurve. Let

f (x) beapositivefunction

definedon someinterval

[a, b]. Wewant to find the area A betweenthe x-axis and the curve

f(x)

for a S x S b. To approximate the area A, slice up the region under the curve into

narrownear-rectangles whose sides are the lines

x = Xi, for points xo,

XI,

X2,

...

, x; between

a and b (Figure 12.1). The area of the slice between x = Xi-I and x = Xi is approximately

f(Ci)(Xi - Xi-I) for any point c, in [Xi-I, Xi], since the valuesof

f(x)

will not varymuch in

a small interval [Xi-I, x;l. The sum of the areas of the small slices is the following

Riemann

sum for

L f(Ci)(Xi - Xi-I).

i=1

For a continuousfunction, thesesumswill approacha limit as maxtx, - Xi-I)

0, and this

limit is the

definite

integral of f over [a, b], denoted

f(x)

dx.

This integralis what we

defineto be the area under the positivefunction

Nowlet

F(x)

be anyindefinite integralof

f(x);

that is,

F'(x)

f(x).

Applythe Mean

Value

Theoremto

F(x)

over each subinterval [Xi-I, Xi], and for each i choose c, E [Xi-I, x;l

such that

F(Xi) -

F(Xi-l)

= F'(Ci)(Xi - Xi-I)

= f(Ci)(Xi - Xi-I).

Notice that if we add all the terms on the left of (12.2), the intermediate terms all cancel, so

L(F(Xi)

F(Xi-I))

F(x

) -

F(xo)

i=1

F(b)

F(a).

Hence, from (12.2) and (12.3) we have

F(b)

F(a)

= L f(Ci)(Xi -

Xi-I),

i=1

(12.3)

(12.4)

Understanding Calculus

Figure 12.1

where

F(x)

is any indefinite integral of

f(x),

and the ci are appropriately chosen. Since the

Riemann sums on the right

(12.4) converge to the definite integral as

m!1X

(Xi - Xi

-I)

----+

F(b) - F(a) = l

f(x)

for any F such that F

'(x)

f(x).

We use the convenient notation

F(x)]:

= F(b) - F(a),

(12.5)

f(x)

F(x)]:

= F(b) - F(a) (12.6)

for any antiderivative F(x). The variable X on the left of (12.6) is a dummy variable that could

be replaced by any other variable without changing the meaning. Thus

f(x)

dx = l

f(t)

dt = l

f(y)

dy = ... . (12.7)

EXAMPLE 12.1

Find the area boundedby the coordinateaxes, the curve y =

and the line x= 2.

Solution

The firststepin anyarea problem(andthis is important) is to graphthe curves(Figure12.2). The area is

21 1

_exdx = _ex =

_[e

3.19.

o 2 2 0 2

Element

of area = f(x) dx

Figure 12.2

The notation

f (x) dx is meant to suggest the limit of sums of terms of the form

f (x) dx. For area problems we regard f (x) and dx as the height and width of a typical

Chapter 12 • The Definite Integral

small rectangular slice of the region, and this provides a good mnemonic device for setting

up area integrals . For example, let us find the area between the parabolas y = x

2 and

y =4x - x

- 2. The first step (and this is still important) is to draw the figure. To graph the

second parabola we complete the square:

4x - x

- 2 =

_(x

- 4x

+4) - 2 +4

=2 - (x - 2)2.

The parabola has the line x = 2 as its axis, and its vertex is (2,2) (Figure 12.3).

Figure 12.3

A typical slice of the region (and we get this information from the figure) is a "rectangle"

whose height is the difference of the y-values on the two curves; namely,

(4x - x

- 2) - (x

- 2) =4x - 2x

. (12.8)

The x-range is from x = 0 to x = 2, which we see from the graphs.

the graphs were

not quite so nice, then we would have to solve the two equations simultaneously to find the

intersections of the curves. The area is, from (12.8),

A =1

(4x - 2x

2)dx

= 2x

~x3]2

3 0

=8-

=~.

3 3

The same slicing technique will work for curves given in the form x =

fey).

For

example, the area bounded by the y-axis and the parabola

9 (

1)2

x=2+y-l=4"-

y-2

(Figure 12.4) is given by

72 Understanding Calculus

- - - y =

-I

y-y2

=2 _ (r- 1

4 • 2

Figure 12.4

In our definition of

f(x)

dx, we assumedthat a < b. It is clear from the definition

that if a < b < c, then

f(x)

dx +1

f(x)

dx =1

f(x)

dx .

It is convenient to makethe following additional agreements: if a < b, we

define

1°

f(x)dx

-1

f(x)dx

(12.9)

and for any

a we define

1°

f(x)

dx =

Withtheseconventions, the identity(12.9)holds for any three numbers a, b, c; for example,

f(x)dx

+ 1

f(x)dx

f(x)dx.

PROBLEMS

Evaluatethe definite integrals.

12.1 1

2x + 7) dx

12.2

(3x

) dx

12.3

5x)dx

12.4

i~2

12.5 1

J 2 +2y dy

12.6 I

cos

Gy)

12.71IX~dX

12.8

sin2x dx

12.9

x l