The Fundamental Theorem of Calculus Monday, February

8/14/2019 The Fundamental Theorem of Calculus Monday, February

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5.3

The Fundamental

Theorem of Calculus

Monday, February 8, 2010

INTEGRALS

In this section, we will learn about:

The Fundamental Theorem of Calculus

and its significance.


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The Fundamental Theorem of Calculus

(FTC) is appropriately named.

It establishes a connection between the twobranches of calculusdifferential calculus andintegral calculus.

FUNDAMENTAL THEOREM OF CALCULUS


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FTC

Differential calculus arose from the tangent

problem.

Integral calculus arose from a seemingly

unrelated problemthe area problem.


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Newtons mentor at Cambridge, Isaac Barrow

(16301677), discovered that these two

problems are actually closely related.

In fact, he realized that differentiation andintegration are inverse processes.

FTC


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The FTC gives the precise inverse

relationship between the derivative

and the integral.

FTC


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It was Newton and Leibniz who exploited this

relationship and used it to develop calculus

into a systematic mathematical method.

In particular, they saw that the FTC enabled themto compute areas and integrals very easily without

having to compute them as limits of sumsas we didin Sections 5.1 and 5.2

FTC


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The first part of the FTC deals with functions

defined by an equation of the form

where fis a continuous function on [a, b]

and xvaries between aand b.

( ) ( )x

ag x f t dt

Equation 1FTC


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Observe that gdepends only on x, which appears

as the variable upper limit in the integral.

If xis a fixed number, then the integralis a definite number.

If we then let xvary, the numberalso varies and defines a function of xdenoted by g(x).

( ) ( )

x

ag x f t dt

( )x

a f t d t

( )x

a f t dt

FTC


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If fhappens to be a positive function, then g(x)

can be interpreted as the area under the graph

of ffrom ato x, where xcan vary from ato b.

Think of gas thearea so far function,

as seen here.

FTC


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If fis the function

whose graph is shown

and ,

find the values of:g(0), g(1), g(2), g(3),

g(4), and g(5).

Then, sketch a rough graph of g.

Example 1

0( ) ( )

x

g x f t dt

FTC


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First, we notice that:

0

0

(0) ( ) 0g f t dt

FTC Example 1


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From the figure, we see that g(1) is

the area of a triangle:

1

0

1

2

(1) ( )

(1 2)

1

g f t dt

Example 1FTC


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To find g(2), we add to g(1) the area of

a rectangle:

2

0

1 2

0 1

(2) ( )

( ) ( )

1 (1 2)

3

g f t dt

f t dt f t dt

Example 1FTC


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We estimate that the area under ffrom 2 to 3

is about 1.3.

So,3

2(3) (2) ( )

3 1.3

4.3

g g f t dt

Example 1FTC


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For t> 3, f(t) is negative.

So, we start subtracting areas, as

follows.

Example 1FTC


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Thus,4

3(4) (3) ( ) 4.3 ( 1.3) 3.0g g f t dt

FTC Example 1

5

4

(5) (4) ( ) 3 ( 1.3) 1.7g g f t dt


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We use these values to sketch the graph

of g. Notice that, because f(t)

is positive for t< 3,

we keep adding areafor t< 3.

So, gis increasing up tox= 3, where it attains

a maximum value. For x> 3, gdecreases

because f(t) is negative.

Example 1FTC


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Notice that g(x) = x, that is, g= f.

In other words, if gis defined as the integral of f

by Equation 1, gturns out to be an antiderivativeof fat least in this case.

FTC


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If we sketch the derivative

of the function g, as in the

first figure, by estimating

slopes of tangents, we geta graph like that of fin the

second figure.

So, we suspect that g= fin Example 1 too.

FTC


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To see why this might be generally true, we

consider a continuous function fwith f(x) 0.

Then, can be interpreted asthe area under the graph of ffrom ato x.

( ) ( )x

a

g x f t dt

FTC


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To compute g(x) from the definition of

derivative, we first observe that, for h> 0,

g(x+ h) g(x) is obtained by subtracting

areas.

It is the areaunder the graph

of ffrom xto x+ h(the gold area).

FTC


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For small h, you can see that this area is

approximately equal to the area of the

rectangle with height f(x) and width h:

So,

FTC

( ) ( ) ( )g x h g x hf x ( ) ( )

( )

g x h g x

h

f x


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Intuitively, we therefore expect that:

The fact that this is true, even when fis notnecessarily positive, is the first part of the FTC(FTC1).

0

( ) ( )'( ) lim ( )

h

g x h g xg x f x

h

FTC


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FTC1

If fis continuous on [a, b], then the function g

defined by

is continuous on [a, b] and differentiable on

(a, b), and g(x) = f(x).

( ) ( )x

ag x f t dt a x b


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In words, the FTC1 says that the derivative

of a definite integral with respect to its upper

limit is the integrand evaluated at the upper

limit.

FTC1


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Using Leibniz notation for derivatives, we can

write the FTC1 as

when fis continuous.

Roughly speaking, Equation 5 says that,if we first integrate fand then differentiatethe result, we get back to the original function f.

( ) ( )x

a

d f t dt f x

dx

Equation 5FTC1


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Find the derivative of the function

As is continuous, the FTC1 gives:

Example 2

2

0( ) 1

x

g x t dt

2( ) 1 f t t

2

'( ) 1g x x

FTC1


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FRESNEL FUNCTION

For instance, consider the Fresnel function

It is named after the French physicist Augustin Fresnel(17881827), famous for his works in optics.

It first appeared in Fresnels theory of the diffraction

of light waves.

More recently, it has been applied to the designof highways.

2

0( ) sin( / 2)

x

S x t dt

Example 3


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FRESNEL FUNCTION

The FTC1 tells us how to differentiate

the Fresnel function:

S(x) = sin(x2/2)

This means that we can apply all the methodsof differential calculus to analyze S.

Example 3


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Find

Here, we have to be careful to use the Chain Rulein conjunction with the FTC1.

4

1 sec

xd

t dtdx

Example 4FTC1


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Let u= x4.

Then,

4

1 1

1

4 3

sec sec

(Chain Rule)

sec (FTC1)

sec( ) 4

x u

u

d dt dt t dt

dx dxd du

sec t dtdu dx

duudx

x x

Example 4FTC1


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In Section 5.2, we computed integrals from

the definition as a limit of Riemann sums

and saw that this procedure is sometimes

long and difficult.

The second part of the FTC (FTC2), which followseasily from the first part, provides us with a much

simpler method for the evaluation of integrals.

FTC1


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FTC2

If fis continuous on [a, b], then

where Fis any antiderivative of f,

that is, a function such that F = f.

( ) ( ) ( )b

af x dx F b F a


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FTC2

The FTC2 states that, if we know an

antiderivative Fof f, then we can evaluate

simply by subtracting the values

of Fat the endpoints of the interval [a, b].

( )b

a f x dx


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FTC2

If v(t) is the velocity of an object and s(t)

is its position at time t, then v(t) = s(t).

So, sis an antiderivative of v.


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FTC2

Evaluate the integral

The function f(x) = x3 is continuous on [-2, 1]and we know from Section 4.9 that an antiderivativeis F(x) = x4.

So, the FTC2 gives:

Example 51

3

2

x dx

13

2

4 41 1

4 4

15

4

(1) ( 2)

1 2

x dx F F


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FTC2

We often use the notation

So, the equation of the FTC2 can be written

as:

Other common notations are and .

( )] ( ) ( )ba

F x F b F a

( ) ( )] where 'b

b

aa

f x dx F x F f

( ) |b

aF x [ ( )]

b

aF x


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FTC2

Find the area under the parabola y= x2

from 0 to 1.

An antiderivative of f(x) = x2 is F(x) = (1/3)x3.

The required area is found using the FTC2:

Example 6

13 3 3

12

00

1 0 1

3 3 3 3

x A x dx


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FTC2

Find the area under the cosine curve

from 0 to b, where 0 b /2.

Since an antiderivative of f(x) = cos xisF(x) = sin x, we have:

Example 7

00

cos sin

sin sin 0

sin

b b

A x dx x

b

b


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FTC2

In particular, taking b= /2, we have

proved that the area under the cosine curve

from 0 to /2 is sin(/2) =1.

Example 7


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FTC2

If we didnt have the benefit of the FTC,

we would have to compute a difficult limit

of sums using either:

Obscure trigonometric identities

A computer algebra system (CAS), as in Section 5.1


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FTC2

What is wrong with this calculation?

31

3

211

1 1 41

1 3 3

xdx

x

Example 8


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FTC2

To start, we notice that the calculation must

be wrong because the answer is negative

but f(x) = 1/x2 0 and Property 6 of integrals

says that when f 0.( ) 0

b

a f x dx

Example 9


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FTC2

The FTC applies to continuous functions.

It cant be applied here because f(x) = 1/x2

is not continuous on [-1, 3].

In fact, fhas an infinite discontinuity at x= 0.

So, does not exist.3

21

1 dxx

Example 9


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FTC

Suppose fis continuous on [a, b].

1.If , then g(x) = f(x).

2. , where Fis

any antiderivative of f, that is, F= f.

( ) ( )x

ag x f t dt

( ) ( ) ( )b

a f x dx F b F a


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INVERSE PROCESSES

As F(x) = f(x), the FTC2 can be rewritten

as:

This version says that, if we take a function F,first differentiate it, and then integrate the result,we arrive back at the original function F.

However, its in the form F(b) - F(a).

'( ) ( ) ( )b

aF x dx F b F a


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SUMMARY

The FTC is unquestionably the most

important theorem in calculus.

Indeed, it ranks as one of the greataccomplishments of the human mind.


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SUMMARY

Before it was discoveredfrom the time

of Eudoxus and Archimedes to that of Galileo

and Fermatproblems of finding areas,

volumes, and lengths of curves were sodifficult that only a genius could meet

the challenge.


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SUMMARY

Now, armed with the systematic method

that Newton and Leibniz fashioned out of

the theorem, we will see in the chapters to

come that these challenging problems areaccessible to all of us.

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The Fundamental Theorem of Calculus Monday, February