INTEGRALSINTEGRALS

5

Suumary

1. Definite Integral

2.FTC1,If , then g’(x) = f(x).

3. FTC2, ,

where F is any antiderivative of f, that is,

F’ = f.

( ) ( )x

ag x f t dt

( ) ( ) ( )b

af x dx F b F a

1

( ) lim ( *)nb

ia ni

f x dx f x x

INTEGRALS

We saw in Section 5.1 that a limit of the form

arises when we compute an area.

We also saw that it arises when we try to find the distance traveled by an object.

1

1 2

lim ( *)

lim[ ( *) ( *) ... ( *) ]

n

in

i

nn

f x x

f x x f x x f x x

Equation 1

DEFINITE INTEGRAL

Then, the definite integral of f from a to b is

provided that this limit exists.

If it does exist, we say f is integrable on [a, b].

1

( ) lim ( *)nb

ia ni

f x dx f x x

Definition 2

In the notation ,

f(x) is called the integrand.

a and b are called the limits of integration; a is the lower limit and b is the upper limit.

For now, the symbol dx has no meaning by itself; is all one symbol. The dx simply indicates

that the independent variable is x.

( )b

af x dx

( )b

af x dx

Note 1( )b

af x dxNOTATION

DEFINITE INTEGRAL

The definite integral is a number.

It does not depend on x.

In fact, we could use any letter in place of x

without changing the value of the integral:

( )b

af x dx

( ) ( ) ( )b b b

a a af x dx f t dt f r dr

Note 2( )b

af x dx

RIEMANN SUM

The sum

that occurs in Definition 2 is called

a Riemann sum.

It is named after the German mathematician Bernhard Riemann (1826–1866).

1

( *)n

ii

f x x

Note 3

RIEMANN SUM

So, Definition 2 says that the definite integral

of an integrable function can be approximated

to within any desired degree of accuracy by

a Riemann sum.

Note 3

RIEMANN SUM

We know that, if f happens to be positive,

the Riemann sum can be interpreted as:

A sum of areas of approximating rectangles

Note 3

RIEMANN SUM

Comparing Definition 2 with the definition

of area in Section 5.1, we see that the definite

integral can be interpreted as:

The area under the curve y = f(x) from a to b

( )b

af x dx

Note 3

RIEMANN SUM

If f takes on both positive and negative values, then the

Riemann sum is:

The sum of the areas of the rectangles that lie above the x-axis and the negatives of the areas of the rectangles that lie below the x-axis

That is, the areas of the gold rectangles minus the areas of the blue rectangles

Note 3

RIEMANN SUM

When we take the limit of such

Riemann sums, we get the situation

illustrated here.

Note 3

© Thomson Higher Education

NET AREA

A definite integral can be interpreted as

a net area, that is, a difference of areas:

A1 is the area of the region above the x-axis and below the graph of f.

A2 is the area ofthe region belowthe x-axis andabovethe graph of f.

1 2( )b

af x dx A A

Note 3

© Thomson Higher Education

INTEGRABLE FUNCTIONS

We have defined the definite integral

for an integrable function.

However, not all functions are integrable.

Note 5

INTEGRABLE FUNCTIONS

The following theorem shows that

the most commonly occurring functions

are, in fact, integrable.

It is proved in more advanced courses.

INTEGRABLE FUNCTIONS

If f is continuous on [a, b], or if f has only

a finite number of jump discontinuities, then

f is integrable on [a, b].

That is, the definite integral exists.( )b

af x dx

Theorem 3

INTEGRABLE FUNCTIONS

If f is integrable on [a, b], then the limit

in Definition 2 exists and gives the same

value, no matter how we choose the sample

points xi*.

PROPERTIES OF THE INTEGRAL

We assume f and g are continuous functions.

1. ( ), where c is any constant

2. ( ) ( ) ( ) ( )

3. ( ) ( ) , where c is any constant

4. ( ) ( ) ( ) ( )

b

a

b b b

a a a

b b

a a

b b b

a a a

c dx c b a

f x g x dx f x dx g x dx

c f x dx c f x dx

f x g x dx f x dx g x dx

COMPARISON PROPERTIES OF THE INTEGRAL

These properties, in which we compare sizes

of functions and sizes of integrals, are true

only if a ≤ b.

6. If ( ) 0 for , then ( ) 0

7. If ( ) ( ) for , then ( ) ( )

8. If ( ) for , then

( ) ( ) ( )

b

a

b b

a a

b

a

f x a x b f x dx

f x g x a x b f x dx g x dx

m f x M a x b

m b a f x dx M b a

The Fundamental Theorem of Calculus

(FTC) is appropriately named.

It establishes a connection between the two branches of calculus—differential calculus and integral calculus.

FUNDAMENTAL THEOREM OF CALCULUS

The first part of the FTC deals with functions

defined by an equation of the form

where f is a continuous function on [a, b]

and x varies between a and b.

( ) ( )x

ag x f t dt

Equation 1FTC

Observe that g depends only on x, which appears as the variable upper limit in the integral.

If x is a fixed number, then the integral is a definite number.

If we then let x vary, the number also varies and defines a function of x denoted by g(x).

( ) ( )x

ag x f t dt

( )x

af t dt

( )x

af t dt

FTC

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

can be interpreted as the area under the

graph of f from a to x, where x can vary from a

to b.

Think of g as the ‘area so far’ function, as seen here.

FTC

FTC1

If f is 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

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

Using Leibniz notation for derivatives, we can

write the FTC1 as

when f is continuous.

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

( ) ( )x

a

df t dt f x

dx

Equation 5FTC1

Find the derivative of the function

As is continuous, the FTC1 gives:

Example 2

2

0( ) 1

xg x t dt

2( ) 1f t t 2'( ) 1g x x

FTC1

A formula of the form

may seem like a strange way of defining

a function.

However, books on physics, chemistry, and statistics are full of such functions.

( ) ( )x

ag x f t dt

FTC1 Example 3

FRESNEL FUNCTION

For instance, consider the Fresnel function

It is named after the French physicist Augustin Fresnel (1788–1827), famous for his works in optics.

It first appeared in Fresnel’s theory of the diffraction of light waves.

More recently, it has been applied to the design of highways.

2

0( ) sin( / 2)

xS x t dt

Example 3

FRESNEL FUNCTION

The FTC1 tells us how to differentiate

the Fresnel function:

S’(x) = sin(πx2/2)

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

Example 3

Find

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

4

1sec

xdt dt

dx

Example 4FTC1

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

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 follows easily from the first part, provides us with a much simpler method for the evaluation of integrals.

FTC1

FTC2

If f is continuous on [a, b], then

where F is any antiderivative of f,

that is, a function such that F’ = f.

( ) ( ) ( )b

af x dx F b F a

FTC2

Let

g’(x) = f(x). But F’(x)= f(x), Hence

F(x) – g(x) = K (K constant)

F(a) –g(a) = K

F(a) – 0 = K => F(a) = K and Hence

F(x) – g(x) = F(a) => F(b)-g(b) = F(a)

F(b)-F(a) = g(b) . Therefore

( ) ( )x

ag x f t dt

Proof

( ) ( ) ( )b

af x dx F b F a g(b)=

FTC2

The FTC2 states that, if we know an

antiderivative F of f, then we can evaluate

simply by subtracting the

values

of F at the endpoints of the interval [a, b].

( )b

af x dx

FTC2

It’s very surprising that , which

was defined by a complicated procedure

involving all the values of f(x) for a ≤ x ≤ b,

can be found by knowing the values of F(x)

at only two points, a and b.

( )b

af x dx

FTC2

At first glance, the theorem may be

surprising.

However, it becomes plausible if we interpret it in physical terms.

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, s is an antiderivative of v.

FTC2

In Section 5.1, we considered an object that

always moves in the positive direction.

Then, we guessed that the area under the

velocity curve equals the distance traveled.

In symbols,

That is exactly what the FTC2 says in this context.

( ) ( ) ( )b

av t dt s b s a

FTC2

Evaluate the integral

The function f(x) = x3 is continuous on [-2, 1] and we know from Section 4.9 that an antiderivative is F(x) = ¼x4.

So, the FTC2 gives:

Example 51 3

2 x dx

1 3

2

4 41 14 4

154

(1) ( 2)

1 2

x dx F F

FTC2

Notice that the FTC2 says that we can use any antiderivative F of f.

So, we may as well use the simplest one, namely F(x) = ¼x4, instead of ¼x4 + 7 or ¼x4 + C.

Example 5

FTC2

We often use the notation

So, the equation of the FTC2 can be written

as:

Other common notations are and .

( )] ( ) ( )baF x F b F a

( ) ( )] where 'b b

aaf x dx F x F f

( ) |baF x [ ( )]baF x

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 31 2

00

1 0 1

3 3 3 3

xA x dx

FTC2

Find the area under the cosine curve

from 0 to b, where 0 ≤ b ≤ π/2.

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

Example 7

00cos sin

sin sin 0

sin

b b

A x dx x

b

b

FTC2

In particular, taking b = π/2, we have

proved that the area under the cosine curve

from 0 to π/2 is sin(π/2) =1.

Example 7

FTC2

When the French mathematician Gilles de

Roberval first found the area under the sine

and cosine curves in 1635, this was a very

challenging problem that required a great deal

of ingenuity.

FTC2

If we didn’t 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

FTC2

It was even more difficult for

Roberval.

The apparatus of limits had not been invented in 1635.

FTC2

However, in the 1660s and 1670s,

when the FTC was discovered by Barrow

and exploited by Newton and Leibniz,

such problems became very easy.

You can see this from Example 7.

FTC2

What is wrong with this calculation?

313

211

1 1 41

1 3 3

x

dxx

Example 8

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.( ) 0b

af x dx

Example 9

FTC2

The FTC applies to continuous functions.

It can’t be applied here because f(x) = 1/x2

is not continuous on [-1, 3].

In fact, f has an infinite discontinuity at x = 0.

So, does not exist.3

21

1dx

x

Example 9

INVERSE PROCESSES

We end this section by

bringing together the two parts

of the FTC.

FTC

Suppose f is continuous on [a, b].

1.If , then g’(x) = f(x).

2. , where F is

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

( ) ( )x

ag x f t dt

( ) ( ) ( )b

af x dx F b F a

SUMMARY

The FTC is unquestionably the most

important theorem in calculus.

Indeed, it ranks as one of the great accomplishments of the human mind.

SUMMARY

Before it was discovered—from the time

of Eudoxus and Archimedes to that of Galileo

and Fermat—problems of finding areas,

volumes, and lengths of curves were so

difficult that only a genius could meet

the challenge.

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 are

accessible to all of us.

Suumary

1. Definite Integral

2.FTC1,If , then g’(x) = f(x).

3. FTC2, ,

where F is any antiderivative of f, that is,

F’ = f.

( ) ( )x

ag x f t dt

( ) ( ) ( )b

af x dx F b F a

1

( ) lim ( *)nb

ia ni

f x dx f x x