Contributed by:

In this section, we will learn about:

The Fundamental Theorem of Calculus and its significance.

The Fundamental Theorem of Calculus and its significance.

1.
5

2.
5.3

The Fundamental

Theorem of Calculus

In this section, we will learn about:

The Fundamental Theorem of Calculus

and its significance.

The Fundamental

Theorem of Calculus

In this section, we will learn about:

The Fundamental Theorem of Calculus

and its significance.

3.
FUNDAMENTAL THEOREM OF CALCULUS

The Fundamental Theorem of Calculus

(FTC) is appropriately named.

It establishes a connection between the two

branches of calculus—differential calculus and

integral calculus.

The Fundamental Theorem of Calculus

(FTC) is appropriately named.

It establishes a connection between the two

branches of calculus—differential calculus and

integral calculus.

4.
Differential calculus arose from the tangent

Integral calculus arose from a seemingly

unrelated problem—the area problem.

Integral calculus arose from a seemingly

unrelated problem—the area problem.

5.
Newton’s mentor at Cambridge, Isaac Barrow

(1630–1677), discovered that these two

problems are actually closely related.

In fact, he realized that differentiation and

integration are inverse processes.

(1630–1677), discovered that these two

problems are actually closely related.

In fact, he realized that differentiation and

integration are inverse processes.

6.
The FTC gives the precise inverse

relationship between the derivative

and the integral.

relationship between the derivative

and the integral.

7.
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 them

to compute areas and integrals very easily without

having to compute them as limits of sums—as we did

in Sections 5.1 and 5.2

relationship and used it to develop calculus

into a systematic mathematical method.

In particular, they saw that the FTC enabled them

to compute areas and integrals very easily without

having to compute them as limits of sums—as we did

in Sections 5.1 and 5.2

8.
FTC Equation 1

The first part of the FTC deals with functions

defined by an equation of the form

x

g ( x) f (t ) dt

a

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

and x varies between a and b.

The first part of the FTC deals with functions

defined by an equation of the form

x

g ( x) f (t ) dt

a

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

and x varies between a and b.

9.
x

g ( x) f (t ) dt

a

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 x

is a definite number. f (t ) dt

a

x

If we then let x vary, the number f (t ) dt

also varies and defines a function aof x denoted by g(x).

g ( x) f (t ) dt

a

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 x

is a definite number. f (t ) dt

a

x

If we then let x vary, the number f (t ) dt

also varies and defines a function aof x denoted by g(x).

10.
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.

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.

11.
FTC Example 1

If f is the function

whose graph is shown

x

and g ( x) f (t ) dt ,

0

find the values of:

g(0), g(1), g(2), g(3),

g(4), and g(5).

Then, sketch a rough graph of g.

If f is the function

whose graph is shown

x

and g ( x) f (t ) dt ,

0

find the values of:

g(0), g(1), g(2), g(3),

g(4), and g(5).

Then, sketch a rough graph of g.

12.
FTC Example 1

First, we notice that:

0

g (0) f (t ) dt 0

0

First, we notice that:

0

g (0) f (t ) dt 0

0

13.
FTC Example 1

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

the area of a triangle:

1

g (1) f (t ) dt

0

(12)

1

2

1

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

the area of a triangle:

1

g (1) f (t ) dt

0

(12)

1

2

1

14.
FTC Example 1

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

a rectangle:

2

g (2) f (t ) dt

0

1 2

f (t ) dt f (t ) dt

0 1

1 (1 2)

3

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

a rectangle:

2

g (2) f (t ) dt

0

1 2

f (t ) dt f (t ) dt

0 1

1 (1 2)

3

15.
FTC Example 1

We estimate that the area under f from 2 to 3

is about 1.3.

3

So, g (3) g (2)

2

f (t ) dt

3 1.3

4.3

We estimate that the area under f from 2 to 3

is about 1.3.

3

So, g (3) g (2)

2

f (t ) dt

3 1.3

4.3

16.
FTC Example 1

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

So, we start subtracting areas, as

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

So, we start subtracting areas, as

17.
FTC Example 1

4

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

3

5

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

4

4

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

3

5

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

4

18.
FTC Example 1

We use these values to sketch the graph

of g.

Notice that, because f(t)

is positive for t < 3,

we keep adding area

for t < 3.

So, g is increasing up to

x = 3, where it attains

a maximum value.

For x > 3, g decreases

because f(t) is negative.

We use these values to sketch the graph

of g.

Notice that, because f(t)

is positive for t < 3,

we keep adding area

for t < 3.

So, g is increasing up to

x = 3, where it attains

a maximum value.

For x > 3, g decreases

because f(t) is negative.

19.
If we take f(t) = t and a = 0, then,

using Exercise 27 in Section 5.2,

we have: 2

x x

g ( x) tdt

0 2

using Exercise 27 in Section 5.2,

we have: 2

x x

g ( x) tdt

0 2

20.
Notice that g’(x) = x, that is, g’ = f.

In other words, if g is defined as the integral of f

by Equation 1, g turns out to be an antiderivative

of f—at least in this case.

In other words, if g is defined as the integral of f

by Equation 1, g turns out to be an antiderivative

of f—at least in this case.

21.
If we sketch the derivative

of the function g, as in the

first figure, by estimating

slopes of tangents, we get

a graph like that of f in the

second figure.

So, we suspect that g’ = f

in Example 1 too.

of the function g, as in the

first figure, by estimating

slopes of tangents, we get

a graph like that of f in the

second figure.

So, we suspect that g’ = f

in Example 1 too.

22.
To see why this might be generally true, we

consider a continuous function f with f(x) ≥ 0.

x

Then, g ( x ) f (t )dt can be interpreted as

a

the area under the graph of f from a to x.

consider a continuous function f with f(x) ≥ 0.

x

Then, g ( x ) f (t )dt can be interpreted as

a

the area under the graph of f from a to x.

23.
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

It is the area

under the graph

of f from x to x + h

(the gold area).

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

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

It is the area

under the graph

of f from x to x + h

(the gold area).

24.
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:

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

So, g ( x h) g ( x)

h

f ( x)

approximately equal to the area of the

rectangle with height f(x) and width h:

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

So, g ( x h) g ( x)

h

f ( x)

25.
Intuitively, we therefore expect that:

g ( x h) g ( x )

g '( x) lim f ( x)

h 0 h

The fact that this is true, even when f is not

necessarily positive, is the first part of the FTC

(FTC1).

g ( x h) g ( x )

g '( x) lim f ( x)

h 0 h

The fact that this is true, even when f is not

necessarily positive, is the first part of the FTC

(FTC1).

26.
If f is continuous on [a, b], then the function g

defined by x

g ( x) f (t )dt a x b

a

is continuous on [a, b] and differentiable on

(a, b), and g’(x) = f(x).

defined by x

g ( x) f (t )dt a x b

a

is continuous on [a, b] and differentiable on

(a, b), and g’(x) = f(x).

27.
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

of a definite integral with respect to its upper

limit is the integrand evaluated at the upper

28.
FTC1 Proof

If x and x + h are in (a, b), then

g ( x h) g ( x )

x h x

f (t )dt f (t )dt

a a

x

f (t)dt

a

x h

x

f (t )dt

x

f (t )dt

a

(Property 5)

x h

f (t )dt

x

If x and x + h are in (a, b), then

g ( x h) g ( x )

x h x

f (t )dt f (t )dt

a a

x

f (t)dt

a

x h

x

f (t )dt

x

f (t )dt

a

(Property 5)

x h

f (t )dt

x

29.
FTC1 Proof—Equation 2

So, for h ≠ 0,

g ( x h) g ( x ) 1 x h

f (t )dt

h h x

So, for h ≠ 0,

g ( x h) g ( x ) 1 x h

f (t )dt

h h x

30.
FTC1 Proof

For now, let us assume that h > 0.

Since f is continuous on [x, x + h], the Extreme Value

Theorem says that there are numbers u and v in

[x, x + h] such that f(u) = m and f(v) = M.

m and M are the absolute

minimum and maximum

values of f on [x, x + h].

For now, let us assume that h > 0.

Since f is continuous on [x, x + h], the Extreme Value

Theorem says that there are numbers u and v in

[x, x + h] such that f(u) = m and f(v) = M.

m and M are the absolute

minimum and maximum

values of f on [x, x + h].

31.
FTC1 Proof

By Property 8 of integrals, we have:

x h

mh f (t ) dt Mh

x

x h

That is, f (u )h

x

f (t ) dt f (v)h

By Property 8 of integrals, we have:

x h

mh f (t ) dt Mh

x

x h

That is, f (u )h

x

f (t ) dt f (v)h

32.
FTC1 Proof

Since h > 0, we can divide this inequality

by h:

1 x h

f (u ) f (t )dt f (v )

h x

Since h > 0, we can divide this inequality

by h:

1 x h

f (u ) f (t )dt f (v )

h x

33.
FTC1 Proof—Equation 3

Now, we use Equation 2 to replace the middle

part of this inequality:

g ( x h) g ( x )

f (u ) f (v )

h

Inequality 3 can be proved in a similar manner

for the case h < 0.

Now, we use Equation 2 to replace the middle

part of this inequality:

g ( x h) g ( x )

f (u ) f (v )

h

Inequality 3 can be proved in a similar manner

for the case h < 0.

34.
FTC1 Proof

Now, we let h → 0.

Then, u → x and v → x, since u and v lie

between x and x + h.

Therefore, lim f (u ) lim f (u ) f ( x)

h 0 u x

and lim f (v) lim f (v) f ( x)

h 0 v x

because f is continuous at x.

Now, we let h → 0.

Then, u → x and v → x, since u and v lie

between x and x + h.

Therefore, lim f (u ) lim f (u ) f ( x)

h 0 u x

and lim f (v) lim f (v) f ( x)

h 0 v x

because f is continuous at x.

35.
FTC1 Proof—Equation 4

From Equation 3 and the Squeeze

Theorem, we conclude that:

g ( x h) g ( x )

g '( x) lim f ( x)

h 0 h

From Equation 3 and the Squeeze

Theorem, we conclude that:

g ( x h) g ( x )

g '( x) lim f ( x)

h 0 h

36.
If x = a or b, then Equation 4 can be

interpreted as a one-sided limit.

Then, Theorem 4 in Section 2.8 (modified

for one-sided limits) shows that g is continuous

on [a, b].

interpreted as a one-sided limit.

Then, Theorem 4 in Section 2.8 (modified

for one-sided limits) shows that g is continuous

on [a, b].

37.
FTC1 Proof—Equation 5

Using Leibniz notation for derivatives, we can

write the FTC1 as d x

f (t )dt f ( x)

dx a

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.

Using Leibniz notation for derivatives, we can

write the FTC1 as d x

f (t )dt f ( x)

dx a

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.

38.
FTC1 Example 2

Find the derivative of the function

x

2

g ( x) 1 t dt

0

As f (t ) 1 t 2 is continuous, the FTC1 gives:

2

g '( x) 1 x

Find the derivative of the function

x

2

g ( x) 1 t dt

0

As f (t ) 1 t 2 is continuous, the FTC1 gives:

2

g '( x) 1 x

39.
A formula of the form x

g ( x) f (t )dt

a

may seem like a strange way of defining

a function.

However, books on physics, chemistry, and

statistics are full of such functions.

g ( x) f (t )dt

a

may seem like a strange way of defining

a function.

However, books on physics, chemistry, and

statistics are full of such functions.

40.
FRESNEL FUNCTION Example 3

For instance, consider the Fresnel function

x

2

S ( x) sin( t / 2) dt

0

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.

For instance, consider the Fresnel function

x

2

S ( x) sin( t / 2) dt

0

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.

41.
FRESNEL FUNCTION Example 3

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.

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.

42.
FRESNEL FUNCTION Example 3

The figure shows the graphs of

f(x) = sin(πx2/2) and the Fresnel function

x

S ( x) f (t )dt

0

A computer was used

to graph S by computing

the value of this integral

for many values of x.

The figure shows the graphs of

f(x) = sin(πx2/2) and the Fresnel function

x

S ( x) f (t )dt

0

A computer was used

to graph S by computing

the value of this integral

for many values of x.

43.
FRESNEL FUNCTION Example 3

It does indeed look as if S(x) is the area

under the graph of f from 0 to x (until x ≈ 1.4,

when S(x) becomes a difference of areas).

It does indeed look as if S(x) is the area

under the graph of f from 0 to x (until x ≈ 1.4,

when S(x) becomes a difference of areas).

44.
FRESNEL FUNCTION Example 3

The other figure shows a larger part

of the graph of S.

The other figure shows a larger part

of the graph of S.

45.
FRESNEL FUNCTION Example 3

If we now start with the graph of S here and

think about what its derivative should look like,

it seems reasonable that S’(x) = f(x).

For instance, S is

increasing when f(x) > 0

and decreasing when

f(x) < 0.

If we now start with the graph of S here and

think about what its derivative should look like,

it seems reasonable that S’(x) = f(x).

For instance, S is

increasing when f(x) > 0

and decreasing when

f(x) < 0.

46.
FRESNEL FUNCTION Example 3

So, this gives a visual confirmation

of the FTC1.

So, this gives a visual confirmation

of the FTC1.

47.
FTC1 Example 4

d x4

Find sec t dt

dx 1

Here, we have to be careful to use the Chain Rule

in conjunction with the FTC1.

d x4

Find sec t dt

dx 1

Here, we have to be careful to use the Chain Rule

in conjunction with the FTC1.

48.
FTC1 Example 4

Let u = x4.

d x 4

d u

sec t dt sec t dt

dx 1 dx 1

d u du

du 1

sec t dt

dx

(Chain Rule)

du

sec u (FTC1)

dx

4 3

sec( x ) 4 x

Let u = x4.

d x 4

d u

sec t dt sec t dt

dx 1 dx 1

d u du

du 1

sec t dt

dx

(Chain Rule)

du

sec u (FTC1)

dx

4 3

sec( x ) 4 x

49.
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.

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.

50.
If f is continuous on [a, b], then

b

f ( x)dx F (b) F (a)

a

where F is any antiderivative of f,

that is, a function such that F’ = f.

b

f ( x)dx F (b) F (a)

a

where F is any antiderivative of f,

that is, a function such that F’ = f.

51.
FTC2 Proof

x

Let g ( x ) f (t ) dt

a

We know from the FTC1 that g’(x) = f(x),

that is, g is an antiderivative of f.

x

Let g ( x ) f (t ) dt

a

We know from the FTC1 that g’(x) = f(x),

that is, g is an antiderivative of f.

52.
FTC2 Proof—Equation 6

If F is any other antiderivative of f on [a, b],

then we know from Corollary 7 in Section 4.2

that F and g differ by a constant

F(x) = g(x) + C

for a < x < b.

If F is any other antiderivative of f on [a, b],

then we know from Corollary 7 in Section 4.2

that F and g differ by a constant

F(x) = g(x) + C

for a < x < b.

53.
FTC2 Proof

However, both F and g are continuous on

[a, b].

Thus, by taking limits of both sides of

Equation 6 (as x → a+ and x → b- ),

we see it also holds when x = a and x = b.

However, both F and g are continuous on

[a, b].

Thus, by taking limits of both sides of

Equation 6 (as x → a+ and x → b- ),

we see it also holds when x = a and x = b.

54.
FTC2 Proof

If we put x = a in the formula for g(x),

we get:

a

g (a ) f (t ) dt 0

a

If we put x = a in the formula for g(x),

we get:

a

g (a ) f (t ) dt 0

a

55.
FTC2 Proof

So, using Equation 6 with x = b and x = a,

we have:

F (b) F (a ) [ g (b) C ] [ g (a) C ]

g (b) g (a )

g (b)

b

f (t )dt

a

So, using Equation 6 with x = b and x = a,

we have:

F (b) F (a ) [ g (b) C ] [ g (a) C ]

g (b) g (a )

g (b)

b

f (t )dt

a

56.
The FTC2 states that, if we know an

antiderivative F of f, then we can evaluate

b

f ( x)dx

a

simply by subtracting the

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

antiderivative F of f, then we can evaluate

b

f ( x)dx

a

simply by subtracting the

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

57.
b

It’s very surprising that f ( x ) dx , which

a

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.

It’s very surprising that f ( x ) dx , which

a

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.

58.
At first glance, the theorem may be

However, it becomes plausible if we interpret it

in physical terms.

However, it becomes plausible if we interpret it

in physical terms.

59.
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.

is its position at time t, then v(t) = s’(t).

So, s is an antiderivative of v.

60.
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.

b

In symbols,

v(t ) dt s(b) s(a)

a

That is exactly what the FTC2 says in this context.

always moves in the positive direction.

Then, we guessed that the area under the

velocity curve equals the distance traveled.

b

In symbols,

v(t ) dt s(b) s(a)

a

That is exactly what the FTC2 says in this context.

61.
FTC2 Example 5

3

x

Evaluate the integral e dx

1

The function f(x) = ex is continuous everywhere

and we know that an antiderivative is F(x) = ex.

3

x

So, the FTC2 gives:

dx F (3) F (1)

1

e

e3 e

3

x

Evaluate the integral e dx

1

The function f(x) = ex is continuous everywhere

and we know that an antiderivative is F(x) = ex.

3

x

So, the FTC2 gives:

dx F (3) F (1)

1

e

e3 e

62.
FTC2 Example 5

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) = ex, instead of ex + 7 or ex + C.

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) = ex, instead of ex + 7 or ex + C.

63.
We often use the notation

b

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

a

So, the equation of the FTC2 can be written

as: b

b

f ( x)dx F ( x)]

a a where F ' f

b

Other common notations are F ( x ) | and [ F ( x)]ba .

a

b

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

a

So, the equation of the FTC2 can be written

as: b

b

f ( x)dx F ( x)]

a a where F ' f

b

Other common notations are F ( x ) | and [ F ( x)]ba .

a

64.
FTC2 Example 6

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:

3 3 3 1

2

1 x 1 0 1

A x dx

0 3 0 3 3 3

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:

3 3 3 1

2

1 x 1 0 1

A x dx

0 3 0 3 3 3

65.
If you compare the calculation in Example 6

with the one in Example 2 in Section 5.1,

you will see the FTC gives a much shorter

with the one in Example 2 in Section 5.1,

you will see the FTC gives a much shorter

66.
FTC2 Example 7

dx 6

3 x

1 6

The given integral is an abbreviation for dx.

3 x

An antiderivative of f(x) = 1/x is F(x) = ln |x|.

As 3 ≤ x ≤ 6, we can write F(x) = ln x.

dx 6

3 x

1 6

The given integral is an abbreviation for dx.

3 x

An antiderivative of f(x) = 1/x is F(x) = ln |x|.

As 3 ≤ x ≤ 6, we can write F(x) = ln x.

67.
FTC2 Example 7

Therefore, 6 1 6

3 x dx ln x]3

ln 6 ln 3

6

ln

3

ln 2

Therefore, 6 1 6

3 x dx ln x]3

ln 6 ln 3

6

ln

3

ln 2

68.
FTC2 Example 8

Find the area under the cosine curve

from 0 to b, where 0 ≤ b ≤ π/2.

As an antiderivative of f(x) = cos x is F(x) = sin x,

we have:

b

A cos x dx sin x]b0 sin b sin 0 sin b

0

Find the area under the cosine curve

from 0 to b, where 0 ≤ b ≤ π/2.

As an antiderivative of f(x) = cos x is F(x) = sin x,

we have:

b

A cos x dx sin x]b0 sin b sin 0 sin b

0

69.
FTC2 Example 8

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

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

70.
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.

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.

71.
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

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

72.
It was even more difficult for

The apparatus of limits had not been invented

in 1635.

The apparatus of limits had not been invented

in 1635.

73.
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 8.

when the FTC was discovered by Barrow

and exploited by Newton and Leibniz,

such problems became very easy.

You can see this from Example 8.

74.
FTC2 Example 9

What is wrong with this calculation?

1 3

3 1 x 1 4

1 x 2

dx 1

1 1 3 3

What is wrong with this calculation?

1 3

3 1 x 1 4

1 x 2

dx 1

1 1 3 3

75.
FTC2 Example 9

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

b

says that a f ( x ) dx 0 when f ≥ 0.

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

b

says that a f ( x ) dx 0 when f ≥ 0.

76.
FTC2 Example 9

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.

31

So, 2 dx does not exist.

1 x

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.

31

So, 2 dx does not exist.

1 x

77.
INVERSE PROCESSES

We end this section by

bringing together the two parts

of the FTC.

We end this section by

bringing together the two parts

of the FTC.

78.
Suppose f is continuous on [a, b].

x

1.If g ( x) f (t ) dt , then g’(x) = f(x).

a

b

2. f ( x) dx F (b) F (a) , where F is

a

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

x

1.If g ( x) f (t ) dt , then g’(x) = f(x).

a

b

2. f ( x) dx F (b) F (a) , where F is

a

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

79.
INVERSE PROCESSES

We noted that the FTC1 can be rewritten

as: d x

f (t ) dt f ( x )

dx a

This says that, if f is integrated and then

the result is differentiated, we arrive back

at the original function f.

We noted that the FTC1 can be rewritten

as: d x

f (t ) dt f ( x )

dx a

This says that, if f is integrated and then

the result is differentiated, we arrive back

at the original function f.

80.
INVERSE PROCESSES

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

as: b

F '( x ) dx F (b ) F ( a )

a

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, it’s in the form F(b) - F(a).

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

as: b

F '( x ) dx F (b ) F ( a )

a

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, it’s in the form F(b) - F(a).

81.
INVERSE PROCESSES

Taken together, the two parts of the FTC

say that differentiation and integration are

inverse processes.

Each undoes what the other does.

Taken together, the two parts of the FTC

say that differentiation and integration are

inverse processes.

Each undoes what the other does.

82.
The FTC is unquestionably the most

important theorem in calculus.

Indeed, it ranks as one of the great

accomplishments of the human mind.

important theorem in calculus.

Indeed, it ranks as one of the great

accomplishments of the human mind.

83.
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.

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.

84.
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.

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.