section 4.3 the fundamental theorem of calculus...the fundamental theorem of calculus example 2:...

1

Section 4.3 The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus establishes connection between two branches of calculus:differential calculus and integral calculus.

Differential calculus arose from the tangent problem.

Integral calculus arose from the area problem.

The theorem gives the precise inverse relationship between the derivative and the integral.

2

The Fundamental Theorem of Calculus

Isaac Barrow (1630-1677) (Newton’s mentor at Cambridge) discovered that these two problems are closely related. He realized that in fact, differentiation and integration are inverse processes.

Newton and Leibniz exploited this relationship and used it to develop calculus into a systematic mathematical method.

In particular, they saw that this theorem enabled them to compute areas and integrals easily without having to compute them as limits of sums – as we did in 4.1 and 4.2.

3


The Fundamental Theorem of Calculus, part 1

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

by , is continuous on [a,b] and

is differentiable on (a,b), and .

g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

4






We will see two examples first, then we will prove it.

g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

5


Example 1: Find the derivative of the function

g (x)=∫1

x

(2+t 4)5dt

6



Solution: is continuous on R as a polynomial function.

g (x)=∫1

x

(2+t 4)5dt

f (t )=(2+t 4)5

7




Therefore, by The Fundamental Theorem of Calculus, part 1,

g (x)=∫1

x

(2+t 4)5dt

f (t )=(2+t 4)5

g ' (x)=f (x)

8





g (x)=∫1

x

(2+t 4)5dt

f (t )=(2+t 4)5

g ' (x)=f (x)=(2+x4)5

9





g (x)=∫1

x

(2+t 4)5dt

f (t )=(2+t 4)5

g ' (x)=f (x)=(2+x4)5

10


Example 2: Find ddx∫5

x2

(t−t2)dt

11


Example 2: Find ddx∫5

x2

(t−t2)dt





g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

12


Example 2: Find

Solution: is a polynomial function, hence continuous on R.

ddx∫5

x2

(t−t2)dt





g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

f (t )=t−t 2

13


Example 2: Find

Solution: is a polynomial function, hence continuous on R.The upper limit of the integral is x2, not x, so we will use the chain rule.

ddx∫5

x2

(t−t2)dt





g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

f (t )=t−t 2

14


Example 2: Find

Solution: let u = x2, then

ddx∫5

x2

(t−t2)dt





g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

ddx∫5

x2

(t−t2)dt= ddu∫5

u

(t−t2)dt ⋅ dudx

=

15


Example 2: Find


ddx∫5

x2

(t−t2)dt





g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

ddx∫5

x2

(t−t2)dt= ddu∫5

u

(t−t2)dt ⋅ dudx

=

=(u−u2) ⋅ dudx

16


Example 2: Find


ddx∫5

x2

(t−t2)dt





g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

ddx∫5

x2

(t−t2)dt= ddu∫5

u

(t−t2)dt ⋅ dudx

=

=(u−u2) ⋅ dudx

=(x2−(x2)2) ⋅d (x2)dx

17


Example 2: Find


ddx∫5

x2

(t−t2)dt





g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

ddx∫5

x2

(t−t2)dt= ddu∫5

u

(t−t2)dt ⋅ dudx

=

=(u−u2) ⋅ dudx

=(x2−(x2)2) ⋅d (x2)dx

=(x2−x4)(2 x )

18


Example 2: Find


ddx∫5

x2

(t−t2)dt





g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

ddx∫5

x2

(t−t2)dt= ddu∫5

u

(t−t2)dt ⋅ dudx

=

=(u−u2) ⋅ dudx

=(x2−(x2)2) ⋅d (x2)dx

=(x2−x4)(2 x )=2 x3−2 x4

19


Example 2: Find


ddx∫5

x2

(t−t2)dt





g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

ddx∫5

x2

(t−t2)dt= ddu∫5

u

(t−t2)dt ⋅ dudx

=

=(u−u2) ⋅ dudx

=(x2−(x2)2) ⋅d (x2)dx

=(x2−x4)(2 x )=2 x3−2 x4

20


Example 2: Find

Solution: let u = x2, then …

Answer:

ddx∫5

x2

(t−t2)dt





g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

ddx∫5

x2

(t−t2)dt=2 x3−2 x4

21





Proof tools: we will use a definition of the derivative,

,

The Extreme Value Theorem and the Squeeze Theorem to show that .

g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

f ' (x)=limh→0

f (x+h)−f (x)h

g ' (x)=f (x)

22





Proof: (1) assume x, x + h (a,b) , h 0, then

g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

g(x+h)−g(x)=

23






g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

g(x+h)−g(x)=∫a

x+h

f (t )dt−∫a

x

f (t)dt=

24






g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

g(x+h)−g(x)=∫a

x+h

f (t )dt−∫a

x

f (t)dt=∫a

x

f (t)dt+∫x

x+h

f (t )dt−

∫a

x

f (t)dt=

25






g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

g(x+h)−g(x)=∫a

x+h

f (t )dt−∫a

x

f (t)dt=∫a

x

f (t)dt+∫x

x+h

f (t )dt−

∫a

x

f (t)dt=∫x

x+h

f (t )dt

26






, so

g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

g(x+h)−g(x)=∫a

x+h

f (t )dt−∫a

x

f (t)dt=∫a

x

f (t)dt+∫x

x+h

f (t )dt−

∫a

x

f (t)dt=∫x

x+h

f (t )dt g(x+h)−g(x)=∫x

x+h

f (t )dt

27






, so

(2) dividing both sides by h:

g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

g(x+h)−g(x)=∫a

x+h

f (t )dt−∫a

x

f (t)dt=∫a

x

f (t)dt+∫x

x+h

f (t )dt−

∫a

x

f (t)dt=∫x

x+h


x+h

f (t )dt

28






, so

(2) dividing both sides by h:

g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

g(x+h)−g(x)=∫a

x+h

f (t )dt−∫a

x

f (t)dt=∫a

x

f (t)dt+∫x

x+h

f (t )dt−

∫a

x

f (t)dt=∫x

x+h


x+h

f (t )dt

g(x+h)−g(x)h

=1h ∫xx+h

f (t )dt

29





Proof:

g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

g(x+h)−g(x)h

=1h ∫xx+h

f (t )dt

30





Proof:

(3) Let’s assume that h > 0

g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

g(x+h)−g(x)h

=1h ∫xx+h

f (t )dt

31





Proof:

(3) Let’s assume that h > 0, then since f is continuous on [x, x+h], by Extreme Value Theorem, there are numbers u,v [x, x+h], such that f(u) = m is the absolute minimum of f on [x, x+h] and f(v) = M is the absolute maximum of f on [x, x+h].

g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

g(x+h)−g(x)h

=1h ∫xx+h

f (t )dt

32





Proof:


g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

g(x+h)−g(x)h

=1h ∫xx+h

f (t )dta bx x+h

33





Proof:


Therefore, and

g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

g(x+h)−g(x)h

=1h ∫xx+h

f (t )dta bx x+h

m≤f (t )≤M ∫x

x+h

mdt≤∫x

x+h

f (t)dt≤∫x

x+h

M dt

34





Proof:

(3) …

g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

g(x+h)−g(x)h

=1h ∫xx+h

f (t )dta bx x+h

∫x

x+h

mdt≤∫x

x+h

f (t)dt≤∫x

x+h

M dt

35





Proof:

(3) …

g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

g(x+h)−g(x)h

=1h ∫xx+h

f (t )dta bx x+h

∫x

x+h

mdt≤∫x

x+h

f (t)dt≤∫x

x+h

M dt

m(x+h−x)≤∫x

x+h

f (t )dt≤M (x+h−x)then by properties of the integrals:

36





Proof:

(3) …

g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

g(x+h)−g(x)h

=1h ∫xx+h

f (t )dta bx x+h

∫x

x+h

mdt≤∫x

x+h

f (t)dt≤∫x

x+h

M dt

m(x+h−x)≤∫x

x+h


dividing both sides by h: m ≤ 1

h ∫xx+h

f (t )dt ≤ M

37





Proof:

(3) …

g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

g(x+h)−g(x)h

=1h ∫xx+h

f (t )dta bx x+h

∫x

x+h

mdt≤∫x

x+h

f (t)dt≤∫x

x+h

M dt

m(x+h−x)≤∫x

x+h


dividing both sides by h: f (u) = m ≤ 1

h ∫xx+h

f (t )dt ≤ M = f (v)

38





Proof:

(3) …

g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

g(x+h)−g(x)h

=1h ∫xx+h

f (t )dta bx x+h

∫x

x+h

mdt≤∫x

x+h

f (t)dt≤∫x

x+h

M dt

m(x+h−x)≤∫x

x+h


dividing both sides by h: f (u) = m ≤ 1

h ∫xx+h

f (t )dt ≤ M = f (v)

39





Proof:

(3) …

g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

g(x+h)−g(x)h

=1h ∫xx+h

f (t )dta bx x+h

∫x

x+h

mdt≤∫x

x+h

f (t)dt≤∫x

x+h

M dt

m(x+h−x)≤∫x

x+h


dividing both sides by h: f (u) = m ≤

g(x+h)−g(x)h

≤ M = f (v)

40





Proof:

(3) …

g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

a bx x+h

∫x

x+h

mdt≤∫x

x+h

f (t)dt≤∫x

x+h

M dt

m(x+h−x)≤∫x

x+h


dividing both sides by h: f (u) ≤

g(x+h)−g(x)h

≤ f (v)

g(x+h)−g(x)h

=1h ∫xx+h

f (t )dt

41





Proof:

(3) …

g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

a bx x+h

∫x

x+h

mdt≤∫x

x+h

f (t)dt≤∫x

x+h

M dt

m(x+h−x)≤∫x

x+h


dividing both sides by h: f (u) ≤

g(x+h)−g(x)h

≤ f (v)

g(x+h)−g(x)h

=1h ∫xx+h

f (t )dt

We can get the same inequality by considering h < 0

42





Proof:

(4) When h 0, then v x and u x , because they lie between x and h. We get:

g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

a bx x+h

limh→0f (u)=lim

u→xf (u)=f (x)

limh→0f (v)=lim

v→ xf (v)=f (x)

, because f is continuous at x

, because f is continuous at x

43





Proof:

(4) ...

g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

a bx x+h

limh→0f (u)=lim

u→xf (u)=f (x)

limh→0f (v)=lim

v→ xf (v)=f (x)

f (u) ≤g(x+h)−g(x)

h≤ f (v)

by the Squeeze Theorem,

limh→0

g(x+h)−g(x)h

=f (x)

44





Proof:

(4) ...

g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

a bx x+h

limh→0f (u)=lim

u→xf (u)=f (x)

limh→0f (v)=lim

v→ xf (v)=f (x)

f (u) ≤g(x+h)−g(x)

h≤ f (v)

by the Squeeze Theorem,

limh→0

g(x+h)−g(x)h

=f (x)

g ' (x) qed

45



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

where is any antiderivative of f , that is, a function such that .

∫a

b

f (x )dx=F (b)−F (a)

FF '=f

46


The Fundamental Theorem of Calculus, parts 1 and 2






g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

∫a

b


F '=fF

47





Proof:




g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

∫a

b


F '=fF

48





Proof: 1) Let




g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

∫a

b


F '=fF

g (x)=∫a

x

f (t)dt

49





Proof: 1) Let 2) from FTC, part 1, we know that , i.e. g is an antiderivative of f




g (x)=∫a

x

f (t)dt a≤x≤b

g ' (x)=f (x)

∫a

b


F '=fF

g ' (x)=f (x)g (x)=∫

a

x

f (t)dt

50





Proof: 1) Let 2) from FTC, part 1 we know that , i.e. g is an antiderivative of f , therefore the most general antiderivative - constant, for

∫a

b


F '=fF

g ' (x)=f (x)

F (x)=g (x)+C ,C a≤x≤b

g (x)=∫a

x

f (t)dt

51





Proof: 1) Let 2) from FTC, part 1 we know that , i.e. g is an antiderivative of f , therefore the most general antiderivative - constant, for 3) Both, g and F are continuous on [a,b], applying lim to both sides of

∫a

b


F '=fF

g ' (x)=f (x)


g (x)=∫a

x

f (t)dt

F (x)=g (x)+C as x→a+ , x→b− :

52






limx→a+

F (x)= limx→a+

(g (x)+C)

∫a

b


F '=fF

g ' (x)=f (x)


g (x)=∫a

x

f (t)dt

F (x)=g (x)+C as x→a+ , x→b− :

53






F (a)= limx→a+

F (x )= limx→a+

(g(x )+C )=g(a)

∫a

b


F '=fF

g ' (x)=f (x)


g (x)=∫a

x

f (t)dt

F (x)=g (x)+C as x→a+ , x→b− :

54






F (a)= limx→a+

F (x )= limx→a+

(g(x )+C )=g(a)=∫a

a

f (t)dt=0

∫a

b


F '=fF

g ' (x)=f (x)


g (x)=∫a

x

f (t)dt

F (x)=g (x)+C as x→a+ , x→b− :

55






F (a)= limx→a+

F (x )= limx→a+

(g(x )+C )=g(a)=∫a

a

f (t)dt=0

∫a

b


F '=fF

g ' (x)=f (x)


g (x)=∫a

x

f (t)dt

F (x)=g (x)+C as x→a+ , x→b− :

limx→b−

F (x )= limx→b−

(g(x)+C)

56






F (a)= limx→a+

F (x )= limx→a+

(g(x )+C )=g(a)=∫a

a

f (t)dt=0

∫a

b


F '=fF

g ' (x)=f (x)


g (x)=∫a

x

f (t)dt

F (x)=g (x)+C as x→a+ , x→b− :

F (b)= limx→b−

F (x)= limx→b−

(g (x)+C)=g(b)

57






F (a)= limx→a+

F (x )= limx→a+

(g(x )+C )=g(a)=∫a

a

f (t)dt=0

∫a

b


F '=fF

g ' (x)=f (x)


g (x)=∫a

x

f (t)dt

F (x)=g (x)+C as x→a+ , x→b− :

F (b)= limx→b−

F (x)= limx→b−

(g (x)+C)=g(b)=∫a

b

f (t)dt

58






F (a)= limx→a+

F (x )= limx→a+

(g(x )+C )=g(a)=∫a

a

f (t)dt=0

∫a

b


F '=fF

g ' (x)=f (x)


g (x)=∫a

x

f (t)dt

F (x)=g (x)+C as x→a+ , x→b− :

F (b)= limx→b−

F (x)= limx→b−

(g (x)+C)=g(b)=∫a

b

f (t)dt

59






F (a)= limx→a+

F (x )= limx→a+

(g(x )+C )=g(a)=∫a

a

f (t)dt=0

∫a

b


F '=fF

g ' (x)=f (x)


g (x)=∫a

x

f (t)dt

F (x)=g (x)+C as x→a+ , x→b− :

F (b)= limx→b−

F (x)= limx→b−

(g (x)+C)=g(b)=∫a

b

f (t)dt

60


Example 1: Evaluate ∫−1

1

x100dx

61


Example 2: Find the area under the curve from 0 to 1.

y=x3

62


Example 3: What is wrong with this equation?

∫0

π

sec2 x dx=tan π −tan 0=0

63


Note the notation:

∫a

b

f (t)dt=F (b)−F (a)=F (t )]ab, where F ' (t )=f (t )

64



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

1) If , then

2) , where is any antiderivative

of f , i.e. .

g (x)=∫a

x

f (t)dt g ' (x)=f (x)

∫a

b


F '=f

F

65


Example 4: Evaluate the definite integral

∫0

4

(4−t)√t dt

66


Example 4: Evaluate the definite integral

∫0

4

(4−t)√t dt


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

1) If , then

2) , where is any antiderivative

of f , i.e. .

g (x)=∫a

x

f (t)dt g ' (x)=f (x)

∫a

b


F '=f

F

67


Example 5: Use a graph to give a rough estimate of the area of the region that is beneath the given curve. Then find the exact area.

y= 3√ x , 0≤x≤27

section 4.3 the fundamental theorem of calculus...the fundamental theorem of calculus example 2:...

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