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# Advanced Calculus (I)

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## Advanced Calculus (I)

WEN-CHINGLIEN

Department of Mathematics National Cheng Kung University

(2)

## 5.3 Fundamental Theorem of Calculus

Theorem (Fundamental Theorem of Calculus) Let [a,b] be nondegenerate and suppose that f : [a, b] →R.

(i)

If f is continuous on [a,b] and F (x ) = Z x

a

f (t)dt, then F ∈ C1[a, b] and

d dx

Z x a

f (t)dt := F0(x ) = f (x )

(3)

## 5.3 Fundamental Theorem of Calculus

Theorem (Fundamental Theorem of Calculus) Let [a,b] be nondegenerate and suppose that f : [a, b] →R.

(i)

If f is continuous on [a,b] and F (x ) = Z x

a

f (t)dt, then F ∈ C1[a, b] and

d dx

Z x a

f (t)dt := F0(x ) = f (x )

(4)

## 5.3 Fundamental Theorem of Calculus

Theorem (Fundamental Theorem of Calculus) Let [a,b] be nondegenerate and suppose that f : [a, b] →R.

(i)

If f is continuous on [a,b] and F (x ) = Z x

a

f (t)dt, then F ∈ C1[a, b] and

d dx

Z x a

f (t)dt := F0(x ) = f (x )

(5)

## 5.3 Fundamental Theorem of Calculus

Theorem (Fundamental Theorem of Calculus) Let [a,b] be nondegenerate and suppose that f : [a, b] →R.

(i)

If f is continuous on [a,b] and F (x ) = Z x

a

f (t)dt, then F ∈ C1[a, b] and

d dx

Z x a

f (t)dt := F0(x ) = f (x )

(6)

Theorem (ii)

If f is differentiable on [a,b] and f0 is integrable on [a,b],

then Z x

a

f0(t)dt = f (x ) − f (a) for each x ∈ [a, b]

(7)

Theorem (ii)

If f is differentiable on [a,b] and f0 is integrable on [a,b],

then Z x

a

f0(t)dt = f (x ) − f (a) for each x ∈ [a, b]

(8)

### Proof:

(i) Let

F (x ) = Z x

a

f (t)dt, x ∈ [a, b].

By symmetry, it suffices to show that if f (x0+) =f (x0)for some x0 ∈ [a, b), then

(11) lim

h→0+

F (x0+h) − F (x0)

h =f (x0)

(9)

### Proof:

(i) Let

F (x ) = Z x

a

f (t)dt, x ∈ [a, b].

By symmetry,it suffices to show that if f (x0+) =f (x0)for some x0 ∈ [a, b), then

(11) lim

h→0+

F (x0+h) − F (x0)

h =f (x0)

(10)

### Proof:

(i) Let

F (x ) = Z x

a

f (t)dt, x ∈ [a, b].

By symmetry, it suffices to show that if f (x0+) =f (x0)for some x0 ∈ [a, b),then

(11) lim

h→0+

F (x0+h) − F (x0)

h =f (x0)

(11)

### Proof:

(i) Let

F (x ) = Z x

a

f (t)dt, x ∈ [a, b].

By symmetry,it suffices to show that if f (x0+) =f (x0)for some x0 ∈ [a, b), then

(11) lim

h→0+

F (x0+h) − F (x0)

h =f (x0)

(12)

### Proof:

(i) Let

F (x ) = Z x

a

f (t)dt, x ∈ [a, b].

By symmetry, it suffices to show that if f (x0+) =f (x0)for some x0 ∈ [a, b),then

(11) lim

h→0+

F (x0+h) − F (x0)

h =f (x0)

(13)

### Proof:

(i) Let

F (x ) = Z x

a

f (t)dt, x ∈ [a, b].

By symmetry, it suffices to show that if f (x0+) =f (x0)for some x0 ∈ [a, b), then

(11) lim

h→0+

F (x0+h) − F (x0)

h =f (x0)

(14)

(see Definition 4.6) Let  > 0 and choose a δ > 0 such that x0 ≤ t < x0+ δ implies |f (t) − f (x0)| < . Fix 0 < h < δ. Notice that by Theorem 5.20,

F (x0+h) − F (x0) =

Z x0+h x0

f (t)dt

and that by Theorem 5.16

f (x0) = 1 h

Z x0+h x0

f (x0)dt.

(15)

(see Definition 4.6)Let  > 0 and choose a δ > 0 such that x0 ≤ t < x0+ δ implies |f (t) − f (x0)| < . Fix 0 < h < δ. Notice that by Theorem 5.20,

F (x0+h) − F (x0) =

Z x0+h x0

f (t)dt

and that by Theorem 5.16

f (x0) = 1 h

Z x0+h x0

f (x0)dt.

(16)

(see Definition 4.6) Let  > 0 and choose a δ > 0 such that x0 ≤ t < x0+ δ implies |f (t) − f (x0)| < . Fix 0 < h < δ. Notice that by Theorem 5.20,

F (x0+h) − F (x0) =

Z x0+h x0

f (t)dt

and that by Theorem 5.16

f (x0) = 1 h

Z x0+h x0

f (x0)dt.

(17)

(see Definition 4.6) Let  > 0 and choose a δ > 0 such that x0 ≤ t < x0+ δ implies |f (t) − f (x0)| < . Fix 0 < h < δ. Notice that by Theorem 5.20,

F (x0+h) − F (x0) =

Z x0+h x0

f (t)dt

and that by Theorem 5.16

f (x0) = 1 h

Z x0+h x0

f (x0)dt.

(18)

(see Definition 4.6) Let  > 0 and choose a δ > 0 such that x0 ≤ t < x0+ δ implies |f (t) − f (x0)| < . Fix 0 < h < δ. Notice that by Theorem 5.20,

F (x0+h) − F (x0) =

Z x0+h x0

f (t)dt

and that by Theorem 5.16

f (x0) = 1 h

Z x0+h x0

f (x0)dt.

(19)

(see Definition 4.6) Let  > 0 and choose a δ > 0 such that x0 ≤ t < x0+ δ implies |f (t) − f (x0)| < . Fix 0 < h < δ. Notice that by Theorem 5.20,

F (x0+h) − F (x0) =

Z x0+h x0

f (t)dt

and that by Theorem 5.16

f (x0) = 1 h

Z x0+h x0

f (x0)dt.

(20)

Therefore,

F (x0+h) − F (x0)

h − f (x0) = 1 h

Z x0+h x0

(f (t) − f (x0))dt.

Since 0 < h < δ, it follows from Theorem 5.22 and the choice of δ that

F (x0+h) − F (x0)

h − f (x0)

≤ 1 h

Z x0+h x0

|f (t) − f (x0)|dt ≤ 

This verifies (11) and the proof of part (i) is complete.

(21)

Therefore,

F (x0+h) − F (x0)

h − f (x0) = 1 h

Z x0+h x0

(f (t) − f (x0))dt.

Since 0 < h < δ,it follows from Theorem 5.22 and the choice of δ that

F (x0+h) − F (x0)

h − f (x0)

≤ 1 h

Z x0+h x0

|f (t) − f (x0)|dt ≤ 

This verifies (11) and the proof of part (i) is complete.

(22)

Therefore,

F (x0+h) − F (x0)

h − f (x0) = 1 h

Z x0+h x0

(f (t) − f (x0))dt.

Since 0 < h < δ, it follows from Theorem 5.22 and the choice of δ that

F (x0+h) − F (x0)

h − f (x0)

≤ 1 h

Z x0+h x0

|f (t) − f (x0)|dt ≤ 

This verifies (11) and the proof of part (i) is complete.

(23)

Therefore,

F (x0+h) − F (x0)

h − f (x0) = 1 h

Z x0+h x0

(f (t) − f (x0))dt.

Since 0 < h < δ,it follows from Theorem 5.22 and the choice of δ that

F (x0+h) − F (x0)

h − f (x0)

≤ 1 h

Z x0+h x0

|f (t) − f (x0)|dt ≤ 

This verifies (11) and the proof of part (i) is complete.

(24)

Therefore,

F (x0+h) − F (x0)

h − f (x0) = 1 h

Z x0+h x0

(f (t) − f (x0))dt.

Since 0 < h < δ, it follows from Theorem 5.22 and the choice of δ that

F (x0+h) − F (x0)

h − f (x0)

≤ 1 h

Z x0+h x0

|f (t) − f (x0)|dt ≤ 

This verifies (11) and the proof of part (i) is complete.

(25)

Therefore,

F (x0+h) − F (x0)

h − f (x0) = 1 h

Z x0+h x0

(f (t) − f (x0))dt.

Since 0 < h < δ, it follows from Theorem 5.22 and the choice of δ that

F (x0+h) − F (x0)

h − f (x0)

≤ 1 h

Z x0+h x0

|f (t) − f (x0)|dt ≤ 

This verifies (11) and the proof of part (i) is complete.

(26)

(ii)

We may suppose that x = b. Let  > 0.Since f0 is

integrable, choose a partition P = {x0,x1, . . . ,xn} of [a,b]

such that

n

X

j=1

f0(tj)(xj − xj−1) − Z b

a

f0(t)dt

< 

for any choice of points tj ∈ [xj−1,xj].

(27)

(ii)

We may suppose that x = b. Let  > 0. Since f0 is

integrable,choose a partition P = {x0,x1, . . . ,xn} of [a,b]

such that

n

X

j=1

f0(tj)(xj − xj−1) − Z b

a

f0(t)dt

< 

for any choice of points tj ∈ [xj−1,xj].

(28)

(ii)

We may suppose that x = b. Let  > 0.Since f0 is

integrable, choose a partition P = {x0,x1, . . . ,xn} of [a,b]

such that

n

X

j=1

f0(tj)(xj − xj−1) − Z b

a

f0(t)dt

< 

for any choice of points tj ∈ [xj−1,xj].

(29)

(ii)

We may suppose that x = b. Let  > 0. Since f0 is

integrable,choose a partition P = {x0,x1, . . . ,xn} of [a,b]

such that

n

X

j=1

f0(tj)(xj − xj−1) − Z b

a

f0(t)dt

< 

for any choice of points tj ∈ [xj−1,xj].

(30)

(ii)

We may suppose that x = b. Let  > 0. Since f0 is

integrable, choose a partition P = {x0,x1, . . . ,xn} of [a,b]

such that

n

X

j=1

f0(tj)(xj − xj−1) − Z b

a

f0(t)dt

< 

for any choice of points tj ∈ [xj−1,xj].

(31)

Use the Mean Value Theorem to choose points

tj ∈ [xj−1,xj]such that f (xj) −f (xj−1) =f0(tj)(xj− xj−1).

It follows by telescoping that

f (b) − f (a) − Z b

a

f0(t)dt

=

n

P

j=1

(f (xj) −f (xj−1)) − Z b

a

f0(t)dt

< .

2

(32)

Use the Mean Value Theorem to choose points

tj ∈ [xj−1,xj]such that f (xj) −f (xj−1) =f0(tj)(xj− xj−1).

It follows by telescoping that

f (b) − f (a) − Z b

a

f0(t)dt

=

n

P

j=1

(f (xj) −f (xj−1)) − Z b

a

f0(t)dt

< .

2

(33)

Use the Mean Value Theorem to choose points

tj ∈ [xj−1,xj]such that f (xj) −f (xj−1) =f0(tj)(xj− xj−1).

It follows by telescoping that

f (b) − f (a) − Z b

a

f0(t)dt

=

n

P

j=1

(f (xj) −f (xj−1)) − Z b

a

f0(t)dt

< .

2

(34)

Use the Mean Value Theorem to choose points

tj ∈ [xj−1,xj]such that f (xj) −f (xj−1) =f0(tj)(xj− xj−1).

It follows by telescoping that

f (b) − f (a) − Z b

a

f0(t)dt

=

n

P

j=1

(f (xj) −f (xj−1)) − Z b

a

f0(t)dt

< .

2

(35)

Use the Mean Value Theorem to choose points

tj ∈ [xj−1,xj]such that f (xj) −f (xj−1) =f0(tj)(xj− xj−1).

It follows by telescoping that

f (b) − f (a) − Z b

a

f0(t)dt

=

n

P

j=1

(f (xj) −f (xj−1)) − Z b

a

f0(t)dt

< .

2

(36)

Theorem (Integration By parts)

Suppose that f,g are differentiable on [a,b] with f0,g0 integrable on [a,b]. Then

Z b a

f0(x )g(x )dx = f (b)g(b) − f (a)g(a) − Z b

a

f (x )g0(x )dx .

(37)

Theorem (Integration By parts)

Suppose that f,g are differentiable on [a,b] with f0,g0 integrable on [a,b]. Then

Z b a

f0(x )g(x )dx = f (b)g(b) − f (a)g(a) − Z b

a

f (x )g0(x )dx .

(38)

Theorem (Change of Variables)

Let φ be continuously differentiable on a closed, nondegenerate interval [a,b]. If

f is continuous on φ([a, b]),

or if φ is strictly increasing on [a,b] and f is integrable on [φ(a), φ(b)], then

Z φ(b) φ(a)

f (t)dt = Z b

a

f (φ(x ))φ0(x )dx .

(39)

Theorem (Change of Variables)

Let φ be continuously differentiable on a closed, nondegenerate interval [a,b]. If

f is continuous on φ([a, b]),

or if φ is strictly increasing on [a,b] and f is integrable on [φ(a), φ(b)], then

Z φ(b) φ(a)

f (t)dt = Z b

a

f (φ(x ))φ0(x )dx .

(40)

## Thank you.

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung