## Advanced Calculus (I)

WEN-CHINGLIEN

Department of Mathematics National Cheng Kung University

## 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 ∈ C^{1}[a, b] and

d dx

Z x a

f (t)dt := F^{0}(x ) = f (x )

## 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 ∈ C^{1}[a, b] and

d dx

Z x a

f (t)dt := F^{0}(x ) = f (x )

## 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 ∈ C^{1}[a, b] and

d dx

Z x a

f (t)dt := F^{0}(x ) = f (x )

## 5.3 Fundamental Theorem of Calculus

**R.**

(i)

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

a

f (t)dt, then
F ∈ C^{1}[a, b] and

d dx

Z x a

f (t)dt := F^{0}(x ) = f (x )

Theorem (ii)

If f is differentiable on [a,b] and f^{0} is integrable on [a,b],

then Z x

a

f^{0}(t)dt = f (x ) − f (a)
for each x ∈ [a, b]

Theorem (ii)

If f is differentiable on [a,b] and f^{0} is integrable on [a,b],

then Z x

a

f^{0}(t)dt = f (x ) − f (a)
for each x ∈ [a, b]

**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)

**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)

**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)

**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)

**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)

**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)

(see Definition 4.6) Let > 0 and choose a δ > 0 such
that x0 ≤ t < x_{0}+ 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 x_{0}+h
x0

f (x0)dt.

(see Definition 4.6)Let > 0 and choose a δ > 0 such
that x0 ≤ t < x_{0}+ 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 x_{0}+h
x0

f (x0)dt.

(see Definition 4.6) Let > 0 and choose a δ > 0 such
that x0 ≤ t < x_{0}+ 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 x_{0}+h
x0

f (x0)dt.

(see Definition 4.6) Let > 0 and choose a δ > 0 such
that x0 ≤ t < x_{0}+ 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 x_{0}+h
x0

f (x0)dt.

_{0}+ 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 x_{0}+h
x0

f (x0)dt.

_{0}+ 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 x_{0}+h
x0

f (x0)dt.

Therefore,

F (x0+h) − F (x0)

h − f (x_{0}) = 1
h

Z x_{0}+h
x_{0}

(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 (x_{0})

≤ 1 h

Z x_{0}+h
x0

|f (t) − f (x_{0})|dt ≤

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

Therefore,

F (x0+h) − F (x0)

h − f (x_{0}) = 1
h

Z x_{0}+h
x_{0}

(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 (x_{0})

≤ 1 h

Z x_{0}+h
x0

|f (t) − f (x_{0})|dt ≤

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

Therefore,

F (x0+h) − F (x0)

h − f (x_{0}) = 1
h

Z x_{0}+h
x_{0}

(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 (x_{0})

≤ 1 h

Z x_{0}+h
x0

|f (t) − f (x_{0})|dt ≤

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

Therefore,

F (x0+h) − F (x0)

h − f (x_{0}) = 1
h

Z x_{0}+h
x_{0}

(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 (x_{0})

≤ 1 h

Z x_{0}+h
x0

|f (t) − f (x_{0})|dt ≤

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

Therefore,

F (x0+h) − F (x0)

h − f (x_{0}) = 1
h

Z x_{0}+h
x_{0}

(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 (x_{0})

≤ 1 h

Z x_{0}+h
x0

|f (t) − f (x_{0})|dt ≤

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

Therefore,

F (x0+h) − F (x0)

h − f (x_{0}) = 1
h

Z x_{0}+h
x_{0}

(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 (x_{0})

≤ 1 h

Z x_{0}+h
x0

|f (t) − f (x_{0})|dt ≤

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

(ii)

We may suppose that x = b. Let > 0.Since f^{0} is

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

such that

n

X

j=1

f^{0}(tj)(xj − xj−1) −
Z b

a

f^{0}(t)dt

<

for any choice of points tj ∈ [x_{j−1},xj].

(ii)

We may suppose that x = b. Let > 0. Since f^{0} is

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

such that

n

X

j=1

f^{0}(tj)(xj − xj−1) −
Z b

a

f^{0}(t)dt

<

for any choice of points tj ∈ [x_{j−1},xj].

(ii)

We may suppose that x = b. Let > 0.Since f^{0} is

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

such that

n

X

j=1

f^{0}(tj)(xj − xj−1) −
Z b

a

f^{0}(t)dt

<

for any choice of points tj ∈ [x_{j−1},xj].

(ii)

We may suppose that x = b. Let > 0. Since f^{0} is

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

such that

n

X

j=1

f^{0}(tj)(xj − xj−1) −
Z b

a

f^{0}(t)dt

<

for any choice of points tj ∈ [x_{j−1},xj].

(ii)

We may suppose that x = b. Let > 0. Since f^{0} is

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

such that

n

X

j=1

f^{0}(tj)(xj − xj−1) −
Z b

a

f^{0}(t)dt

<

for any choice of points tj ∈ [x_{j−1},xj].

Use the Mean Value Theorem to choose points

tj ∈ [xj−1,xj]such that f (xj) −f (xj−1) =f^{0}(tj)(xj− xj−1).

It follows by telescoping that

f (b) − f (a) − Z b

a

f^{0}(t)dt

=

n

P

j=1

(f (xj) −f (xj−1)) − Z b

a

f^{0}(t)dt

< .

2

Use the Mean Value Theorem to choose points

tj ∈ [xj−1,xj]such that f (xj) −f (xj−1) =f^{0}(tj)(xj− xj−1).

It follows by telescoping that

f (b) − f (a) − Z b

a

f^{0}(t)dt

=

n

P

j=1

(f (xj) −f (xj−1)) − Z b

a

f^{0}(t)dt

< .

2

Use the Mean Value Theorem to choose points

tj ∈ [xj−1,xj]such that f (xj) −f (xj−1) =f^{0}(tj)(xj− xj−1).

It follows by telescoping that

f (b) − f (a) − Z b

a

f^{0}(t)dt

=

n

P

j=1

(f (xj) −f (xj−1)) − Z b

a

f^{0}(t)dt

< .

2

Use the Mean Value Theorem to choose points

tj ∈ [xj−1,xj]such that f (xj) −f (xj−1) =f^{0}(tj)(xj− xj−1).

It follows by telescoping that

f (b) − f (a) − Z b

a

f^{0}(t)dt

=

n

P

j=1

(f (xj) −f (xj−1)) − Z b

a

f^{0}(t)dt

< .

2

Use the Mean Value Theorem to choose points

tj ∈ [xj−1,xj]such that f (xj) −f (xj−1) =f^{0}(tj)(xj− xj−1).

It follows by telescoping that

f (b) − f (a) − Z b

a

f^{0}(t)dt

=

n

P

j=1

(f (xj) −f (xj−1)) − Z b

a

f^{0}(t)dt

< .

2

Theorem (Integration By parts)

Suppose that f,g are differentiable on [a,b] with f^{0},g^{0}
integrable on [a,b]. Then

Z b a

f^{0}(x )g(x )dx = f (b)g(b) − f (a)g(a) −
Z b

a

f (x )g^{0}(x )dx .

Theorem (Integration By parts)

Suppose that f,g are differentiable on [a,b] with f^{0},g^{0}
integrable on [a,b]. Then

Z b a

f^{0}(x )g(x )dx = f (b)g(b) − f (a)g(a) −
Z b

a

f (x )g^{0}(x )dx .

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 ))φ^{0}(x )dx .

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 ))φ^{0}(x )dx .