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 ∈ C1[a, b] and
d dx
Z x a
f (t)dt := F0(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 ∈ C1[a, b] and
d dx
Z x a
f (t)dt := F0(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 ∈ C1[a, b] and
d dx
Z x a
f (t)dt := F0(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 ∈ C1[a, b] and
d dx
Z x a
f (t)dt := F0(x ) = f (x )
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]
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]
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 < 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.
(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.
(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.
(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.
(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.
(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.
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.
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.
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.
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.
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.
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.
(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].
(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].
(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].
(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].
(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].
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
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
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
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
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
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 .
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 .
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 .
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 .
