Volume by Slicing
The Mean Value Theorem for Integrals
Mathematics 53
Institute of Mathematics (UP Diliman)
For today
1 Volumes of Solids by Slicing
2 Mean Value Theorem for Integrals
For today
1 Volumes of Solids by Slicing
2 Mean Value Theorem for Integrals
Volume of a Solid by Slicing
Acylinderis a solid bounded by two congruent plane regionsB1andB2lying in
parallel planes and by a lateral surface formed by all parallel line segments having
as endpoints corresponding points on the boundaries ofB1andB2. If the line
segments are perpendicular toB1andB2, the cylinder is called aright cylinder.
Volume of a cylinder = Area of a cross-section×Height
Volume of a Solid by Slicing
Acylinderis a solid bounded by two congruent plane regionsB1andB2lying in
parallel planes and by a lateral surface formed by all parallel line segments having
as endpoints corresponding points on the boundaries ofB1andB2. If the line
segments are perpendicular toB1andB2, the cylinder is called aright cylinder.
Volume of a cylinder
= Area of a cross-section×Height
Volume of a Solid by Slicing
Acylinderis a solid bounded by two congruent plane regionsB1andB2lying in
parallel planes and by a lateral surface formed by all parallel line segments having
as endpoints corresponding points on the boundaries ofB1andB2. If the line
segments are perpendicular toB1andB2, the cylinder is called aright cylinder.
Volume of a cylinder = Area of a cross-section×Height
Volumes of Solids by Slicing
LetSbe the following solid:
Sis bounded by two parallel planes perpendicular to thex-axis atx=aand
x=b.
The area of a cross-section ofSin a plane perpendicular to thex-axis at an
arbitraryxin[a,b]is given by a continuous function A(x).
Volumes of Solids by Slicing
LetSbe the following solid:
Sis bounded by two parallel planes perpendicular to thex-axis atx=aand
x=b.
The area of a cross-section ofSin a plane perpendicular to thex-axis at an
arbitraryxin[a,b]is given by a continuous function A(x).
Volumes of Solids by Slicing
LetSbe the following solid:
Sis bounded by two parallel planes perpendicular to thex-axis atx=aand
x=b.
The area of a cross-section ofSin a plane perpendicular to thex-axis at an
arbitraryxin[a,b]is given by a continuous function A(x).
Volume of a Solid by Slicing
Divide the interval[a,b]intonsubintervals of equal length∆x
.
Letxi∗be an arbitrary point in theith subinterval.
Volume of theith cylinder:
Vi=
A(x∗i)∆x
Volume of a Solid by Slicing
Divide the interval[a,b]intonsubintervals of equal length∆x.
Letxi∗be an arbitrary point in theith subinterval.
Volume of theith cylinder:
Vi=
A(x∗i)∆x
Volume of a Solid by Slicing
Divide the interval[a,b]intonsubintervals of equal length∆x.
Letxi∗be an arbitrary point in theith subinterval.
Volume of theith cylinder:
Vi=
A(x∗i)∆x
Volume of a Solid by Slicing
Divide the interval[a,b]intonsubintervals of equal length∆x.
Letxi∗be an arbitrary point in theith subinterval.
Volume of theith cylinder:
Vi=
A(x∗i)∆x
Volume of a Solid by Slicing
Divide the interval[a,b]intonsubintervals of equal length∆x.
Letxi∗be an arbitrary point in theith subinterval.
Volume of theith cylinder:
Vi=
A(x∗i)∆x
Volume of a Solid by Slicing
Divide the interval[a,b]intonsubintervals of equal length∆x.
Letxi∗be an arbitrary point in theith subinterval.
Volume of theith cylinder:
Vi=A(x∗i)∆x
Volume of a Solid by Slicing
Approximate the volume of the solid:
V≈ n
∑
i=1
Vi= n
∑
i=1
A(x∗i)∆x
Letn→∞to get the volume of the solid:
V= lim n→∞
n
∑
i=1
A(x∗i)∆x
The volume of the solid is given by
V=
Z b
a A(x)dx.
Volume of a Solid by Slicing
Approximate the volume of the solid:
V≈ n
∑
i=1
Vi= n
∑
i=1
A(x∗i)∆x
Letn→∞to get the volume of the solid:
V= lim n→∞
n
∑
i=1
A(x∗i)∆x
The volume of the solid is given by
V=
Z b
a A(x)dx.
Volume of a Solid by Slicing
Approximate the volume of the solid:
V≈ n
∑
i=1
Vi= n
∑
i=1
A(x∗i)∆x
Letn→∞to get the volume of the solid:
V= lim n→∞
n
∑
i=1
A(x∗i)∆x
The volume of the solid is given by
V=
Z b
a A(x) dx.
Volume of a Solid by Slicing
V=
Z b
a A dw,
where
[a,b]is the intervalIcovered by the region
Ais the area of a cross-section at an arbitrary point in the intervalI
dw=dxif the solid is sliced vertically,dw=dyif the solid is sliced horizontally
Volume of a Solid by Slicing
V=
Z b
a A dw,
where
[a,b]is the intervalIcovered by the region
Ais the area of a cross-section at an arbitrary point in the intervalI
dw=dxif the solid is sliced vertically,dw=dyif the solid is sliced horizontally
Volume of a Solid by Slicing
V=
Z b
a A dw,
where
[a,b]is the intervalIcovered by the region
Ais the area of a cross-section at an arbitrary point in the intervalI
dw=dxif the solid is sliced vertically,dw=dyif the solid is sliced horizontally
Volume of a Solid by Slicing
V=
Z b
a A dw,
where
[a,b]is the intervalIcovered by the region
Ais the area of a cross-section at an arbitrary point in the intervalI
dw=dxif the solid is sliced vertically,dw=dyif the solid is sliced horizontally
Example
Find the volume of the solid shown below (each cross-section is a circle).
Interval: I= [0, 1]
Radius of the cross-section atx:
r(x) = 2 √
x−√x
2 =
√
x
2
Area of the cross-section atx:
A(x) =π[r(x)]2 = πx
4 Volume: V = Z 1 0 πx 4 dx
= πx 2 8
x=1
x=0 = π
8cubic units
Example
Find the volume of the solid shown below (each cross-section is a circle).
Interval:
I= [0, 1]
Radius of the cross-section atx:
r(x) = 2 √
x−√x
2 =
√
x
2
Area of the cross-section atx:
A(x) =π[r(x)]2 = πx
4 Volume: V = Z 1 0 πx 4 dx
= πx 2 8
x=1
x=0 = π
8cubic units
Example
Find the volume of the solid shown below (each cross-section is a circle).
Interval: I= [0, 1]
Radius of the cross-section atx:
r(x) = 2 √
x−√x
2 =
√
x
2
Area of the cross-section atx:
A(x) =π[r(x)]2 = πx
4 Volume: V = Z 1 0 πx 4 dx
= πx 2 8
x=1
x=0 = π
8cubic units
Example
Find the volume of the solid shown below (each cross-section is a circle).
Interval: I= [0, 1]
Radius of the cross-section atx:
r(x) =
2√x−√x
2 =
√
x
2
Area of the cross-section atx:
A(x) =π[r(x)]2 = πx
4
Volume:
V =
Z 1
0
πx
4 dx
= πx 2
8
x=1
x=0 = π
8cubic units
Example
Find the volume of the solid shown below (each cross-section is a circle).
Interval: I= [0, 1]
Radius of the cross-section atx:
r(x) = 2 √
x−√x
2 =
√
x
2
Area of the cross-section atx:
A(x) =π[r(x)]2 = πx
4 Volume: V = Z 1 0 πx 4 dx
= πx 2 8
x=1
x=0 = π
8cubic units
Example
Find the volume of the solid shown below (each cross-section is a circle).
Interval: I= [0, 1]
Radius of the cross-section atx:
r(x) = 2 √
x−√x
2 =
√
x
2
Area of the cross-section atx:
A(x) =
π[r(x)]2 = πx
4 Volume: V = Z 1 0 πx 4 dx
= πx 2 8
x=1
x=0 = π
8cubic units
Example
Find the volume of the solid shown below (each cross-section is a circle).
Interval: I= [0, 1]
Radius of the cross-section atx:
r(x) = 2 √
x−√x
2 =
√
x
2
Area of the cross-section atx:
A(x) =π[r(x)]2
= πx 4 Volume: V = Z 1 0 πx 4 dx
= πx 2 8
x=1
x=0 = π
8cubic units
Example
Find the volume of the solid shown below (each cross-section is a circle).
Interval: I= [0, 1]
Radius of the cross-section atx:
r(x) = 2 √
x−√x
2 =
√
x
2
Area of the cross-section atx:
A(x) =π[r(x)]2 = πx
4 Volume: V = Z 1 0 πx 4 dx
= πx 2 8
x=1
x=0 = π
8cubic units
Example
Find the volume of the solid shown below (each cross-section is a circle).
Interval: I= [0, 1]
Radius of the cross-section atx:
r(x) = 2 √
x−√x
2 =
√
x
2
Area of the cross-section atx:
A(x) =π[r(x)]2 = πx
4 Volume: V = Z 1 0 πx 4 dx
= πx 2 8
x=1
x=0 = π
8cubic units
Example
Find the volume of the solid shown below (each cross-section is a circle).
Interval: I= [0, 1]
Radius of the cross-section atx:
r(x) = 2 √
x−√x
2 =
√
x
2
Area of the cross-section atx:
A(x) =π[r(x)]2 = πx
4 Volume: V = Z 1 0 πx 4 dx
= πx 2 8
x=1
x=0 = π
8cubic units
Example
Find the volume of the solid shown below (each cross-section is a circle).
Interval: I= [0, 1]
Radius of the cross-section atx:
r(x) = 2 √
x−√x
2 =
√
x
2
Area of the cross-section atx:
A(x) =π[r(x)]2 = πx
4 Volume: V = Z 1 0 πx 4 dx
= πx 2 8
x=1
x=0 = π
8cubic units
Example
Find the volume of the solid whose base is the region in the first quadrant
bounded by the graph ofy=1− x2
4 and the coordinate axes, and whose
cross-sections perpendicular to thexy-plane are squares.
x y
Example
Find the volume of the solid whose base is the region in the first quadrant
bounded by the graph ofy=1− x2
4 and the coordinate axes, and whose
cross-sections perpendicular to thexy-plane are squares.
x y
Example
Find the volume of the solid whose base is the region in the first quadrant
bounded by the graph ofy=1− x2
4 and the coordinate axes, and whose
cross-sections perpendicular to thexy-plane are squares.
x y
Example
Find the volume of the solid whose base is the region in the first quadrant
bounded by the graph ofy=1− x2
4 and the coordinate axes, and whose
cross-sections perpendicular to thexy-plane are squares.
x y
Example
Find the volume of the solid whose base is the region in the first quadrant
bounded by the graph ofy=1− x2
4 and the coordinate axes, and whose
cross-sections perpendicular to thexy-plane are squares.
x y
Example
Find the volume of the solid whose base is the region in the first quadrant
bounded by the graph ofy=1− x2
4 and the coordinate axes, and whose
cross-sections perpendicular to thexy-plane are squares.
x y
Example
Find the volume of the solid whose base is the region in the first quadrant
bounded by the graph ofy=1− x2
4 and the coordinate axes, and whose
cross-sections perpendicular to thexy-plane are squares.
x y
Example
Find the volume of the solid whose base is the region in the first quadrant
bounded by the graph ofy=1− x42 and the coordinate axes, and whose
cross-sections perpendicular to thex-axis are squares.
x y
x
y=1−x2
4
(0, 1)
(2, 0)
1−x
2
4
A(x) =
1− x 2
4 2
Example
Find the volume of the solid whose base is the region in the first quadrant
bounded by the graph ofy=1− x42 and the coordinate axes, and whose
cross-sections perpendicular to thex-axis are squares.
x y
x
y=1−x2
4
(0, 1)
(2, 0)
1−x
2
4
A(x) =
1− x 2
4 2
Example
Find the volume of the solid whose base is the region in the first quadrant
bounded by the graph ofy=1− x42 and the coordinate axes, and whose
cross-sections perpendicular to thex-axis are squares.
x y
x
y=1−x2
4
(0, 1)
(2, 0)
1−x
2
4
A(x) =
1− x 2
4 2
Example
Find the volume of the solid whose base is the region in the first quadrant
bounded by the graph ofy=1− x42 and the coordinate axes, and whose
cross-sections perpendicular to thex-axis are squares.
x y
x
y=1−x2
4
(0, 1)
(2, 0)
1−x
2
4
A(x) =
1− x 2
4 2
Example
Find the volume of the solid whose base is the region in the first quadrant
bounded by the graph ofy=1− x42 and the coordinate axes, and whose
cross-sections perpendicular to thex-axis are squares.
x y
x
y=1−x2
4
(0, 1)
(2, 0)
1−x
2
4
A(x) =
1− x 2
4 2
Example
Find the volume of the solid whose base is the region in the first quadrant
bounded by the graph ofy=1− x2
4 and the coordinate axes, and whose
cross-sections perpendicular to thex-axis are squares.
V=
Z 2
0
1−x
2 4 2 dx = Z 2 0 1− x
2 2 + x4 16 dx =
x− x 3 6 + x5 80 2 0 = 16
15 cubic units
Example
Find the volume of the solid whose base is the region in the first quadrant
bounded by the graph ofy=1− x2
4 and the coordinate axes, and whose
cross-sections perpendicular to thex-axis are squares.
V=
Z 2
0
1−x
2 4 2 dx = Z 2 0 1−x
2 2 + x4 16 dx =
x− x 3 6 + x5 80 2 0 = 16
15 cubic units
Example
Find the volume of the solid whose base is the region in the first quadrant
bounded by the graph ofy=1− x2
4 and the coordinate axes, and whose
cross-sections perpendicular to thex-axis are squares.
V=
Z 2
0
1−x
2 4 2 dx = Z 2 0 1−x
2 2 + x4 16 dx =
x− x 3 6 + x5 80 2 0 = 16
15 cubic units
Example
Show that the volume of a square pyramid with heighthand base of side lengths
is 1
3s
2hcubic units.
x y
Each cross-section parallel to the base is also a square.
Example
Show that the volume of a square pyramid with heighthand base of side lengths
is 1
3s
2hcubic units.
x y
Each cross-section parallel to the base is also a square.
Example
Show that the volume of a square pyramid with heighthand base of side lengths
is 1
3s
2hcubic units.
x y
Each cross-section parallel to the base is also a square.
Example
Show that the volume of a square pyramid with heighthand base of side lengths
is 1
3s
2hcubic units.
x y
Each cross-section parallel to the base is also a square.
Example
Show that the volume of a square pyramid with heighthand base of side lengths
is 1
3s
2hcubic units.
x y
Each cross-section parallel to the base is also a square.
Example
Show that the volume of a square pyramid with heighthand base of side lengths
is 1
3s
2hcubic units.
x y
Each cross-section parallel to the base is also a square.
Example
Show that the volume of a square pyramid with heighthand base of side lengths
is 1
3s
2hcubic units.
x y
(x,y)
(h,s
2)
2y
By similar triangles,
y x =
s 2
h =⇒ y=
s
2h ·x
Example
Show that the volume of a square pyramid with heighthand base of side lengths
is 1
3s
2hcubic units.
x y
(x,y)
(h,s
2)
2y
By similar triangles,
y x =
s 2
h =⇒ y=
s
2h ·x
Example
Show that the volume of a square pyramid with heighthand base of side lengths
is 1
3s
2hcubic units.
x y
(x,y)
(h,s
2)
2y
By similar triangles,
y x =
s 2
h =⇒ y=
s
2h ·x
Example
Show that the volume of a square pyramid with heighthand base of side lengths
is 1
3s
2hcubic units.
x y
(x,y)
(h,s
2)
2y
By similar triangles,
y x =
s 2
h =⇒ y=
s
2h ·x
Example
Show that the volume of a square pyramid with heighthand base of side lengths
is 1
3s
2hcubic units.
x y
(x,y)
(h,s
2)
2y
By similar triangles,
y x =
s 2
h
=⇒ y= s
2h ·x
Example
Show that the volume of a square pyramid with heighthand base of side lengths
is 1
3s
2hcubic units.
x y
(x,y)
(h,s
2)
2y
By similar triangles,
y x =
s 2
h =⇒ y=
s
2h ·x
Example
Show that the volume of a square pyramid with heighthand base of side lengths
is 1
3s
2hcubic units.
A(x) = (2y)2=
2· s
2h ·x 2
= s 2
h2 ·x 2
V=
Z h
0
s2 h2·x
2dx
=
s2 h2 ·
x3 3 h 0 = s 2
h2·
h3 3 −0
= 1 3s
2hcubic units
Example
Show that the volume of a square pyramid with heighthand base of side lengths
is 1
3s
2hcubic units.
A(x) = (2y)2= 2· s 2h ·x
2
= s 2
h2 ·x 2
V=
Z h
0
s2 h2·x
2dx
=
s2 h2 ·
x3 3 h 0 = s 2
h2·
h3 3 −0
= 1 3s
2hcubic units
Example
Show that the volume of a square pyramid with heighthand base of side lengths
is 1
3s
2hcubic units.
A(x) = (2y)2= 2· s 2h ·x
2 = s
2
h2 ·x 2
V=
Z h
0
s2 h2·x
2dx
=
s2 h2 ·
x3 3 h 0 = s 2
h2·
h3 3 −0
= 1 3s
2hcubic units
Example
Show that the volume of a square pyramid with heighthand base of side lengths
is 1
3s
2hcubic units.
A(x) = (2y)2= 2· s 2h ·x
2 = s
2
h2 ·x 2
V=
Z h
0
s2 h2·x
2dx
=
s2 h2 ·
x3 3 h 0 = s 2
h2·
h3 3 −0
= 1 3s
2hcubic units
Example
Show that the volume of a square pyramid with heighthand base of side lengths
is 1
3s
2hcubic units.
A(x) = (2y)2= 2· s 2h ·x
2 = s
2
h2 ·x 2
V=
Z h
0
s2 h2·x
2dx
=
s2 h2 ·
x3 3 h 0 = s 2
h2·
h3 3 −0
= 1 3s
2hcubic units
Example
Show that the volume of a square pyramid with heighthand base of side lengths
is 1
3s
2hcubic units.
A(x) = (2y)2= 2· s 2h ·x
2 = s
2
h2 ·x 2
V=
Z h
0
s2 h2·x
2dx
=
s2 h2 ·
x3 3 h 0 = s 2
h2·
h3 3 −0
= 1 3s
2hcubic units
Example
Show that the volume of a square pyramid with heighthand base of side lengths
is 1
3s
2hcubic units.
A(x) = (2y)2= 2· s 2h ·x
2 = s
2
h2 ·x 2
V=
Z h
0
s2 h2·x
2dx
=
s2 h2 ·
x3 3 h 0 = s 2
h2·
h3 3 −0
= 1 3s
2hcubic units
For today
1 Volumes of Solids by Slicing
2 Mean Value Theorem for Integrals
Some Theorems
Theorem
If the functions f andgare integrable on[a,b], and if f(x)≥g(x)for allxin
[a,b],
then
Z b
a f (x)dx
≥
Z b
a g (x)dx.
Some Theorems
Theorem
If the functions f andgare integrable on[a,b], and if f(x)≥g(x)for allxin
[a,b], then
Z b
a f (x)dx
≥
Z b
a g (x)dx.
Some Theorems
Theorem
If the functions f andgare integrable on[a,b], and if f(x)≥g(x)for allxin
[a,b], then
Z b
a f (x)dx
≥
Z b
a g (x)dx.
Some Theorems
Theorem
If the functions f andgare integrable on[a,b], and if f(x)≥g(x)for allxin
[a,b], then
Z b
a f
(x)dx≥
Z b
a g (x)dx.
Some Theorems
Theorem
Suppose f is continuous on the closed interval[a,b]. Ifmis the absolute minimum
function value andMis the absolute maximum function value of f on[a,b], then
m(b−a)
≤
Z b
a f(x)dx
≤
M(b−a).
Some Theorems
Theorem
Suppose f is continuous on the closed interval[a,b]. Ifmis the absolute minimum
function value andMis the absolute maximum function value of f on[a,b], then
m(b−a)
≤
Z b
a f (x)dx
≤
M(b−a).
Some Theorems
Theorem
Suppose f is continuous on the closed interval[a,b]. Ifmis the absolute minimum
function value andMis the absolute maximum function value of f on[a,b], then
m(b−a)
≤
Z b
a f (x)dx
≤
M(b−a).
Some Theorems
Theorem
Suppose f is continuous on the closed interval[a,b]. Ifmis the absolute minimum
function value andMis the absolute maximum function value of f on[a,b], then
m(b−a)≤
Z b
a f
(x)dx≤M(b−a).
The Mean Value Theorem for Integrals
Theorem
If the function f is continuous on the closed interval[a,b], then
there exists a
numbercin[a,b]such that
Z b
a f(x)dx= f(c)(b−a).
The Mean Value Theorem for Integrals
Theorem
If the function f is continuous on the closed interval[a,b], then there exists a
numbercin[a,b]such that
Z b
a f
(x)dx= f(c)(b−a).
The Mean Value Theorem for Integrals
Theorem
If the function f is continuous on the closed interval[a,b], then there exists a
numbercin[a,b]such that
Z b
a f
(x)dx= f(c)(b−a).
Proof of MVTI
Suppose f is continuous on[a,b].
By the Extreme Value Theorem, f attains an absolute minimummand an
absolute maximumMon[a,b]. By a theorem earlier,
m≤
Z b
a f (x)dx
b−a ≤M.
By the Intermediate Value Theorem, there existsc∈[a,b]such that
F0(c) = F(b)−F(a) b−a .
f(c) =
Z b
a f(x)dx
b−a .
Thus,
Z b
a f(x)dx= f(c)(b−a).
Proof of MVTI
Suppose f is continuous on[a,b].
By the Extreme Value Theorem, f attains an absolute minimummand an
absolute maximumMon[a,b].
By a theorem earlier,
m≤
Z b
a f (x)dx
b−a ≤M.
By the Intermediate Value Theorem, there existsc∈[a,b]such that
F0(c) = F(b)−F(a) b−a .
f(c) =
Z b
a f(x)dx
b−a .
Thus,
Z b
a f(x)dx= f(c)(b−a).
Proof of MVTI
Suppose f is continuous on[a,b].
By the Extreme Value Theorem, f attains an absolute minimummand an
absolute maximumMon[a,b]. By a theorem earlier,
m≤
Z b
a f (x)dx
b−a ≤M.
By the Intermediate Value Theorem, there existsc∈[a,b]such that
F0(c) = F(b)−F(a) b−a .
f(c) =
Z b
a f(x)dx
b−a .
Thus,
Z b
a f(x)dx= f(c)(b−a).
Proof of MVTI
Suppose f is continuous on[a,b].
By the Extreme Value Theorem, f attains an absolute minimummand an
absolute maximumMon[a,b]. By a theorem earlier,
m(b−a)≤
Z b
a f(x)dx≤M(b−a)
m≤
Z b
a f (x)dx
b−a ≤M.
By the Intermediate Value Theorem, there existsc∈[a,b]such that
F0(c) = F(b)−F(a) b−a .
f(c) =
Z b
a f(x)dx
b−a .
Thus,
Z b
a f(x)dx= f(c)(b−a).
Proof of MVTI
Suppose f is continuous on[a,b].
By the Extreme Value Theorem, f attains an absolute minimummand an
absolute maximumMon[a,b]. By a theorem earlier,
m≤
Z b
a f (x)dx
b−a ≤M.
By the Intermediate Value Theorem, there existsc∈[a,b]such that
F0(c) = F(b)−F(a) b−a .
f(c) =
Z b
a f(x)dx
b−a .
Thus,
Z b
a f(x)dx= f(c)(b−a).
Proof of MVTI
Suppose f is continuous on[a,b].
By the Extreme Value Theorem, f attains an absolute minimummand an
absolute maximumMon[a,b]. By a theorem earlier,
m≤
Z b
a f (x)dx
b−a ≤M.
By the Intermediate Value Theorem,
there existsc∈[a,b]such that
F0(c) = F(b)−F(a) b−a .
f(c) =
Z b
a f(x)dx
b−a .
Thus,
Z b
a f(x)dx= f(c)(b−a).
Proof of MVTI
Suppose f is continuous on[a,b].
By the Extreme Value Theorem, f attains an absolute minimummand an
absolute maximumMon[a,b]. By a theorem earlier,
m≤
Z b
a f (x)dx
b−a ≤M.
By the Intermediate Value Theorem, there existsc∈[a,b]such that
F0(c) = F(b)−F(a) b−a .
f(c) =
Z b
a f(x)dx
b−a .
Thus,
Z b
a f(x)dx= f(c)(b−a).
Proof of MVTI
Suppose f is continuous on[a,b].
By the Extreme Value Theorem, f attains an absolute minimummand an
absolute maximumMon[a,b]. By a theorem earlier,
m≤
Z b
a f (x)dx
b−a ≤M.
By the Intermediate Value Theorem, there existsc∈[a,b]such that
F0(c) = F(b)−F(a) b−a .
f(c)
=
Z b
a f(x)dx
b−a .
Thus,
Z b
a f(x)dx= f(c)(b−a).
Proof of MVTI
Suppose f is continuous on[a,b].
By the Extreme Value Theorem, f attains an absolute minimummand an
absolute maximumMon[a,b]. By a theorem earlier,
m≤
Z b
a f (x)dx
b−a ≤M.
By the Intermediate Value Theorem, there existsc∈[a,b]such that
F0(c) = F(b)−F(a) b−a .
f(c) =
Z b
a f(x)dx
b−a .
Thus,
Z b
a f(x)dx= f(c)(b−a).
Proof of MVTI
Suppose f is continuous on[a,b].
By the Extreme Value Theorem, f attains an absolute minimummand an
absolute maximumMon[a,b]. By a theorem earlier,
m≤
Z b
a f (x)dx
b−a ≤M.
By the Intermediate Value Theorem, there existsc∈[a,b]such that
F0(c) = F(b)−F(a) b−a .
f(c) =
Z b
a f(x)dx
b−a .
Thus,
Z b
a f(x)dx= f(c)(b−a).
Proof of MVTI
Suppose f is continuous on[a,b].
By the Extreme Value Theorem, f attains an absolute minimummand an
absolute maximumMon[a,b]. By a theorem earlier,
m≤
Z b
a f (x)dx
b−a ≤M.
By the Intermediate Value Theorem, there existsc∈[a,b]such that
F0(c) = F(b)−F(a) b−a .
f(c) =
Z b
a f(x)dx
b−a .
Thus,
Z b
a f(x)dx= f(c)(b−a).
Example
Since f(x) =x2is continuous on[0, 2], there existscin[0, 2]satisfying
Z 2
0 x
2dx = c2(2−0) Try to findc.
x3 3
x=2
x=0
= 2c2
8
3 = 2c 2
4
3 = c 2
c = ±2
√ 3 3
Butc∈[0, 2], soc= 2
√ 3 3 .
Remark:In general, the value ofcis not unique.
Example
Since f(x) =x2is continuous on[0, 2], there existscin[0, 2]satisfying
Z 2
0 x 2dx
= c2(2−0) Try to findc.
x3 3
x=2
x=0
= 2c2
8
3 = 2c 2
4
3 = c 2
c = ±2
√ 3 3
Butc∈[0, 2], soc= 2
√ 3 3 .
Remark:In general, the value ofcis not unique.
Example
Since f(x) =x2is continuous on[0, 2], there existscin[0, 2]satisfying
Z 2
0 x
2dx = c2(2−0)
Try to findc.
x3 3
x=2
x=0
= 2c2
8
3 = 2c 2
4
3 = c 2
c = ±2
√ 3 3
Butc∈[0, 2], soc= 2
√ 3 3 .
Remark:In general, the value ofcis not unique.
Example
Since f(x) =x2is continuous on[0, 2], there existscin[0, 2]satisfying
Z 2
0 x
2dx = c2(2−0) Try to findc.
x3 3
x=2
x=0
= 2c2
8
3 = 2c 2
4
3 = c 2
c = ±2
√ 3 3
Butc∈[0, 2], soc= 2
√ 3 3 .
Remark:In general, the value ofcis not unique.
Example
Since f(x) =x2is continuous on[0, 2], there existscin[0, 2]satisfying
Z 2
0 x
2dx = c2(2−0) Try to findc.
x3 3
x=2
x=0
= 2c2
8
3 = 2c 2
4
3 = c 2
c = ±2
√ 3 3
Butc∈[0, 2], soc= 2
√ 3 3 .
Remark:In general, the value ofcis not unique.
Example
Since f(x) =x2is continuous on[0, 2], there existscin[0, 2]satisfying
Z 2
0 x
2dx = c2(2−0) Try to findc.
x3 3
x=2
x=0
= 2c2
8
3 = 2c 2
4
3 = c 2
c = ±2
√ 3 3
Butc∈[0, 2], soc= 2
√ 3 3 .
Remark:In general, the value ofcis not unique.
Example
Since f(x) =x2is continuous on[0, 2], there existscin[0, 2]satisfying
Z 2
0 x
2dx = c2(2−0) Try to findc.
x3 3
x=2
x=0
= 2c2
8
3 = 2c 2
4
3 = c 2
c = ±2
√ 3 3
Butc∈[0, 2], soc= 2
√ 3 3 .
Remark:In general, the value ofcis not unique.
Example
Since f(x) =x2is continuous on[0, 2], there existscin[0, 2]satisfying
Z 2
0 x
2dx = c2(2−0) Try to findc.
x3 3
x=2
x=0
= 2c2
8
3 = 2c 2
4
3 = c 2
c = ±2
√ 3 3
Butc∈[0, 2], soc= 2
√ 3 3 .
Remark:In general, the value ofcis not unique.
Example
Since f(x) =x2is continuous on[0, 2], there existscin[0, 2]satisfying
Z 2
0 x
2dx = c2(2−0) Try to findc.
x3 3
x=2
x=0
= 2c2
8
3 = 2c 2
4
3 = c 2
c = ±2
√ 3 3
Butc∈[0, 2], so
c= 2 √
3 3 .
Remark:In general, the value ofcis not unique.
Example
Since f(x) =x2is continuous on[0, 2], there existscin[0, 2]satisfying
Z 2
0 x
2dx = c2(2−0) Try to findc.
x3 3
x=2
x=0
= 2c2
8
3 = 2c 2
4
3 = c 2
c = ±2
√ 3 3
Butc∈[0, 2], soc= 2
√ 3 3 .
Remark:In general, the value ofcis not unique.
Example
Since f(x) =x2is continuous on[0, 2], there existscin[0, 2]satisfying
Z 2
0 x
2dx = c2(2−0) Try to findc.
x3 3
x=2
x=0
= 2c2
8
3 = 2c 2
4
3 = c 2
c = ±2
√ 3 3
Butc∈[0, 2], soc= 2
√ 3 3 .
Remark:In general, the value ofcis not unique.
2 4
2√3 3 4
3
2 4
2√3 3 4
3
2 4
2√3 3
4 3
2 4
2√3 3 4
3
A Comparison
Theorem (Mean Value Theorem)
If f is cont. on[a,b]and diff. on(a,b), then there existsc∈(a,b)such that
f0(c) = f(b)−f(a) b−a .
Theorem (Mean Value Theorem for Integrals)
If f is cont. on[a,b], then there existsc∈[a,b]such that
f(c) =
Z b
a f(x)dx
b−a .
Average Value
The value of f(c)in the Mean Value Theorem is called theaverage valueof f
on the interval[a,b].
It is a generalization of the arithmetic mean of a finite set of numbers.
Definition
If the function f is integrable on[a,b], theaverage valueof f on[a,b]is
fave=
Z b
a f (x)dx
b−a .
Average Value
The value of f(c)in the Mean Value Theorem is called theaverage valueof f
on the interval[a,b].
It is a generalization of the arithmetic mean of a finite set of numbers.
Definition
If the function f is integrable on[a,b], theaverage valueof f on[a,b]is
fave=
Z b
a f (x)dx
b−a .
Average Value
The value of f(c)in the Mean Value Theorem is called theaverage valueof f
on the interval[a,b].
It is a generalization of the arithmetic mean of a finite set of numbers.
Definition
If the function f is integrable on[a,b], theaverage valueof f on[a,b]is
fave=
Z b
a f (x)dx
b−a .
Average Value
Examples
1 The average value of f(x) =x2on[0, 2]is
fave=
Z 2
0 x 2dx
2−0 = 8 3 2 =
4 3.
2 The average value of f(x) =cosxon0,π
2
is
fave=
Z π
2
0 cosx dx
π
2 −0 =
(sinx)
x=π2
x=0
π
2
= 1−π0 2
= 2
π.
Average Value
Examples
1 The average value of f(x) =x2on[0, 2]is
fave =
Z 2
0 x 2dx
2−0 = 8 3 2 =
4 3.
2 The average value of f(x) =cosxon0,π
2
is
fave=
Z π
2
0 cosx dx
π
2 −0 =
(sinx)
x=π2
x=0
π
2
= 1−π0 2
= 2
π.
Average Value
Examples
1 The average value of f(x) =x2on[0, 2]is
fave =
Z 2
0 x 2dx
2−0 = 8 3 2 =
4 3.
2 The average value of f(x) =cosxon0,π
2
is
fave=
Z π
2
0 cosx dx
π
2 −0 =
(sinx)
x=π2
x=0
π
2
= 1−π0 2
= 2
π.
Average Value
Examples
1 The average value of f(x) =x2on[0, 2]is
fave =
Z 2
0 x 2dx
2−0
= 8 3 2 = 4 3.
2 The average value of f(x) =cosxon0,π
2
is
fave=
Z π
2
0 cosx dx
π
2 −0 =
(sinx)
x=π2
x=0
π
2
= 1−π0 2
= 2
π.
Average Value
Examples
1 The average value of f(x) =x2on[0, 2]is
fave =
Z 2
0 x 2dx
2−0 = 8 3 2
= 4 3.
2 The average value of f(x) =cosxon0,π
2
is
fave=
Z π
2
0 cosx dx
π
2 −0 =
(sinx)
x=π2
x=0
π
2
= 1−π0 2
= 2
π.
Average Value
Examples
1 The average value of f(x) =x2on[0, 2]is
fave =
Z 2
0 x 2dx
2−0 = 8 3 2 =
4 3.
2 The average value of f(x) =cosxon0,π
2
is
fave=
Z π
2
0 cosx dx
π
2 −0 =
(sinx)
x=π2
x=0
π
2
= 1−π0 2
= 2
π.
Average Value
Examples
1 The average value of f(x) =x2on[0, 2]is
fave =
Z 2
0 x 2dx
2−0 = 8 3 2 =
4 3.
2 The average value of f(x) =cosxon0,π
2
is
fave=
Z π
2
0 cosx dx
π
2 −0 =
(sinx)
x=π2
x=0
π
2
= 1−π0 2
= 2
π.
Average Value
Examples
1 The average value of f(x) =x2on[0, 2]is
fave =
Z 2
0 x 2dx
2−0 = 8 3 2 =
4 3.
2 The average value of f(x) =cosxon0,π
2
is
fave =
Z π
2
0 cosx dx
π
2 −0 =
(sinx)
x=π2
x=0
π
2
= 1−π0 2
= 2
π.
Average Value
Examples
1 The average value of f(x) =x2on[0, 2]is
fave =
Z 2
0 x 2dx
2−0 = 8 3 2 =
4 3.
2 The average value of f(x) =cosxon0,π
2
is
fave =
Z π
2
0 cosx dx
π
2 −0 =
(sinx)
x=π2
x=0
π
2
= 1−π0 2
= 2
π.
Average Value
Examples
1 The average value of f(x) =x2on[0, 2]is
fave =
Z 2
0 x 2dx
2−0 = 8 3 2 =
4 3.
2 The average value of f(x) =cosxon0,π
2
is
fave =
Z π
2
0 cosx dx
π
2 −0
=
(sinx)
x=π2
x=0
π
2
= 1−π0 2
= 2
π.
Average Value
Examples
1 The average value of f(x) =x2on[0, 2]is
fave =
Z 2
0 x 2dx
2−0 = 8 3 2 =
4 3.
2 The average value of f(x) =cosxon0,π
2
is
fave =
Z π
2
0 cosx dx
π
2 −0 =
(sinx)
x=π2
x=0
π
2
= 1−π0 2
= 2
π.
Average Value
Examples
1 The average value of f(x) =x2on[0, 2]is
fave =
Z 2
0 x 2dx
2−0 = 8 3 2 =
4 3.
2 The average value of f(x) =cosxon0,π
2
is
fave =
Z π
2
0 cosx dx
π
2 −0 =
(sinx)
x=π2
x=0
π
2
= 1−π0 2
= 2
π.
Average Value
Examples
1 The average value of f(x) =x2on[0, 2]is
fave =
Z 2
0 x 2dx
2−0 = 8 3 2 =
4 3.
2 The average value of f(x) =cosxon0,π
2
is
fave =
Z π
2
0 cosx dx
π
2 −0 =
(sinx)
x=π2
x=0
π
2
= 1−π0 2
= 2
π.
Theorem
If f is cont. on[a,b], then there existsc∈[a,b]such that
f(c) =
Z b
a f(x)dx
b−a .
The MVTI states that if f is cont. on[a,b], then there is ac∈[a,b]such that f(c)
is equal to the average value of f on[a,b].
Theorem
If f is cont. on[a,b], then there existsc∈[a,b]such that
f(c) =
Z b
a f(x)dx
b−a .
The MVTI states that if f is cont. on[a,b], then there is ac∈[a,b]such that f(c)
is equal to the average value of f on[a,b].
2 4
2√3 3 4
3
Proof of the First Fundamental Theorem of Calculus
Theorem
Let f be a function continuous on[a,b]and letxbe any number in[a,b]. IfFis
the function defined byF(x) =
Z x
a f
(t)dt, then
F0(x) = f(x).
F0(x) = lim
∆x→0
F(x+∆x)−F(x)
∆x
Note that
F(x+∆x)−F(x)
∆x =
Z x+∆x
a f
(t)dt−
Z x
a f (t)dt ∆x
=
Z x+∆x
x f
(t)dt
∆x .
Proof of the First Fundamental Theorem of Calculus
Theorem
Let f be a function continuous on[a,b]and letxbe any number in[a,b]. IfFis
the function defined byF(x) =
Z x
a f
(t)dt, then
F0(x) = f(x).
F0(x) =
lim
∆x→0
F(x+∆x)−F(x)
∆x
Note that
F(x+∆x)−F(x)
∆x =
Z x+∆x
a f
(t)dt−
Z x
a f (t)dt ∆x
=
Z x+∆x
x f
(t)dt
∆x .
Proof of the First Fundamental Theorem of Calculus
Theorem
Let f be a function continuous on[a,b]and letxbe any number in[a,b]. IfFis
the function defined byF(x) =
Z x
a f
(t)dt, then
F0(x) = f(x).
F0(x) = lim
∆x→0
F(x+∆x)−F(x)
∆x
Note that
F(x+∆x)−F(x)
∆x =
Z x+∆x
a f
(t)dt−
Z x
a f (t)dt ∆x
=
Z x+∆x
x f
(t)dt
∆x .
Proof of the First Fundamental Theorem of Calculus
Theorem
Let f be a function continuous on[a,b]and letxbe any number in[a,b]. IfFis
the function defined byF(x) =
Z x
a f
(t)dt, then
F0(x) = f(x).
F0(x) = lim
∆x→0
F(x+∆x)−F(x)
∆x
Note that
F(x+∆x)−F(x)
∆x =
Z x+∆x
a f
(t)dt−
Z x
a f (t)dt ∆x
=
Z x+∆x
x f
(t)dt
∆x .
Proof of the First Fundamental Theorem of Calculus
Theorem
Let f be a function continuous on[a,b]and letxbe any number in[a,b]. IfFis
the function defined byF(x) =
Z x
a f
(t)dt, then
F0(x) = f(x).
F0(x) = lim
∆x→0
F(x+∆x)−F(x)
∆x
Note that
F(x+∆x)−F(x)
∆x =
Z x+∆x
a f
(t)dt−
Z x
a f (t)dt ∆x
=
Z x+∆x
x f
(t)dt
∆x .
Proof of the First Fundamental Theorem of Calculus
Theorem
Let f be a function continuous on[a,b]and letxbe any number in[a,b]. IfFis
the function defined byF(x) =
Z x
a f
(t)dt, then
F0(x) = f(x).
F0(x) = lim
∆x→0
F(x+∆x)−F(x)
∆x
Note that
F(x+∆x)−F(x)
∆x =
Z x+∆x
a f
(t)dt−
Z x
a f (t)dt ∆x
=
Z x+∆x
x f
(t)dt
∆x .
Proof of the First Fundamental Theorem of Calculus
Theorem
Let f be a function continuous on[a,b]and letxbe any number in[a,b]. IfFis
the function defined byF(x) =
Z x
a f
(t)dt, then
F0(x) = f(x).
F0(x) = lim
∆x→0
F(x+∆x)−F(x)
∆x
Note that
F(x+∆x)−F(x)
∆x =
Z x+∆x
a f
(t)dt−
Z x
a f (t)dt
∆x
=
Z x+∆x
x f
(t)dt
∆x .
Proof of the First Fundamental Theorem of Calculus
Theorem
Let f be a function continuous on[a,b]and letxbe any number in[a,b]. IfFis
the function defined byF(x) =
Z x
a f
(t)dt, then
F0(x) = f(x).
F0(x) = lim
∆x→0
F(x+∆x)−F(x)
∆x
Note that
F(x+∆x)−F(x)
∆x =
Z x+∆x
a f
(t)dt−
Z x
a f (t)dt ∆x
=
Z x+∆x
x f
(t)dt
∆x .
Proof of the First Fundamental Theorem of Calculus
Theorem
Let f be a function continuous on[a,b]and letxbe any number in[a,b]. IfFis
the function defined byF(x) =
Z x
a f
(t)dt, then
F0(x) = f(x).
F0(x) = lim
∆x→0
F(x+∆x)−F(x)
∆x
Note that
F(x+∆x)−F(x)
∆x =
Z x+∆x
a f
(t)dt−
Z x
a f (t)dt ∆x
=
Z x+∆x
x f
(t)dt
∆x .
Consider f(t)on[x,x+∆x].
Since f is continuous, by the Mean Value Theorem for Integrals, there exists
c∈[x,x+∆x]such that
Z x+∆x
x f
(t)dt= f(c)
(x+∆x)−x
= f(c)∆x.
Thus,
F0(x) = lim
∆x→0
F(x+∆x)−F(x)
∆x
= lim
∆x→0
Z x+∆x
x f
(t)dt ∆x
= lim
∆x→0
f(c)∆x ∆x = lim
∆x→0f(c).
Here,cdepends onxand∆x.
Consider f(t)on[x,x+∆x].
Since f is continuous, by the Mean Value Theorem for Integrals,
there exists
c∈[x,x+∆x]such that
Z x+∆x
x f
(t)dt= f(c)
(x+∆x)−x
= f(c)∆x.
Thus,
F0(x) = lim
∆x→0
F(x+∆x)−F(x)
∆x
= lim
∆x→0
Z x+∆x
x f
(t)dt ∆x
= lim
∆x→0
f(c)∆x ∆x = lim
∆x→0f(c).
Here,cdepends onxand∆x.
Consider f(t)on[x,x+∆x].
Since f is continuous, by the Mean Value Theorem for Integrals, there exists
c∈[x,x+∆x]such that
Z x+∆x
x f
(t)dt= f(c)
(x+∆x)−x
= f(c)∆x.
Thus,
F0(x) = lim
∆x→0
F(x+∆x)−F(x)
∆x
= lim
∆x→0
Z x+∆x
x f
(t)dt ∆x
= lim
∆x→0
f(c)∆x ∆x = lim
∆x→0f(c).
Here,cdepends onxand∆x.
Consider f(t)on[x,x+∆x].
Since f is continuous, by the Mean Value Theorem for Integrals, there exists
c∈[x,x+∆x]such that
Z x+∆x
x f
(t)dt=
f(c)
(x+∆x)−x
= f(c)∆x.
Thus,
F0(x) = lim
∆x→0
F(x+∆x)−F(x)
∆x
= lim
∆x→0
Z x+∆x
x f
(t)dt ∆x
= lim
∆x→0
f(c)∆x ∆x = lim
∆x→0f(c).
Here,cdepends onxand∆x.
Consider f(t)on[x,x+∆x].
Since f is continuous, by the Mean Value Theorem for Integrals, there exists
c∈[x,x+∆x]such that
Z x+∆x
x f
(t)dt= f(c)
(x+∆x)−x =
f(c)∆x.
Thus,
F0(x) = lim
∆x→0
F(x+∆x)−F(x)
∆x
= lim
∆x→0
Z x+∆x
x f
(t)dt ∆x
= lim
∆x→0
f(c)∆x ∆x = lim
∆x→0f(c).
Here,cdepends onxand∆x.
Consider f(t)on[x,x+∆x].
Since f is continuous, by the Mean Value Theorem for Integrals, there exists
c∈[x,x+∆x]such that
Z x+∆x
x f
(t)dt= f(c)
(x+∆x)−x
= f(c)∆x.
Thus,
F0(x) = lim
∆x→0
F(x+∆x)−F(x)
∆x
= lim
∆x→0
Z x+∆x
x f
(t)dt ∆x
= lim
∆x→0
f(c)∆x ∆x = lim
∆x→0f(c).
Here,cdepends onxand∆x.
Consider f(t)on[x,x+∆x].
Since f is continuous, by the Mean Value Theorem for Integrals, there exists
c∈[x,x+∆x]such that
Z x+∆x
x f
(t)dt= f(c)
(x+∆x)−x
= f(c)∆x.
Thus,
F0(x) = lim
∆x→0
F(x+∆x)−F(x)
∆x
= lim
∆x→0
Z x+∆x
x f
(t)dt ∆x
= lim
∆x→0
f(c)∆x ∆x = lim
∆x→0f(c).
Here,cdepends onxand∆x.
Consider f(t)on[x,x+∆x].
Since f is continuous, by the Mean Value Theorem for Integrals, there exists
c∈[x,x+∆x]such that
Z x+∆x
x f
(t)dt= f(c)
(x+∆x)−x
= f(c)∆x.
Thus,
F0(x) = lim
∆x→0
F(x+∆x)−F(x)
∆x
= lim
∆x→0
Z x+∆x
x f(t)dt
∆x
= lim
∆x→0
f(c)∆x ∆x = lim
∆x→0f(c).
Here,cdepends onxand∆x.
Consider f(t)on[x,x+∆x].
Since f is continuous, by the Mean Value Theorem for Integrals, there exists
c∈[x,x+∆x]such that
Z x+∆x
x f
(t)dt= f(c)
(x+∆x)−x
= f(c)∆x.
Thus,
F0(x) = lim
∆x→0
F(x+∆x)−F(x)
∆x
= lim
∆x→0
Z x+∆x
x f(t)dt
∆x
= lim
∆x→0
f(c)∆x ∆x
= lim
∆x→0f(c).
Here,cdepends onxand∆x.
Consider f(t)on[x,x+∆x].
Since f is continuous, by the Mean Value Theorem for Integrals, there exists
c∈[x,x+∆x]such that
Z x+∆x
x f
(t)dt= f(c)
(x+∆x)−x
= f(c)∆x.
Thus,
F0(x) = lim
∆x→0
F(x+∆x)−F(x)
∆x
= lim
∆x→0
Z x+∆x
x f(t)dt
∆x
= lim
∆x→0
f(c)∆x ∆x = lim
∆x→0f(c).
Here,cdepends onxand∆x.
Note that
x≤c≤x+∆x,
and
lim
∆x→0x=∆limx→0(x+∆x) =x,
so by the Squeeze Theorem, lim
∆x→0c=x.
Hence, by continuity of f,
F0(x) = lim
∆x→0f(c) = f
lim
∆x→0c
= f(x).
Note that
x≤c≤x+∆x,
and
lim
∆x→0x=∆limx→0(x+∆x) =x,
so by the Squeeze Theorem, lim
∆x→0c=x.
Hence, by continuity of f,
F0(x) = lim
∆x→0f(c) = f
lim
∆x→0c
= f(x).
Note that
x≤c≤x+∆x,
and
lim
∆x→0x=∆limx→0(x+∆x) =x,
so by the Squeeze Theorem, lim
∆x→0c=x.
Hence, by continuity of f,
F0(x) = lim
∆x→0f(c) = f
lim
∆x→0c
= f(x).
Note that
x≤c≤x+∆x,
and
lim
∆x→0x=∆limx→0(x+∆x) =x,
so by the Squeeze Theorem, lim
∆x→0c=x.
Hence, by continuity of f,
F0(x) = lim
∆x→0f(c) = f
lim
∆x→0c
= f(x).
Note that
x≤c≤x+∆x,
and
lim
∆x→0x=∆limx→0(x+∆x) =x,
so by the Squeeze Theorem, lim
∆x→0c=x.
Hence, by continuity of f,
F0(x) = lim
∆x→0f(c) =
f lim
∆x→0c
= f(x).
Note that
x≤c≤x+∆x,
and
lim
∆x→0x=∆limx→0(x+∆x) =x,
so by the Squeeze Theorem, lim
∆x→0c=x.
Hence, by continuity of f,
F0(x) = lim
∆x→0f(c) = f
lim
∆x→0c
= f(x).
Note that
x≤c≤x+∆x,
and
lim
∆x→0x=∆limx→0(x+∆x) =x,
so by the Squeeze Theorem, lim
∆x→0c=x.
Hence, by continuity of f,
F0(x) = lim
∆x→0f(c) = f
lim
∆x→0c
= f(x).
Exercise
1 Find the average value of f(x) =sin 3xon[0,π].
2 The acceleration function of a particle moving along a horizontal line is2t−3
m/s2and its initial velocity is2m/s. Find its average velocity during the first4
seconds.
3 Find the volume of the solid whose base is the unit circle if each cross-section
perpendicular to they-axis is an isosceles right triangle with one leg lying on
the base of the solid.
