The Mean Value Theorem for Integrals
Mathematics 11: Lecture 38
Dan Sloughter
Furman University
November 28, 2007
Dan Sloughter (Furman University)
The Mean Value Theorem for Integrals
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Theorem
I
If f is continuous on [a, b], then there exists a number c in (a, b) such
that
Z b
f (x)dx = f (c)(b − a).
a
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The Mean Value Theorem for Integrals
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Proof
I
Suppose f is continuous on [a, b] and let
Z x
F (x) =
f (t)dt.
a
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The Mean Value Theorem for Integrals
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Proof
I
Suppose f is continuous on [a, b] and let
Z x
F (x) =
f (t)dt.
a
I
Then F is continuous on [a, b] and differentiable on (a, b), so, by the
Mean Value Theorem,
F (b) − F (a)
= F 0 (c) = f (c)
b−a
for some c in (a, b).
Dan Sloughter (Furman University)
The Mean Value Theorem for Integrals
November 28, 2007
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Proof
I
Suppose f is continuous on [a, b] and let
Z x
F (x) =
f (t)dt.
a
I
Then F is continuous on [a, b] and differentiable on (a, b), so, by the
Mean Value Theorem,
F (b) − F (a)
= F 0 (c) = f (c)
b−a
for some c in (a, b).
I
The result follows since
Z
F (b) − F (a) =
b
f (t)dt
a
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The Mean Value Theorem for Integrals
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Geometric interpretation
I
Note: the theorem says that the definite integral is exactly equal to
the signed area of a rectangle with base of length b − a and height
f (c).
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Geometric interpretation
I
Note: the theorem says that the definite integral is exactly equal to
the signed area of a rectangle with base of length b − a and height
f (c).
I
For this reason, we call f (c) the average value of f on [a, b].
Dan Sloughter (Furman University)
The Mean Value Theorem for Integrals
November 28, 2007
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Geometric interpretation
I
Note: the theorem says that the definite integral is exactly equal to
the signed area of a rectangle with base of length b − a and height
f (c).
I
For this reason, we call f (c) the average value of f on [a, b].
I
Note: we do not have to find c to find the average value of f . The
average value of f on [a, b] is simply
1
b−a
Dan Sloughter (Furman University)
Z
b
f (x)dx.
a
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Example
I
Since
Z
0
4
4
64
1 3 x dx = x = ,
3 3
2
0
the average value of f (x) =
x2
on [0, 4] is
64
3 = 64 = 16 .
4−0
12
3
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The Mean Value Theorem for Integrals
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Example
I
Since
Z
0
4
4
64
1 3 x dx = x = ,
3 3
2
0
the average value of f (x) =
x2
on [0, 4] is
64
3 = 64 = 16 .
4−0
12
3
I
Note: this average value occurs when x =
Dan Sloughter (Furman University)
√4 .
3
The Mean Value Theorem for Integrals
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Example
I
If P = (x, 3x 2 ) is a point on the curve y = 3x 2 , then the distance
from P to the origin is
q
q
p
p
2
2
2
2
4
x + (3x ) = x + 9x = x 2 (1 + 9x 2 ) = |x| 1 + 9x 2 .
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The Mean Value Theorem for Integrals
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Example
I
If P = (x, 3x 2 ) is a point on the curve y = 3x 2 , then the distance
from P to the origin is
q
q
p
p
2
2
2
2
4
x + (3x ) = x + 9x = x 2 (1 + 9x 2 ) = |x| 1 + 9x 2 .
I
As x goes from 0 to 2, the average distance from P to the origin is
Z
1 2 p
A=
x 1 + 9x 2 dx.
2 0
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The Mean Value Theorem for Integrals
November 28, 2007
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Example
I
If P = (x, 3x 2 ) is a point on the curve y = 3x 2 , then the distance
from P to the origin is
q
q
p
p
2
2
2
2
4
x + (3x ) = x + 9x = x 2 (1 + 9x 2 ) = |x| 1 + 9x 2 .
I
As x goes from 0 to 2, the average distance from P to the origin is
Z
1 2 p
A=
x 1 + 9x 2 dx.
2 0
I
√
To find an antiderivative of x 1 + 9x 2 , we first note that it must be
3
something like (1 + 9x 2 ) 2 .
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The Mean Value Theorem for Integrals
November 28, 2007
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Example
I
If P = (x, 3x 2 ) is a point on the curve y = 3x 2 , then the distance
from P to the origin is
q
q
p
p
2
2
2
2
4
x + (3x ) = x + 9x = x 2 (1 + 9x 2 ) = |x| 1 + 9x 2 .
I
As x goes from 0 to 2, the average distance from P to the origin is
Z
1 2 p
A=
x 1 + 9x 2 dx.
2 0
I
√
To find an antiderivative of x 1 + 9x 2 , we first note that it must be
3
something like (1 + 9x 2 ) 2 .
I
But
Dan Sloughter (Furman University)
p
3
d
(1 + 9x 2 ) 2 = 27x 1 + 9x 2 .
dx
The Mean Value Theorem for Integrals
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Example (cont’d)
I
Hence
Z
x
Dan Sloughter (Furman University)
p
1 + x 2 dx =
3
1
(1 + 9x 2 ) 2 + c.
27
The Mean Value Theorem for Integrals
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Example (cont’d)
I
Hence
Z
x
I
p
1 + x 2 dx =
3
1
(1 + 9x 2 ) 2 + c.
27
And so
1
A=
2
Z
0
2
√
2
p
1
37 37 − 1
2 32 2
x 1 + 9x dx = (1 + 9x ) =
≈ 4.1493.
54
54
0
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