##
Mean Value Theorems for Integrals

Let *f* (*x*) be continuous on [*a*, *b*]. Set

*F*(

*x*) =

*f* (

*t*)

*dt* .

The Fundamental Theorem of Calculus implies
*F* '(*x*) = *f* (*x*). The
Mean Value Theorem implies the existence of
*c* (*a*, *b*) such
that
=

*F* '(

*c*), or equivalently

*F*(

*b*) -

*F*(

*a*) =

*F* '(

*c*)

*b* -

*a*
which implies
This is known as the **First Mean Value Theorem for
Integrals**. The point *f* (*c*) is called **the average value of
***f* (*x*) on [*a*, *b*].
As the name "First Mean Value Theorem" seems to imply, there is
also a Second Mean Value Theorem for Integrals:

**Second Mean Value Theorem for Integrals.** Let *f* (*x*) and
*g*(*x*) be continuous on [*a*, *b*]. Assume that *g*(*x*) is positive,
i.e.
*g*(*x*) 0 for any
*x* [*a*, *b*]. Then there exists
*c* (*a*, *b*) such that

*f* (

*t*)

*g*(

*t*)

*dt* =

*f* (

*c*)

*g*(

*t*)

*dt* .

The number *f* (*c*) is called the *g*(*x*)-weighted average of *f* (*x*) on the interval [*a*, *b*].
As an application one may define the **Center of Mass** of
one-dimensional non-homogeneous objects such as a metal rod. If
the object is homogeneous and lying on the *x*-axis from *x* = *a* to
*x* = *b*, then its center of mass is simply the midpoint

^{ . }
If, on the other hand, the object is not homogeneous with
(*x*) being the density function, then the total mass *M*
is
*M* =

(

*x*)

*dx* .

The density-weighted average *x*_{c} is defined by
*x*(

*x*)

*dx* =

*x*_{c}(

*x*)

*dx* =

*x*_{c}*M*,

or equivalently
The point *x*_{c} is the center of mass of the object.
**Example.** A rod of length *L* is placed on the *x*-axis from
*x* = 0 to *x* = *L*. Assume that the density
(*x*) of the rod
is proportional to the distance from the *x* = 0 endpoint of the
rod. Let us find the total mass *M* and the center of mass *x*_{c}
of the rod. We have
(*x*) = *kx*, for some constant *k* > 0. We have

and
*x*_{c} =

*x*(

*x*)

*dx* =

*kx*^{2}*dx* =

*k* =

*L* .

If the rod was homogeneous, then the center of mass would be at
the midpoint of the rod. Now it is closer to the endpoint *x* = *L*.
This is not surprising since there is more mass at this end.
**Exercise 1.** Find the average value of

*f* (*x*) = sec^{2}(*x*)

for
*x* [0,/4].

**Answer.**

**Exercise 2.** Find the average value of

*f* (

*x*) =

for
*x* [0, 4].

**Answer.**

**Exercise 3.** Assume that *f* (*x*) be continuous and increasing
on [*a*, *b*]. Compare

*f* (

*a*)(

*b* -

*a*) and

*f* (

*x*)

*dx* .

**Answer.**

**
**

**
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