## 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

f (t)dt = f (c)b - a  .

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 xc is defined by

x(x)dx = xc(x)dx = xcM,

or equivalently

xc = x(x)dx  .

The point xc 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 xc of the rod. We have (x) = kx, for some constant k > 0. We have

M = (x)dx = k x dx = k =

and

xc = x(x)dx = kx2dx = 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.

f (x) = sec2(x)

for x [0,/4].

f (x) =

for x [0, 4].

f (a)(b - a)    and    f (x)dx  .

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