The Mean Value Theorem for Integrals

If  f (x ) is a continuous function on [a,b] then there is a number c in [a,b] such as,

                                   ∫^b_af ( x ) dx = f (c ) (b - a )

Note as well that one way to think of this theorem is the following.  Firstly rewrite the result as,

                               1/( b - a)  ∫^b_af ( x ) dx =f(c)

and from this we can illustrates that this theorem is telling us that there is a number a < c < b such that f_avg  = f (c ) . Or, in other terms, if f (x) is continuous function then somewhere within [a,b] the function will take on its average value.

Let's take a rapid look at an example using this theorem.

Example:  Find out the number c which satisfies the Mean Value Theorem for Integrals for the function  f ( x ) =x² + 3x + 2 within the interval [1,4]

 Solution

Firstly let's notice that the function is a polynomial and therefore is continuous on the given interval. It means that we can use the Mean Value Theorem.  Therefore, let's do that.

∫₁ ⁴ x2+3x+2dx = (c²+3c+2)(4-1)

( (1/3)x² + (3/2) x² +2x |₁⁴ =  3(c²  + 3c + 2)

                                      = 99/2 = 3c² + 9c + 6

                                  0 = 3c² + 9c - (87/2)

It is a quadratic equation which we can solve out.  Using the quadratic formula we obtain the following two solutions,

c = (-3 +√67)/2 = 2.593

c = (-3 -√67)/2 = -5.593

Obviously the second number is not within the interval and therefore that isn't the one that we're after. However, the first is in the interval and therefore that's the number we desire.

Note as well that it is possible for both numbers to be in the interval therefore don't expect only one to be in the interval.