# Average Value

Definition : Rolle's Theorem

f is a continuous function defined on the closed interval [a,b], is differentiable on the open interval (a,b), and f(a)=f(b)

Then, there exists a value "c" in the interval (a,b)such that f'(c)=0

If f is a continuous function defined on the interval [a,b]

The average of f(x) is : Average =${\displaystyle {\frac {1}{b-a}}}$baf(x)dx

Example 1 :

Find the average value of f(x) = ${\displaystyle e^{x}}$ for 0 greater than equal to x and x less than equal to 2.
Solution :
Average = ${\displaystyle {\frac {1}{2-0}}}$20f(x)dx
=${\displaystyle {\frac {1}{2}}}$20${\displaystyle e^{x}}$dx
=${\displaystyle {\frac {1}{2}}}$ [${\displaystyle e^{x}}$]20
=${\displaystyle {\frac {1}{2}}}$ [${\displaystyle e^{2}-e^{0}}$]
=${\displaystyle {\frac {1}{2}}}$ [${\displaystyle e^{2}-1}$]

Example 2 :

Find the average value of f(x) = ${\displaystyle sqrt{x}}$ given the interval 0 greater than equal to x and x less than equal to 9.
Solution :
Average = ${\displaystyle {\frac {1}{9-0}}}$90${\displaystyle sqrt{x}}$dx
=${\displaystyle {\frac {1}{9}}}$90${\displaystyle sqrt{x}}$dx
=${\displaystyle {\frac {1}{9}}}$ (${\displaystyle {\frac {2}{3}}x^{\frac {3}{2}}}$) |90
=${\displaystyle {\frac {1}{9}}}$ (${\displaystyle {\frac {2}{3}}9^{\frac {3}{2}}}$ - 0)
=${\displaystyle {\frac {1}{9}}}$ (${\displaystyle {\frac {2}{3}}}$ x 27)
=2

Example 3 :

Find the consumer's surplus for p = 50 - 0.06${\displaystyle x^{2}}$, the demand curve, with x = 20.
Solution :
The price is:
B = 50 - 0.06${\displaystyle (20)^{2}}$
= 50 - 24 = 26
The consumer's surplus is:
200${\displaystyle [(50-0.06x^{2})-26]}$dx
=∫200${\displaystyle (24-0.06x^{2})}$dx
=${\displaystyle (24x-0.02x^{3})}$ |200
=24(20) - ${\displaystyle 0.02(20)^{3}}$
= 480 - 160
= 320