Average Value

This article is part of the MathHelp Tutoring Wiki

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 =$\frac{1}{b-a}$∫^b_af(x)dx

Example 1 :

Find the average value of f(x) = $e^x$ for 0 greater than equal to x and x less than equal to 2.

Solution :

Average = $\frac{1}{2-0}$∫²₀f(x)dx

=$\frac{1}{2}$ ∫²₀$e^x$dx

=$\frac{1}{2}$ [$e^x$]²₀

=$\frac{1}{2}$ [$e^2-e^0$]

=$\frac{1}{2}$ [$e^2-1$]

---

Example 2 :

Find the average value of f(x) = $sqrtx$ given the interval 0 greater than equal to x and x less than equal to 9.

Solution :

Average = $\frac{1}{9-0}$∫⁹₀$sqrtx$dx

=$\frac{1}{9}$ ∫⁹₀$sqrtx$dx

=$\frac{1}{9}$ ($\frac{2}{3}{x}^{\frac{3}{2}}$) |⁹₀

=$\frac{1}{9}$ ($\frac{2}{3}{9}^{\frac{3}{2}}$ - 0)

=$\frac{1}{9}$ ($\frac{2}{3}$ x 27)

=2

---

Example 3 :

Find the consumer's surplus for p = 50 - 0.06$x^2$, the demand curve, with x = 20.

Solution :

The price is:

B = 50 - 0.06$(20{)}^2$

= 50 - 24 = 26

The consumer's surplus is:

∫²⁰₀$[(50-0.06x^2)-26]$dx

=∫²⁰₀$(24-0.06x^2)$dx

=$(24x-0.02x^3)$ |²⁰₀

=24(20) - $0.02(20{)}^3$

= 480 - 160

= 320

---

• Back to Integral Calculus
• Back to MathHelp