Average Value

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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) =

sqrt{x}

given the interval 0 greater than equal to x and x less than equal to 9.

Solution :

Average =

{\frac {1}{9-0}}

∫⁹₀

sqrt{x}

dx

=

{\frac {1}{9}}

∫⁹₀

sqrt{x}

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

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