User:Egm6341.spring-11.5.D/HW1

Find

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\int _a^bf(x) w(x)dx=f(\zeta ) \int_a^b w(x) \, dx $$
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Solution
'''Solved via combination of Wikipedia http://en.wikipedia.org/wiki/Mean_value_theorem#Mean_value_theorems_for_integration and Egm6341.s10.team3.sa/HW1 http://en.wikiversity.org/w/index.php?title=User:Egm6341.s10.team3.sa/HW1&oldid=524038#Prove_IMVT'''

First, since w(x) is arbitrary and non negative we will set it equal to I

$$ I=\int_a^b w(x) \, dx$$

$$\displaystyle

\int_a^b w(x) f(x) \, dx=f(\zeta ) I $$

Next we will set constants at the extreme values of $$ f(x)\in [a,b]$$

$$\displaystyle M=\max (f(x)),m=\min (f(x)) $$ Therefore, from the Extreme Value Theorem

http://en.wikipedia.org/wiki/Extreme_value_theorem

$$\displaystyle

m I \leq \int_a^b w(x) f(x) \, dx\leq M  I $$ Evaluating and subsituting in from above

$$\displaystyle m \leq \frac{1}{ I } \int_a^b w(x) f(x) \, dx\leq M $$

$$Z=\frac{1}{ I } \int_a^b w(x) f(x) \, dx$$ $$\displaystyle \min (f(x))\leq f(\zeta )\leq \max (f(x)) $$

Since f(x)is continous from M to m there exists $$ \zeta \in [a,b] at   f(\zeta)=Z $$


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$$\therefore f(\zeta)=\frac{1}{\int_a^b w(x) \, dx} \int_a^b w(x) f(x) \, dx$$
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Egm6341.spring-11.5.D 19:30, 24 January 2011 (UTC)

If w(x) is strictly negative

$$ f(\zeta)=\frac{1}{(-1) \cdot \int_a^b w(x) \, dx} \cdot (-1) \cdot \int_a^b w(x) f(x) \, dx$$


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$$\therefore f(\zeta)=\frac{1}{\int_a^b w(x) \, dx} \int_a^b w(x) f(x) \, dx$$


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lecture hyperlink
http://clesm.mae.ufl.edu/~vql/videos/nm1.s11.mtg29.wmv

http://clesm.ippd.ufl.edu/~vql/videos/nm1.s11.mtg29.wmv

http://clesm.mae.ufl.edu/~vql/videos/nm1.s11.mtg30.wmv

http://clesm.ippd.ufl.edu/~vql/videos/nm1.s11.mtg30.wmv