User talk:Egm6341.spring-11.5.D

= Problem Homework 1 problem 4 Prove the Integral mean value theorem =

Given

 * {| style="width:100%" border="0" align="left"

\int _a^bf(x) w(x)dx=f(\zeta ) \int_a^b w(x) \, dx $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

where f(x) and w(x) are conitnous nonnegative functions $$\epsilon [a,b]$$

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

$$ \int_a^b 1 f(x) \, dx=f(\zeta ) \int_a^b 1 \, dx $$

$$ \int_a^b f(x) \, dx=(b-a) f(\zeta ) $$

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


 * $$\displaystyle

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

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

$$\displaystyle m (b-a)\leq (b-a) f(\zeta )\leq M (b-a) $$

$$\displaystyle \min (f(x))\leq f(\zeta )\leq \max (f(x)) $$ Egm6341.spring-11.5.D 02:42, 22 January 2011 (UTC)