User:Egm6341.s11.team2.franklin/hwk1

=Problem 1.1=

Solution
=Problem 1.2=

Solution
=Problem 1.3 Proving the Integral Mean Value Theorom=

''' I solved this problem without referring to S10 homework. '''

Find
Prove the Integral Mean Value Theorem (IMVT) for

a) $$\displaystyle w\left(x\right) \geq 1, \forall x \in \left[a,b\right]$$ and

b) $$\displaystyle w\left(x\right) < 1, \forall x \in \left[a,b\right]$$

using same technique used in lecture 4 to prove IVMT for $$\displaystyle w\left(x\right) = 1 $$.

Given
IVMT states
 * {| style="width:100%" border="0"

\int_a^b{w\left(x\right)f\left(x\right)}dx = f\left(\xi\right)\int_a^b{w\left(x\right)}dx, \forall x \in \left[a,b\right]. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.3.1)
 * style= |
 * }

a
Let $$\displaystyle w\left(x\right) \geq 0, \forall x \in \left[a,b\right]. $$

Consider $$\displaystyle w\left(x\right) f\left(x\right) $$ continuos on $$\displaystyle\left[a,b\right] .$$

Let


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

m:=min f\left(x\right), \forall x \in \left[a,b\right] $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.3.2)
 * style= |
 * }

and


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

M:=max f\left(x\right), \forall x \in \left[a,b\right] , $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.3.3)
 * style= |
 * }

such that


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

m \leq f\left(x\right) \leq M, \forall x \in \left[a,b\right]. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.3.4)
 * style= |
 * }

Multiplying Eq. 1.3.3 by $$\displaystyle w\left(x\right) $$


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

w\left(x\right) m \leq w\left(x\right) f\left(x\right) \leq  w\left(x\right) M, \forall x \in \left[a,b\right]. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.3.4)
 * style= |
 * }

Integrating Eq. 1.3.4 over $$\displaystyle \left[a,b\right] $$


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

\int_a^b{w\left(x\right) m}dx \leq \int_a^b{w\left(x\right) f\left(x\right)}dx \leq \int_a^b{ w\left(x\right) M}dx. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.3.5)
 * style= |
 * }

$$\displaystyle m $$ and $$\displaystyle M $$ are both constants and can be pulled out of the integrals.


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

m \int_a^b{w\left(x\right)}dx \leq \int_a^b{w\left(x\right) f\left(x\right)}dx \leq M \int_a^b{ w\left(x\right) }dx. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.3.6)
 * style= |
 * }

Let


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

\alpha := \int_a^b{ w\left(x\right) }dx. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.3.7)
 * style= |
 * }

Divide Eq 1.3.6 by $$\displaystyle \alpha $$


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

m \leq \underbrace{\frac{1}{\alpha} \int_a^b{w\left(x\right) f\left(x\right)}dx}_{Z:=} \leq M. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.3.8)
 * style= |
 * }

By the Intermediate Value Theorem (IVT), $$\displaystyle \exists \xi  \in \left[a,b\right] $$ such that $$\displaystyle f\left(\xi\right)=Z. $$


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

f\left(\xi\right)=\frac{1}{\alpha} \int_a^b{w\left(x\right) f\left(x\right)}dx $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.3.9)
 * style= |
 * }

Therefore,


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

$$ \int_a^b{w\left(x\right)f\left(x\right)}dx = f\left(\xi\right)\int_a^b{w\left(x\right)}dx, w\left(x\right) \geq 0. $$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:100%" border="0"|
 * $$\,(Eq. 1..3.10)
 * style= |
 * }

b
Let $$\displaystyle w\left(x\right) < 0, \forall x \in \left[a,b\right]. $$

Consider $$\displaystyle w\left(x\right) f\left(x\right) $$ continuos on $$\displaystyle\left[a,b\right] .$$

Let


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

m:=min f\left(x\right), \forall x \in \left[a,b\right] $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.3.11)
 * style= |
 * }

and


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

M:=max f\left(x\right), \forall x \in \left[a,b\right] , $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.3.12)
 * style= |
 * }

such that


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

m \leq f\left(x\right) \leq M, \forall x \in \left[a,b\right]. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.3.13)
 * style= |
 * }

Multiplying Eq. 1.3.3 by $$\displaystyle w\left(x\right) $$


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

$$ $$
 * $$\displaystyle
 * w\left(x\right)| m \leq |w\left(x\right))| f\left(x\right) \leq  |w\left(x\right)| M, \forall x \in \left[a,b\right].
 * w\left(x\right)| m \leq |w\left(x\right))| f\left(x\right) \leq  |w\left(x\right)| M, \forall x \in \left[a,b\right].
 * $$\displaystyle (Eq. 1.3.14)
 * style= |
 * }

Integrating Eq. 1.3.4 over $$\displaystyle \left[a,b\right] $$


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

\int_a^b{|w\left(x\right)| m}dx \leq \int_a^b{|w\left(x\right)| f\left(x\right)}dx \leq \int_a^b{ |w\left(x\right)| M}dx. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.3.15)
 * style= |
 * }

$$\displaystyle m $$ and $$\displaystyle M $$ are both constants and can be pulled out of the integrals.


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

m \int_a^b{|w\left(x\right)|}dx \leq \int_a^b{|w\left(x\right)| f\left(x\right)}dx \leq M \int_a^b{ |w\left(x\right)| }dx. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 1.3.16)
 * style= |
 * }

Let


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

\alpha := \int_a^b{ |w\left(x\right)| }dx. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 1.3.17)
 * style= |
 * }

Divide Eq 1.3.6 by $$\displaystyle \alpha $$


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

m \leq \underbrace{\frac{1}{\alpha} \int_a^b{|w\left(x\right)| f\left(x\right)}dx}_{Z:=} \leq M. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 1.3.18)
 * style= |
 * }

By the Intermediate Value Theorem (IVT), $$\displaystyle \exists \xi  \in \left[a,b\right] $$ such that $$\displaystyle f\left(\xi\right)=Z. $$


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

f\left(\xi\right)=\frac{1}{\alpha} \int_a^b{w\left(x\right) f\left(x\right)}dx. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 1.3.19)
 * style= |
 * }

Since $$ \displaystyle w\left(x\right) < 0 ,$$


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

f\left(\xi\right)=\frac{1}{ \cancel{\left(-1\right)} \int_a^b{ w\left(x\right) }dx} \cancel{\left(-1\right)} \int_a^b{w\left(x\right) f\left(x\right)}dx. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 1.3.20)
 * style= |
 * }

Therefore,


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

$$ \int_a^b{w\left(x\right)f\left(x\right)}dx = f\left(\xi\right)\int_a^b{w\left(x\right)}dx, w\left(x\right) < 0. $$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:100%" border="0"|
 * <p style="text-align:right;">$$\,(Eq. 1..3.21)
 * style= |
 * }

Contributing Members

 * Solved and posted by Egm6341.s11.team2.franklin 22:22, 23 January 2011 (UTC)

=Problem 1.4=