User:Egm6341.s11.team4.fields/HW1

=Problem 1.3 IMVT Proof=

Given
IMVT: $$\int_{a}^{b}{w(x)f(x)dx} = f(\xi)\int_{a}^{b}{w(x)dx}\,\!$$

Find
a.) Prove the Integral Mean Value Theorem (IMVT) for the non-negative function $$w(\cdot)$$

b.) Prove the IMVT for $$w(x) \neq 0 \,\,\,\, \forall \,x \in [a,b]$$

Solution
a.)

$$\int_{a}^{b}{w(x)f(x)dx} = f(\xi)\int_{a}^{b}{w(x)dx}\,\!$$

Where w(x) and f(x) are continuous on [a,b], w(x) is non-negative and &xi; is a point on [a,b].

For the case where w(x) = 0, both integrals = 0 and we are left with 0 = 0

The rest of this proof is for w(x) > 0

Since w(x) > 0, $$\int_{a}^{b}{w(x)dx}  >  0$$

Let $$\,m := min f(x) \,\,\forall x \in [a,b]$$

Let $$\,M := max f(x) \,\,\forall x \in [a,b]$$

It follows that:  $$\,m\,\le  \,f(x) \,\le\, M\,\, \forall x \in [a,b]$$

We now multiply through by w(x) to get $$\,m w(x)\,\le \,f(x) w(x) \,\le\, M w(x)\,\, \forall x \in [a,b]$$

Which leads to: $$\,m \int_{a}^{b}{w(x)dx}\,\le  \, \int_{a}^{b}{f(x)}{w(x)dx}\,\le\, M \int_{a}^{b}{w(x)dx}\,\, \forall x \in [a,b]$$

We can now divide through by $$\int_{a}^{b}{w(x)dx}$$

To get $$\,m \,\le \, \frac{\int_{a}^{b}{f(x)}{w(x)dx}}{\int_{a}^{b}{w(x)dx}}\,\le\, M \, \forall x \in [a,b]$$

Since f(x) is continuous, there must be at least one value &xi; on the interval [a,b]; such that $$f(\xi) = \frac{\int_{a}^{b}{f(x)}{w(x)dx}}{\int_{a}^{b}{w(x)dx}}$$

Plugging this value into the original statement $$\int_{a}^{b}{w(x)f(x)dx} = f(\xi)\int_{a}^{b}{w(x)dx} = \frac{\int_{a}^{b}{f(x)}{w(x)dx}}{\cancel{\int_{a}^{b}{w(x)dx}}}\cancel{\int_{a}^{b}{w(x)dx}}  =  \int_{a}^{b}{w(x)f(x)dx}$$

proves the theorem.

b.)

The proof for w(x) > 0 is contained in part a. above.

What follows is the proof for the IMVT with w(x) < 0.

$$\int_{a}^{b}{w(x)f(x)dx} = f(\xi)\int_{a}^{b}{w(x)dx}\,\!$$

Where w(x) and f(x) are continuous on [a,b], w(x) < 0 and &xi; is a point on [a,b].

Since w(x) < 0, $$\int_{a}^{b}{w(x)dx}  <  0$$

Let $$\,m := min f(x) \,\,\forall x \in [a,b]$$

Let $$\,M := max f(x) \,\,\forall x \in [a,b]$$

It follows that:  $$\,m\,\le  \,f(x) \,\le\, M\,\, \forall x \in [a,b]$$

We now multiply through by w(x) to get $$\,m w(x)\,\ge \,f(x) w(x) \,\ge\, M w(x)\,\, \forall x \in [a,b]$$

Which leads to: $$\,m \int_{a}^{b}{w(x)dx}\,\ge  \, \int_{a}^{b}{f(x)}{w(x)dx}\,\ge\, M \int_{a}^{b}{w(x)dx}\,\, \forall x \in [a,b]$$

We can now divide through by $$\int_{a}^{b}{w(x)dx}$$

To get $$\,m \,\le \, \frac{\int_{a}^{b}{f(x)}{w(x)dx}}{\int_{a}^{b}{w(x)dx}}\,\le\, M \, \forall x \in [a,b]$$

Since f(x) is continuous, there must be at least one value &xi; on the interval [a,b]; such that $$f(\xi) = \frac{\int_{a}^{b}{f(x)}{w(x)dx}}{\int_{a}^{b}{w(x)dx}}$$

Plugging this value into the original statement $$\int_{a}^{b}{w(x)f(x)dx} = f(\xi)\int_{a}^{b}{w(x)dx} = \frac{\int_{a}^{b}{f(x)}{w(x)dx}}{\cancel{\int_{a}^{b}{w(x)dx}}}\cancel{\int_{a}^{b}{w(x)dx}}  =  \int_{a}^{b}{w(x)f(x)dx}$$

proves the theorem.