Week 10 3-21, 3-22, 3-24, 3-25, 2016

Included: Monotonicity. Linearity. Additivity. Bounded Constraint. Continuity of Integral Function for integrable functions. Continuous Functions are integrable. Fundamental Theorem (Derivative form) for Continuous Functions. FTofC (Evaluation form) for Continuous Functions. Mean Value Theorem for Integrals for Continuous Functions.

3-22.

Recall the definition of the Darboux Integral for `f` a bounded function on `[a,b]`.

Suppose `P =\{ a=x_0<x_1<...<x_{n-1}<x_n=b\}; I_k = [x_{k-1},x_k]; Delta x_k=x_k-x_{k-1}` for `k=1,2,...,n`.

`M_k = lub\{f(x): x in I_k\}; m_k= glb\{f(x): x in I_k\}`

` U(P)=\sum_{k=1}^n M_k cdot Delta x_k ;L(P)=\sum_{k=1}^n m_k cdot Delta x_k ;` and

`int_U f= glb\{U(P)\}; \int_L f= glb\{L(P)\}`.

Then

Note: We have shown tha

Also

`\int_a^b alpha f + g = alpha \int_a^b f + \int_a^b g`.

Excerpt from Spivak, Calculus, Ch 13

Theorem [Additivity]

Proof: [Converse proof omitted]

EOP

Theorem:[Boundedness] Suppose `f ` is integrable on `[a,b]` and there exist m and M so that for all `x in [a,b]` `m \le f(x) \le M`. Then `m(b-a) \le \int_a^b f \le M(b-a).

Proof:

Now consider `F` on the interval `[c, c+h]`, of length `h>0`.

Then `F(c+h) = \int_a^{c+h} f ` and using additivity we have `F(c+h) - F(c) = \int_c^{c+h} f `.

By the previous result we have

`-Mh \le \int_c^{c+h}f
\le Mh` or

(1) `-Mh \le F(c+h) - F(c) \le Mh`

Now consider `F` on the interval `[c+h, c]`, where `h`<0, and length `-h>0`.

Then using the additive property we have `F(c) = \int _a^c f = \int _a^{c+h} f + \int_{c+h}^c f` and so

`F(c+h)-F(c) = -\int_{c+h}^c f.`

Using boundedness we have then that `-M(-h) \le \int_{c+h}^c f \le M(-h)` or `M(-h) \ge -\int_{c+h}^c f \ge M(h)`.

Thus we have when `h <0` that `Mh \le F(c+h) - F(c) \le M(-h)`. In either case we see that as `h to 0` the squeeze theorem can apply , showing that `F(c+h)-F(c) to 0` or `lim_{x to c} f(x) = f(c)`, and thus `F` is continuous at `x=c` for all `c in [a,b]`.**EOP**

(1) `-Mh \le F(c+h) - F(c) \le Mh`

Now consider `F` on the interval `[c+h, c]`, where `h`<0, and length `-h>0`.

Then using the additive property we have `F(c) = \int _a^c f = \int _a^{c+h} f + \int_{c+h}^c f` and so

`F(c+h)-F(c) = -\int_{c+h}^c f.`

Using boundedness we have then that `-M(-h) \le \int_{c+h}^c f \le M(-h)` or `M(-h) \ge -\int_{c+h}^c f \ge M(h)`.

Thus we have when `h <0` that `Mh \le F(c+h) - F(c) \le M(-h)`. In either case we see that as `h to 0` the squeeze theorem can apply , showing that `F(c+h)-F(c) to 0` or `lim_{x to c} f(x) = f(c)`, and thus `F` is continuous at `x=c` for all `c in [a,b]`.