# Properties of Integrals

1. Feb 17, 2009

### jdz86

1. The problem statement, all variables and given/known data

A.) Prove: $$m_{i}$$(f) + $$m_{i}$$(g) $$\leq$$ $$m_{i}$$(f+g) $$\leq$$ $$M_{i}$$(f+g) $$\leq$$ $$M_{i}$$(f) + $$M_{i}$$(g)

B.) Prove: $$M_{i}$$($$\alpha$$f) = {$$\alpha$$$$M_{i}$$(f) $$\alpha$$ $$\succ$$ 0 or $$\alpha$$$$m_{i}$$(f) $$\alpha$$ $$\prec$$ 0
and
$$m_{i}$$($$\alpha$$f) = {$$\alpha$$$$m_{i}$$(f) $$\alpha$$ $$\succ$$ 0 or $$\alpha$$$$M_{i}$$(f) $$\alpha$$ $$\prec$$ 0

C.) Take P = $$P_{1}$$ = {a,b}, and write $$m_{i}$$ = m, $$M_{i}$$ = M (i=1). Give an example for each of the following:
(a) m(f) + m (g) $$\prec$$ m(f+g)
(b) M(f+g) $$\prec$$ M(f) + M(g)
(c) both (a) and (b) hold

2. Relevant equations

P = {$$x_{0}$$,...,$$x_{n}$$} and is a partition of [a,b]
$$M_{i}$$ = {sup {f(x) : x $$\in$$ [$$x_{i-1}$$, $$x_{i}$$]}}
$$m_{i}$$ = {inf {f(x) : x $$\in$$ [$$x_{i-1}$$, $$x_{i}$$]}}
$$\alpha$$ is just any multiplier

3. The attempt at a solution

For A.) and B.) i was going to take different sets and show it that way, but it will be a long drawn out process, i can't put my finger on the "shortcut/abbreviated" version. Any suggestions on what method to use??

And for C.) (a) f = [-2,-1) g = [-1,0]
(b) f = [0,1) g = [1,2]
(c) ??