here another problem i can't seem to understand this problem here: If f is increasing on the interval [a,b], which of the following must be true? I. FnInt (f(x),x,a,b) f(a)(b-a) III. FnInt (f(x),x,a,b) \ 0

Question

here another problem i can't seem to understand this problem here: If f is increasing on the interval [a,b], which of the following must be true? I. FnInt (f(x),x,a,b) f(a)(b-a) III. FnInt (f(x),x,a,b) \> 0

anonymous · Answer

i want to say III is correct because if the function is increasing it would be positive right?

agent0smith · Answer

Area under a curve can still be negative, even if the function is increasing... eg y=x is increasing on the interval -1 to 0, but area is still negative. http://www.wolframalpha.com/input/?i=integral+-1+to+0+x+dx

anonymous · Answer

oh okay, so that eliminates III, i was trying to figure out the two out, but i found out thank you. i guess i was brain farting.

agent0smith · Answer

What did you get for the other two?

anonymous · Answer

so f(x)=x, a=1, and b=2 (something i made up) basically started plugging in. and what i found is I is true. Work:
I.
1.5≤2
II.
$$1.5 \ge 1$$
tell me if im wrong.

anonymous · Answer

...and i just realized the second one is correct too

agent0smith · Answer

Yeah that's kinda what I had in mind too, that both I and II were true, it just seems hard to prove it... I don't think I've seen something like that before, where it's 

$$\Large \int\limits_{a}^{b} f(x) dx < f(b)(b-a) $$ (assuming that's what it is... it's kinda hard to read what you posted)

anonymous · Answer

yeah your right. i put integral in the form that you would on a TI calculator