Given a function and a closed interval such that
for all .
- Prove is continuous at every point .
- If is integrable on prove
- If is any point prove
- Proof. Let be any point in . Then, for any let . Then
Thus, for any , we have whenever . Hence,
Therefore, is continuous at every point
- Proof. Since is a constant (the value of the function evaluated at the constant ), we have
Therefore, we can compute,
The final inequality follows since, using the monotone property of the integral,
Then, since by our hypothesis on , we have
- Proof. The proof proceeds similarly to that of part (b),
Then, since , we break the integral into two pieces (to deal with the fact that for and for ):
But now, we know implies , so we have the inequality,
In part c, observe that |x – c| <= |x – a| for x in [a, b] and all the lengthy calculations are not needed.
That’s not the case. For like x=a+1 and c=a+3.
Since I had trouble to understand the last step in part c, I share my approach to it:
Thx!
I think you have a mistake in part (c): in the last inequality you have
c(c-(b+a)) <= -ab.
For example for a=2, b=5 and c=1 we get
c(c-(b+a)) = -6
-ab = -10.
ah sorry, my example is wrong since c<a
You’ve got a sign error in part (c). It should be -c(b-c) in the 3rd line of the 2nd section. It changes a little how you get back to the end inequality.
Awesome, thanks! Working on a fix now… you’re right though, it cascades into the rest of the proof.
I think it should be fixed now. Thanks for letting me know.