- #1

sony

- 104

- 0

1.

0 if 0<=x<1 or 1<x<=2

f(x) = 1 if x=1

(by "<=" i mean less than or equal)

I'm supposed to show that it is Riemann integrable on that interval.

They chose P to be: {0, 1-e/3, 1+e/3,2}

L(f,P)=0 (I think I understand this, its just the sum of zero?)

They then get U(f,p)= 0(1-e/3) +1(2e/3) +0(2-(1+e/3))=2e/3 <-- I have no idea how they get this!

And how do they chose that specific partition?

(I get that f is integrable since U(f,P)-L(f,P)=2e/3<e

(I've used e as epsilon :P)

Question 2:

1 if x is rational

f(x)= 0 if x is irrational

I have no idea how they get L(f,P) to be 0 and U(f,P) to be 1!

Thanks!