# Show that f is integrable

1. Jan 20, 2009

### snipez90

1. The problem statement, all variables and given/known data
Let f be a function on [0,3] so that f(x) = 1, for x in [0,3); 17, for x = 3. Show that f is integrable on [0,3] and the value of the integral is 3.

2. Relevant equations
Upper and lower sums

3. The attempt iat a solution
I think this is a valid proof. Let e > 0. Define a partition P = {0, 3 - e/32, 3}.

Then U(f,P) = 1*(3 - e/32) + 17*(e/32) = 3 + e/2. Clearly, L(f,P) = 3. Hence, U(f,P) - L(f,P) = e/2 < e. Hence f is integrable.

Now $$L(f,P) \leq 3 \leq U(f,P)$$. Since f is integrable, there is only one number between all the lower and upper sums, hence the value of the integral is 3.

2. Jan 20, 2009

### Dick

I think that works just fine.

3. Jan 21, 2009

### snipez90

Hmmm, cool. The only real change I made to this in the actual proof is letting e' = min{1,e}, in case someone wanted to choose a morbidly large epsilon.