Is f Integrable on [0,3] with Integral Value of 3?

[snipez90]

Homework Statement

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.

Homework 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.

 

[Dick]

I think that works just fine.

 

[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.