(adsbygoogle = window.adsbygoogle || []).push({}); 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 [tex]L(f,P) \leq 3 \leq U(f,P)[/tex]. Since f is integrable, there is only one number between all the lower and upper sums, hence the value of the integral is 3.

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# Homework Help: Show that f is integrable

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