Prove a function that differs in finitely many points has the same integral as another

I'll show a proof if the two functions differ for only one point. Then, you can repeat the same argument in the neighbourhood of every other point.

So, suppose that we have

We can write

We can write the exact same thing for , and we have (as far as we know)

So, we only need to prove that the term where f and g are not the same function can be made arbitrarily small.

Since f and g are integrable on an interval, they are bounded. Let's call and the maximum of f and g, respectively, in the interval . We have

The same goes for g(x). So, You have

which tends to 0 as

answer: a

step-by-step explanation:

if xy ~ yz therefore xy would be the same length as yz

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