Here's a fun integral question I made for a competition a few years ago. I was inspired by this much easier question:
Warmup: Let f(x) be a continuous real-valued function on [0,1]. Calculate: \[\lim_{n\to\infty}\int_0^1x^nf(x)dx\]
Now for the real question: Let f(x) be a continuous real-valued function on [0,1]. Calculate \[\lim_{n\to\infty} n\int_0^1x^nf(x)dx\]

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- JamesJ

The first one is clearly zero.
The second one .... one sec

- JamesJ

f(1)

- anonymous

Didn't take too long. I meant to say "calculate with proof" but I suppose that is trivial once you know the answers.

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## More answers

- JamesJ

Yes, I know. But I can write the proofs for these.
These integrals are actually a kind of 'primitive' form of the Laplace transform.

- JamesJ

we usually have e^(-st) as the function we're integrating against. If we write x = e^-s, then that's how we get back to these integrals.

- anonymous

Makes sense, never real thought about that connection too much. Very clever!
Want a harder question (that should take longer than 10 seconds for you to solve)?

- JamesJ

Talking of writing proofs though, writing a good proof for that first integral isn't to obvious, I don't think.

- JamesJ

It seems to me you need to write it as
\[ \int_0^{1-\delta} + \int_{1-\delta}^1 \]
Showing the first piece goes to zero is straightforward as the function \( g(x) = x^n f(x) \) converges uniformly to the zero function on \( [0,1-\delta] \). Then you just need to deal with the the other bit.

- JamesJ

Anyway. What's the other question?

- anonymous

I don't think all of that machinery is too necessary. Since the function is continuous on a closed interval, it must be bounded. Then some manipulations followed by a squeeze theorem will give you the result.
Anyhoo, I'll type up the next question (an inequality) real quick

- JamesJ

Yes, you don't need the uniform convergence. But I'd still break it up into the two pieces.

- JamesJ

...which isn't to say there isn't another way to skin that cat. Just the way I'd do it.

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