Convergence, differentiable, integrable, sequence of functions

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Homework Statement



For k = 1,2,\ldots define f_k : \mathbb{R} \to \mathbb{R}
by f_k(x) = \sqrt{k} x^k (1 - x). Does \{ f_k \} converge? In
what sense? Is the limit integrable? Differentiable?

Homework Equations





The Attempt at a Solution



I don't know how to approach this question. How can I determine if the sequence converges? What are the theorems to dertermine if the limit is differentiable or integrable?
 
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http://en.wikipedia.org/wiki/Uniform_convergence

In particular, note the difference between uniform convergence and pointwise convergence under Definition: Notes (this answers the convergence in what sense question). Look under Applications to see the theorems that guarantee that the limit function is differentiable or integrable.
 
Well, look at f_k carefully: it has exactly two real roots no matter what k is: zero and 1. What is happening outside of the interval [0,1]?

Inside the interval, it's a little more interesting. On the interval [0,1], for arbitrary k > 0 where does f_k achieve its maximum value, and what is that maximum value?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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