4a)
$ 0 \leq sin^n(x)\leq 1 $ on $ [0, \pi] $ and 1 is integrable on $ [0,\pi] $, so by Leb. Dom. Conv. Thm.:
$ lim_n \int_0^{\pi}sin^n(x)dx= \int_0^{\pi}lim_n \ sin^n(x)dx = \int_0^{\pi} \chi_{\{ \frac{\pi}{2}\} } = 0 $.
b) Note that if $ x \in [\frac{\pi}{4},\frac{3\pi}{4}] $ then $ 1 \leq 2sin(x) \Rightarrow 2^nsin^n(x)\leq 2^{n+1}sin^{n+1}(x) $.
Then using MCT,
$ lim_n \int_0^\pi 2^nsin^n(x) dx \geq lim _n \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} 2^nsin^n(x) dx = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} lim _n \ [2sin(x)]^n dx = \infty. $ (since $ [2sin(x)] > 1 $).
So $ lim_n \int_0^\pi 2^nsin^n(x) dx = \infty $.
--Wardbc 15:39, 8 July 2008 (EDT)
