http://us.f6.yahoofs.com/hkblog/_9eQ...AiL9KBfQqz3hPo

how to solve (b) ......?

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- Nov 5th 2009, 12:30 AMkinproblem of Integral proof
http://us.f6.yahoofs.com/hkblog/_9eQ...AiL9KBfQqz3hPo

how to solve (b) ......? - Nov 5th 2009, 12:39 AMchisigma
You have to take into account that for $\displaystyle |x|<1$ is...

$\displaystyle \frac{x^{4}\cdot (1-x)^{4}}{1-x^{2}}= (1-x)^{4}\cdot (x^{4} + x^{6} + x^{8} + \dots)$

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Nov 5th 2009, 01:02 AMkin
- Nov 5th 2009, 01:32 AMchisigma
- Nov 5th 2009, 02:40 AMLaurent
It seems that you're not expected to use a) to evaluate the integral. The simplest way anyway is probably to expand the numerator, and perform the Euclidian division of $\displaystyle X^4(1-X)^4$ by $\displaystyle 1+X^2$ (like with integers, you know?), and get something like $\displaystyle X^4(1-X)^4=Q(X)(1+X^2)+aX+b$ hence your integral equals $\displaystyle \int_0^1 Q(x)dx + a\int_0^1 \frac{x}{1+x^2}dx + b\int_0^1\frac{dx}{1+x^2}$. The first term is easy, the second term has antiderivative $\displaystyle \frac{1}{2}\log(1+x^2)$, and the third one relates to arctan.

For the second part, you probably have to write $\displaystyle 1\leq 1+x^2\leq 2$ hence $\displaystyle \frac{1}{2}x^4(1-x)^4\leq \frac{x^4(1-x)^4}{1+x^2}\leq x^4(1-x)^4$ and integrate. - Nov 5th 2009, 04:45 AMkin
- Nov 5th 2009, 04:48 AMkin
- Nov 5th 2009, 06:21 AMdebanik1
which book are these problems from?

- Nov 5th 2009, 06:37 AMkin