problem of Integral proof

**kin** · November 5th 2009, 12:30 AM

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

**chisigma** · November 5th 2009, 12:39 AM

You have to take into account that for $|x|<1$ is...

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

Kind regards

$\chi$ $\sigma$

**kin** · November 5th 2009, 01:02 AM

Originally Posted by chisigma

You have to take into account that for $|x|<1$ is...

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

Kind regards

$\chi$ $\sigma$

it's $\frac{x^{4}\cdot (1-x)^{4}}{1+x^{2}}$ not (1-x),,,,,,,,,

**chisigma** · November 5th 2009, 01:32 AM

Originally Posted by kin

it's $\frac{x^{4}\cdot (1-x)^{4}}{1+x^{2}}$ not (1-x)...

Very sorry!

...

... in such a case is...

$\frac{x^{4}\cdot (1-x)^{4}}{1+x^{2}}= (1-x)^{4}\cdot (x^{4} - x^{6} + x^{8} -...)$ , $|x|<1$

Kind regards

$\chi$ $\sigma$

**Laurent** · November 5th 2009, 02:40 AM

Originally Posted by kin

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

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 $X^4(1-X)^4$ by $1+X^2$ (like with integers, you know?), and get something like $X^4(1-X)^4=Q(X)(1+X^2)+aX+b$ hence your integral equals $\int_0^1 Q(x)dx + a\int_0^1 \frac{x}{1+x^2}dx + b\int_0^1\frac{dx}{1+x^2}$ . The first term is easy, the second term has antiderivative $\frac{1}{2}\log(1+x^2)$ , and the third one relates to arctan.

For the second part, you probably have to write $1\leq 1+x^2\leq 2$ hence $\frac{1}{2}x^4(1-x)^4\leq \frac{x^4(1-x)^4}{1+x^2}\leq x^4(1-x)^4$ and integrate.

**kin** · November 5th 2009, 04:45 AM

Originally Posted by chisigma

Very sorry!

...

... in such a case is...

$\frac{x^{4}\cdot (1-x)^{4}}{1+x^{2}}= (1-x)^{4}\cdot (x^{4} - x^{6} + x^{8} -...)$ , $|x|<1$

Kind regards

$\chi$ $\sigma$

but i still dont understand your point...
how does it help to solve the integral..?

**kin** · November 5th 2009, 04:48 AM

Originally Posted by Laurent

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 $X^4(1-X)^4$ by $1+X^2$ (like with integers, you know?), and get something like $X^4(1-X)^4=Q(X)(1+X^2)+aX+b$ hence your integral equals $\int_0^1 Q(x)dx + a\int_0^1 \frac{x}{1+x^2}dx + b\int_0^1\frac{dx}{1+x^2}$ . The first term is easy, the second term has antiderivative $\frac{1}{2}\log(1+x^2)$ , and the third one relates to arctan.

For the second part, you probably have to write $1\leq 1+x^2\leq 2$ hence $\frac{1}{2}x^4(1-x)^4\leq \frac{x^4(1-x)^4}{1+x^2}\leq x^4(1-x)^4$ and integrate.

thank u very much ,
this is a straightforward method....
are there any ways of combining result of (a) to solve it quickly...?

**debanik1** · November 5th 2009, 06:21 AM

which book are these problems from?

**kin** · November 5th 2009, 06:37 AM

Originally Posted by debanik1

which book are these problems from?

they are not from books....just from my tutorial notes...

anyway, i have solved it.

thanks for help , guys...

Thread: problem of Integral proof

LinkBack

Thread Tools

Display

problem of Integral proof

Similar Math Help Forum Discussions

Integral proof

Integral proof problem

Proof that Upper Riemann Integral > Lower Riemann Integral...?

integral proof

integral proof

Search Tags