# find maclaurin series

• May 19th 2009, 02:24 PM
cm3pyro
find maclaurin series
integral of 1/(2+x^3)dx bounds:0 to .5
• May 19th 2009, 03:03 PM
skeeter
Quote:

Originally Posted by cm3pyro
integral of 1/(2+x^3)dx bounds:0 to .5

$\frac{1}{2+x^3} = \frac{1}{2} \cdot \frac{1}{1 - \left(-\frac{x^3}{2}\right)} = \frac{1}{2}\left(1 - \frac{x^3}{2} + \frac{x^6}{4} - \frac{x^9}{8} + ... \right)$

integrate ...

$\frac{1}{2}\left[x - \frac{x^4}{8} + \frac{x^7}{28} - \frac{x^{10}}{80} + ... \right]_0^{0.5}$

using the first three terms should get you within 0.00001 of the actual value
• May 19th 2009, 03:10 PM
cm3pyro
Quote:

Originally Posted by skeeter
$\frac{1}{2+x^3} = \frac{1}{2} \cdot \frac{1}{1 - \left(-\frac{x^3}{2}\right)} = \frac{1}{2}\left(1 - \frac{x^3}{2} + \frac{x^6}{4} - \frac{x^9}{8} + ... \right)$

integrate ...

$\frac{1}{2}\left[x - \frac{x^4}{8} + \frac{x^7}{28} - \frac{x^{10}}{80} + ... \right]_0^{0.5}$

using the first three terms should get you within 0.00001 of the actual value

i need a maclaurin series
• May 19th 2009, 04:18 PM
skeeter
Quote:

Originally Posted by cm3pyro
i need a maclaurin series

what do you think I posted?