# definite integral, w/ power series.

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• Apr 19th 2007, 08:48 PM
rcmango
definite integral, w/ power series.
have to Use power series to approximate the definite integral to six decimal places.

INT 0_0.2 1/(1 + x^5) dx

its definite from 0 to 0.2 showed above there.

appreciate any help for this one. integrals are difficult for me with series especially.

thankyou.
• Apr 19th 2007, 10:11 PM
CaptainBlack
Quote:

Originally Posted by rcmango
have to Use power series to approximate the definite integral to six decimal places.

INT 0_0.2 1/(1 + x^5) dx

its definite from 0 to 0.2 showed above there.

appreciate any help for this one. integrals are difficult for me with series especially.

thankyou.

1/(1+x^5) = 1 - x^5 + x^10 - R(x)

with remainder R(x)>0, less than 0.2^15 on (0, 0.2)

Therefore:

I = INT 0_0.2 1/(1 + x^5) dx = integral_{x=0 to 0.2} 1 - x^5 + x^10 - R(x) dx

..................................= 0.2 - (0.2)^6/6 + (0.2)^11/11 + epsilon

where |epsilon| < (0.2)^16.

So I ~= 0.194444 to six decimal places

RonL
• Apr 19th 2007, 11:13 PM
rcmango
thankyou.