Thread: definite integral, w/ power series.

1. 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.

2. 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

3. thankyou.