Question

0.5
∫(x^3)/(10+4x^5)dx
0

I need to integrate this using series (Maclaurin/Taylor if im not mistaken). I think i should reduce the function to the (1+x)^m = 1+mx+m(m-1)/2! *x^2 + m(m-1)(m-2)/3! * x^3 + ... form, and then simply integrate, but the problem is, i dont know how to do that.

Answer 1

Ok, I've done the definite integral in Wolfram wich value is 0.001553... so I can check if McLaurin's is fine. If your function is x^3/(4x^5+10) and you compute McLaurin expansion you get x^3/10 - x^8 /25 + O(x^13) (where O(x^n) is the truncation error). So if you compute the integral of x^3/10 - x^8 /25 between 0 an 0.5 you get 0.0016, quite close

Answer 2

Thanks for taking time to look at my problem :) could you please explain how to transform function x^3/(4x^5+10) to something that could be integrated using series (probably to thiat you just wrote: x^3/10 - x^8 /25 + O(x^13).
I basicaly need to do this: http://www.youtube.com/watch?v=3ZOS69YTxQ8 to that function of mine :) hope it makes sense.

Answer 3

Okei, I'm going to put it as a series:

\[\frac{ 1 }{ 1-x }=\sum_{n=0}^{\infty}x^n\]

Answer 4

We want to put your function in a way you can write it using the above so:

\[\frac{ x^3 }{ 10 + 4x^5 }=x^3(\frac{ 1 }{ 10+4x^5 })=\]
\[=\frac{ x^3 }{ 10 }(\frac{ 1 }{ 1-(\frac{ -4x^5 }{ 10 }) })\]

Answer 5

So we can write the function in terms of the serie as:

\[\frac{ x^3 }{ 10 }\sum_{0}^{\infty}(\frac{ -4x^5 }{ 10 })^n=\]

\[=\frac{ x^3 }{ 10}\sum_{0}^{\infty}(-1)^n(\frac{ 4 }{ 10 })^n(x^5)^n=\]

Answer 6

If you put all together: