Can You Convert an Integral Into an Infinite Series?

[maxpowers_00]

hi, i need a little help calculating the infinite series sorry if it seems confusing, but i don't know how to put in the sigma or intergral symbols i did my best to make it clear:

i am sopposed to express (integral from 0 to 1) of 2dx/[(3x^4)+16] as a sum of an infinite series here's what i did:

the (integral from 0 to 1) of 2dx/[(3x^4) + 16]
i found that the series of 2dx/[(3x^4)+16] = 1/8* (sigma from n=0 to infinity) of (-1)^n *(3x^4/16)^n

i then pulled out the constants and got

1/8 * (sigma from n=0 to infinity) (-1)^n * (3/16)^n (integral from 0 to 1) (x^4n)

after evaluating the integral i got for my infinit series

1/8 (sigma from n=0 to infinity) (-1)^n (3/16)^n [1^(4n-1)/4n-1]

i just wanted to know if this seemed like it was the right way.
thanks

 

[turin]

i just wanted to know if this seemed like it was the right way.

Yes. I didn't look at the details, but you seem to have the correct approach.

 

[M98Ranger]

Yes, your approach seems to be correct. You have correctly expressed the given integral as a sum of an infinite series using the formula for the geometric series. However, there is a small error in your final answer. The correct expression for the infinite series would be:

1/8 * (sigma from n=0 to infinity) (-1)^n * (3/16)^n [1^(4n+1)/4n+1]

The exponent should be 4n+1 instead of 4n-1. Other than that, your method is correct. Keep up the good work!