- #1
maxpowers_00
- 5
- 0
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
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