Integral, x = 0 to 0.5, of ln(1 + x^4), dx: use power series

[huntsiekfind]

Need help again. Use power series to approximate the following integral to six decimal places

.5
?ln(1+x^4)dx
0

 

[Deleted member 4993]

Re: Integral

DUPLICATE POST:

http://www.sosmath.com/CBB/viewtopic.ph ... c32980c72f

Need help again. Use power series to approximate the following integral to six decimal places

.5
?ln(1+x^4)dx
0

Use Taylor series to expand ln(1+x^4)

Please show your work, indicating exactly where you are stuck, so that we know where to begin to help you.

 

[huntsiekfind]

Re: Integral

Need help again. Use power series to approximate the following integral to six decimal places

.5
?ln(1+x^4)dx
0

I have scanned a copy of my work, but can't get it to send it. Is there another way?

 

[BigGlenntheHeavy]

Re: Integral

$\displaystyle \int_{0}^{1/2}ln(1+x^{4})dx \ = \ .006144515218, \ (Trusty \ TI-89).$

$\displaystyle Series \ for \ elementary \ function \ ln(x) :$

$\displaystyle ln(x) \ = \ (x-1)-\frac{(x-1)^{2}}{2}+\frac{(x-1)^{3}}{3}-\frac{(x-1)^{4}}{4}+...+\frac{(-1)^{n-1}(x-1)^{n}}{n}+...$

$\displaystyle Replacing \ x \ with \ (1+x^{4}) \ in \ the \ series \ for \ ln(x), \ we \ have;$

$\displaystyle \int_{0}^{1/2}ln(1+x^{4})dx \ = \ \int_{0}^{1/2}[x^{4}-\frac{x^{8}}{2}+\frac{x^{12}}{3}-\frac{x^{16}}{4}+\frac{x^{20}}{5}+...]dx$

$\displaystyle = \ \frac{x^{5}}{5}-\frac{x^{9}}{18}+\frac{x^{13}}{39}-\frac{x^{17}}{68}+\frac{x^{21}}{105}+...]_{0}^{1/2}$

$\displaystyle = \ \frac{.5^{5}}{5}-\frac{.5^{9}}{18}+\frac{.5^{13}}{39}-\frac{.5^{17}}{68}+\frac{.5^{21}}{105}+... \ = \ .006144515408$