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Hi could anyone help me with this problem. Use series to approximate the value of the following function to two decimal places. Integrate from 1 to 0 $\sqrt{1+x^4}$. I tried to differentiate the expression but I still could not change it to a form that allows me to express it as a power series.

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  • $\begingroup$ I don't get it, $\int_0^1 \sqrt{1+x^4}\;dx$ is a number, not a function. $\endgroup$ – Daniel R Apr 25 '14 at 14:09
  • $\begingroup$ Sorry I edited the question $\endgroup$ – yswong Apr 25 '14 at 14:13
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Well, to start with, consider the function $$f(t)=\sqrt{1+t}=(1+t)^{1/2},$$ and find its Maclaurin series. It shouldn't be difficult to find an explicit formula for the $n$th derivative of $f$ for all $n\ge0,$ and the coefficients of the Maclaurin series will fall right out of that formula.

Now, $\sqrt{1+x^4}=f(x^4),$ so you'll easily get a power series for your function that way, which you'll want to integrate from $0$ to $1$. Can you take it from there?

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HINT :

The series expansion of $(1+y)^{\large\frac12}$ is $$ (1+y)^{\large\frac12}=1+\frac12 y- \frac18 y^2+\frac1{16} y^3-\frac5{128} y^4+\frac7{256} y^5-\cdots $$ then take $y=x^4$ and integrate it.

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