Answer
$I=0.24488789$
Work Step by Step
The Taylor approximation for degree $n$ at $x=p$ can be written as:
$f(x)=f(p)+f'(p)(x-p)+\dfrac{1}{2}f""(p)(x-p)^2+.......\dfrac{1}{n!}f^n(p)(x-p)^n$----
We have: $f(x)=\dfrac{x^8}{24}-\dfrac{x^6}{6}+\dfrac{x^4}{2}-x^2+1$
Integrate it to obtain: $f'(x)=\dfrac{x^9}{216}-\dfrac{x^7}{42}+\dfrac{x^5}{10}-\dfrac{x^3}{3}+x$
Now take integral.
$I=\int_0^{0.25}f(x) \ dx \\=\int_0^{0.25} (\dfrac{x^8}{24}-\dfrac{x^6}{6}+\dfrac{x^4}{2}-x^2+1) \ dx \\=|\dfrac{x^9}{216}-\dfrac{x^7}{42}+\dfrac{x^5}{10}-\dfrac{x^3}{3}+x|_0^{0.25}$
By using a calculator, we get:
$I=0.24488789$
