Answer
$0.69580993$
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: $f(x)=\dfrac{2x^8}{3}-2x^4+1$
Integrate it to obtain: $f'(x)=\dfrac{2x^9}{27}-\dfrac{2x^5}{5}+x$
Now take integral.
$I=\int_{-0.35}^{0.35} (\dfrac{2x^8}{3}-2x^4+1) \ dx \\=|\dfrac{2x^9}{27}-\dfrac{2x^5}{5}+x|_{-0.35}^{0.35}$
By using a calculator, we get:
$I=0.69580993$
