Answer
$0.060056733$
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: $f(x)=\dfrac{-x^7}{7}+\dfrac{x^5}{5}-\dfrac{x^3}{3}+x$
Integrate it to obtain: $f'(x)=\dfrac{-x^8}{56}+\dfrac{x^6}{30}-\dfrac{x^4}{12}+\dfrac{x^2}{2}$
Now take integral.
$I=\int_{0}^{0.35} f(x) \ dx\\=\int_{0}^{0.35} \dfrac{-x^7}{7}+\dfrac{x^5}{5}-\dfrac{x^3}{3}+x\ dx \\=|\dfrac{-x^8}{56}+\dfrac{x^6}{30}-\dfrac{x^4}{12}+\dfrac{x^2}{2}|_{0}^{0.35}$
By using a calculator, we get:
$I=0.060056733$
