Answer
$0.020380071$
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: $f(x)=\dfrac{-x^8}{4}+\dfrac{x^6}{3}-\dfrac{x^4}{2}+x^2$
Integrate it to obtain: $f'(x)=\dfrac{-x^9}{36}+\dfrac{x^7}{21}-\dfrac{x^5}{10}+\dfrac{x^3}{3}$
Now take integral.
$I=\int_{0}^{0.4} f(x) \ dx\\=\int_{0}^{0.4} (\dfrac{-x^8}{4}+\dfrac{x^6}{3}-\dfrac{x^4}{2}+x^2) \ dx \\=|\dfrac{-x^9}{36}+\dfrac{x^7}{21}-\dfrac{x^5}{10}+\dfrac{x^3}{3}|_{0}^{0.4}$
By using a calculator, we get:
$I=0.020380071$
