Answer
$I=0.0026663619$
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: $f(x)=x^2-\dfrac{x^6}{6}$
Integrate it to obtain: $f'(x)=\dfrac{-x^3(x^4-14)}{42}$
Now take integral.
$I=\int_0^{0.2}f(x) \ dx \\=\int_0^{0.2} (x^2-\dfrac{x^6}{6}) \ dx \\=|\dfrac{-x^3(x^4-14)}{42}|_0^{0.2}$
By using a calculator, we get:
$I=0.0026663619$
