Answer
$0.382$
Work Step by Step
The Taylor approximation for degree $n$ at $x=p$ can be written as:
$f(x)=f(p)+f'(p)(x-p)+\dfrac{1}{2}f""(p)(x-p)^2+.......\dfrac{1}{n!}f^n(p)(x-p)^n$----
and $\cos x=1-\dfrac{(x)^2}{2!}+\dfrac{(x)^4}{4!}+......$
Now take integral.
$I=\int_{0}^{1} x \cos x \ dx \\=\int_{0}^{1} x (1-\dfrac{(x)^2}{2!}+\dfrac{(x)^4}{4!}+....) \ dx \\=\int_0^1 (x-\dfrac{x^3}{2!}+\dfrac{x^5}{4!}+.......) \ dx $
By using a calculator, we get:
$I=[\dfrac{x^2}{2}-\dfrac{x^4}{8}+\dfrac{x^6}{144}]_0^1 =0.382$
