Answer
$0.015$
Work Step by Step
The Taylor approximation for degree $n$ at $x=p$ can be written as:
$f(x)=f(p)+f'(p)(x-p)+\dfrac{1}{2}f""(p)(x-p)^2+.......\dfrac{1}{n!}f^n(p)(x-p)^n$----
Our aim is to solve the integral.
$I=\int_{0}^{1} x^2 \tan^{-1} x \ dx \\=\int_{0}^{1} x^2 (x-\dfrac{x^3}{3}+\dfrac{x^5}{5}+....) \ dx \\=\int_0^1 (x^3-\dfrac{x^5}{3}+\dfrac{x^7}{5}+.......) \ dx $
By using a calculator, we get:
$I=[\dfrac{x^4}{4}-\dfrac{x^6}{18}+\dfrac{x^8}{40}]_0^{1/2} =0.015$
