I have $f(x) = x^{5}$ with center $c=5$. I am trying to use a Taylor series with three terms and include an "error" term to account for the rest.
My error term is as follows: $\int_{x}^{2} (x-c)f'''(x)dx => \int_{x}^{2} (x-2) 60x^{2} => -80 - \frac{x^{4}}{4} + \frac{2}{3}x^{3}$.
Therefore, the Taylor series with three terms + the error term would be:
$x^{5} + 5x^{4}(x-2) + 10x^{3}(x-2)^{2} - 80 - \frac{x^{4}}{4} + \frac{2}{3}x^{3}$.
So now, I am trying to re-express the error term in the integral mean value form (i.e., $\int_{a}^{b} g(x)dx=(b-a)g(\xi)$). I am a little confused how to go about this.
Thanks in advance for your guidance!
Given that the error term is $-80 - \frac{x^{4}}{4} + \frac{2}{3}x^{3}$, we can reexpress it using the Integral Mean Value Theorem by doing such:
$(2-x)g(\xi)$, where $x < \xi < 2$ (2 being the center).
If you wanted to estimate 1.8, just plug 1.8 in for $x$.