Parameterized Integration - Part 3

$A = \int_{0}^{1} (x-1)^2 f(x) \ dx$ Integrating by parts:

$A = \frac{(x-1)^3}{3}f(x) \biggr \rvert_0^1 - \frac{1}{3}\int_{0}^{1} (x-1)^3f'(x) \ dx$ $A = \frac{(x-1)^3}{3} \left(\int_{0}^{x} \mathrm{e}^{-y^2+2y} \ dy\right) \biggr \rvert_0^1 - \frac{1}{3}\int_{0}^{1} (x-1)^3 \ \mathrm{e}^{-x^2+2x} \ dx$

The above is obtained by substituting $f(x)$ and applying the Leibnitz rule of differentiating an integral. Plugging in the limits of integration simplifies the integral to:

$A = 0- \frac{\mathrm{e}}{3}\int_{0}^{1}(x-1)^3 \ \mathrm{e}^{-(x-1)^2} \ dx$ $A = \frac{\mathrm{e}}{3}\int_{0}^{1} (1-x)^3 \ \mathrm{e}^{-(1-x)^2} \ dx$ $A = \frac{\mathrm{e}}{3}\int_{0}^{1} x^3 \ \mathrm{e}^{-x^2} \ dx$

$\because \int_{0}^{a} f(x) \ dx = \int_{0}^{a} f(a-x) \ dx$

Taking $x^2 = z$ transforms the integral to:

$\implies 10000A = \frac{10000\mathrm{e}}{6}\int_{0}^{1} z \mathrm{e}^{-z} \ dz$

Integrating by parts gives the required closed form solution. The answer is:

$\boxed{\lfloor 10000 A \rfloor = 1197}$