Equal Areas 2

$\lim_{n \rightarrow \infty} \sum_{j = 1}^{n} (\dfrac{j}{n})^n x^{n - j} = \sum_{j = 0}^{n - 1} (1 - \dfrac{j}{n})^{n} x^{j} = \sum_{n = 0}^{\infty} (\dfrac{x}{e})^n = \dfrac{e}{e - x}$ on $|x| < e$

$\implies f(x) = \lim_{n \rightarrow \infty} \dfrac{\sum_{j = 1}^{n} x^{j}}{\sum_{j = 1}^{n} (\dfrac{j}{n})^n x^{n - j}} = (\dfrac{x}{1 - x})(\dfrac{e - x}{e}) = \dfrac{x(e - x)}{e(1 - x)}$ on $|x| < 1$ .

$\lim_{x \rightarrow 0} f(x) =\lim_{x \rightarrow 0} g(x) = 0 \implies c = 0$ and $f(e - 2) = g(e - 2) = \dfrac{2(e - 2)}{e(3 - e)} \implies$ $\boxed{(e - 2)a + b = \dfrac{2}{e(3 - e)}}$

$\int_{0}^{e - 2} g(x) = \int_{0}^{e - 2} f(x) = \dfrac{1}{e}\int_{0}^{e - 2} (x + \dfrac{e - 1}{1 - x} - (e - 1)) \:\ dx = \dfrac{1}{2e}(x^2 - 2(e - 1)(\ln(1 - x) + x))|_{0}^{e - 2} =$

$\dfrac{1}{2e}(2(e - 1)\ln(\dfrac{1}{3 - e}) -e(e - 2))$

$\implies \dfrac{1}{2e}(2(e - 1)\ln(\dfrac{1}{3 - e}) -e(e - 2)) = \int_{0}^{e - 2} g(x) \:\ dx = \dfrac{a(e - 2)^3}{3} + \dfrac{b(e - 2)^2}{2} \implies$

$\boxed{2(e - 2)a + 3b = \dfrac{3}{e(e - 2)^2}(2(e - 1)\ln(\dfrac{1}{3 - e}) - e(e - 2))}$

and

$\boxed{(e - 2)a + b = \dfrac{2}{e(3 - e)}}$

Solving the system we obtain:

$a = \dfrac{3(e - 1)}{e(e - 3)(e - 2)^3}((e - 2)(e - 4) - 2(3 - e)\ln(3 - e))$

and

$b = -\dfrac{1}{e(e - 3)(e - 2)^2}((e - 2)(3e^2 - 13e + 8) - 6(e - 1)(3 - e)\ln(3 - e))$

$\int_{2 - e}^{0} f(x) \:\ dx = \dfrac{1}{2e}(x^2 - 2(e - 1)(\ln(1 - x) + x))|_{2 - e}^{0} =$ $-\dfrac{1}{2e}((3e - 4)(e - 2) - 2(e - 1)\ln(e - 1)) \approx \boxed{-.206757934}$

$\int_{2 - e}^{0} g(x) \:\ dx = \dfrac{1}{6}(2ax^3 + 3bx^2)|_{2 - e}^{0} = \dfrac{1}{6}(2a(e - 2)^3 - 3b(e - 2)^2) =$

$\dfrac{1}{2e(e - 3)}((e - 2)(5e^2 - 23e + 16) - 10(e - 1)(3 - e)\ln(3 - e)) \approx \boxed{.48658699}$

$\implies \int_{2 - e}^{0} g(x) - f(x) \:\ dx \approx \boxed{0.69334499}$