Area Mania

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

Let $g(x) = ax^2 + bx + c$ .

$f(1) = g(1) \implies \boxed{a + b + c = \dfrac{e}{e - 1}}$

$f(e) = g(e) \implies \boxed{e^2a + eb + c = \dfrac{e^2}{e^2 - 1}}$

$\int_{1}^{e} f(x) dx = \int_{1}^{e} 1 + \dfrac{1}{ex - 1} dx = x + \dfrac{1}{e}\ln(ex - 1)|_{1}^{e} = \dfrac{e(e - 1) + \ln(e + 1)}{e} = \int_{1}^{e} g(x) = \dfrac{ax^3}{3} + \dfrac{bx^2}{2} + cx|_{1}^{e} =$

$\dfrac{1}{3}(e - 1)(e^2 + e - 1)a + \dfrac{1}{2}(e - 1)(e + 1)b + (e - 1)c \implies$

$\boxed{2(e^2 + e + 1)a + 3(e + 1)b + 6c = 6(\dfrac{e(e - 1) + \ln(e + 1)}{(e - 1)e})}$

$\implies$

$\boxed{a + b + c = \dfrac{e}{e - 1}}$

$\boxed{e^2a + eb + c = \dfrac{e^2}{e^2 - 1}}$

$\boxed{2(e^2 + e + 1)a + 3(e + 1)b + 6c = 6(\dfrac{e(e - 1) + \ln(e + 1)}{(e - 1)e})}$

Solving the system we obtain:

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

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

$c = \dfrac{e^4 - e^3 + 3e^2 + 6e - 6(1 + e)\ln(1 + e)}{(e - 1)^3 (1 + e)}$

$\int_{6}^{12} g(x) dx = 504a + 54b + 6c =$

$\dfrac{1512(e^2 + 2e - 2(1 + e)\ln(1 + e)) - 108(2e^3 + 4e^2 + 3e - 3(e + 1)^2\ln(1 + e)) + 6e(e^4 - e^3 + 3e^2 + 6e - 6(1 + e)\ln(1 + e))}{(e - 1)^3 (1 + e) e} =$

$\dfrac{6e^5 - 6e^4 - 198e^3 + 1116e^2 + 2700e + 36(8e - 75)(e + 1)\ln(e + 1)}{(e - 1)^3 (1 + e) e}$

$\approx \dfrac{2809.984144}{57.276689} = \boxed{54.800421}$

and

$\int_{6}^{12} f(x) dx = 12 + \dfrac{1}{e}\ln(12e - 1) - 6 - \dfrac{1}{e}\ln(6e - 1) =$ $6 + \dfrac{1}{e}\ln(\dfrac{12e - 1}{6e - 1}) \approx \boxed{6.266817}$

$\implies \int_{6}^{12} g(x) - f(x) dx = 54.800421 - 6.266817 = \boxed{48.533604}$ .