Area Problem 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 $|x| < e$

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

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

$\implies (\dfrac{\sum_{j = 1}^{n} (-1)^{n - j} (\dfrac{j}{n})^n x^{n - j})}{\sum_{j = 1}^{n} (-1)^{j + 1} x^{j}}) = \boxed{\dfrac{e(1 + x)}{x(e + x)}}$ on $|x| < 1$

$\implies m(x) = \dfrac{(e - x)(1 + x)}{(1 - x)(e + x)}$ .

$\lim_{x \rightarrow 0} m(x) =\lim_{x \rightarrow 0} p(x) = 1 \implies c = 1$ and $m(2 - e) = p(2 - e) = 3 - e \implies \boxed{(2 - e)a + b = 1}$ .

For $\int_{2 - e}^{0} m(x) dx$ :

$m(x) = \dfrac{(e - x)(1 + x)}{(1 - x)(e + x)} = \dfrac{x^2 - (e - 1)x - e}{x^2 + (e - 1)x - e} =$ $1 + \dfrac{2(e - 1)x}{(1 - x)(e + 1)}$

Using partial fractions we obtain:

$(A - B)x + (eA + B) = 2(e - 1)x \implies$

$A - B = 2(e - 1)$

$eA + B = 0$

$\implies A = \dfrac{2(e - 1)}{1 + e}$ and $B = \dfrac{-2e(e - 1)}{1 + e} \implies$ $m(x) = 1 + (\dfrac{2(e - 1)}{1 + e}) * (\dfrac{1}{1 - x} - \dfrac{e}{e + x}) \implies$

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

$\implies \dfrac{-e^2 + e - 2 + 2(e - 1)\ln((e - 1)2^e)}{1 + e} = \dfrac{1}{6}(2ax^3 + 3bx^2 + 6x)|_{2 - e}^{0} \implies$ $2(2 - e)^2a + 3(2 - e) + 6 = \dfrac{6(-e^2 + e - 2 + 2(e - 1)\ln((e - 1)2^e))}{(e + 1)(e - 2)} \implies$

$\boxed{2(2 - e)a + 3b = \dfrac{12(e^2 -e - (e - 1)\ln((e - 1)2^e))}{(e + 1)(e - 2)^2}}$

and

$\boxed{(2 - e)a + b = 1}$

Solving the system we obtain:

$b = \dfrac{-2e^3 + 18e^2 - 12e - 8 - 12(e - 1)\ln((e - 1)2^e)}{(e + 1)(e - 2)^2}$

and

$a = \dfrac{-3e^3 + 21e^2 - 12e - 12 - 12(e - 1)\ln((e - 1)2^e)}{(e + 1)(e - 2)^3}$

For $\int_{0}^{e - 2} m(x) - p(x) dx$ .

$\int_{0}^{e - 2} m(x) dx = x - \dfrac{2(e - 1)}{1 + e}(\ln((1 - x)(e + x)^e)|_{0}^{e - 2} =$ $e - 2 - \dfrac{2(e - 1)}{1 + e}\ln((3 - e)(2e - 2)^e) + \dfrac{2(e - 1)e}{1 + e} =$

$\boxed{\dfrac{3e^2 - 3e - 2 - 2(e - 1)\ln((3 - e)(2e - 2)^e)}{1 + e} \approx 1.3000753}$

and

$\int_{0}^{e - 2} p(x) dx = \dfrac{1}{6}(2ax^3 + 3bx^2 + 6x)|_{0}^{e - 2} =$ $\dfrac{-6e^3 + 42e^2 - 24e - 24 - 24(e - 1)\ln((e - 1)2^e) + -6e^3 + 54e^2 - 36e - 24 - 36(e - 1)\ln((e - 1)2^e) + 6e^2 - 6e - 12}{6(e + 1)}$

$= \boxed{\dfrac{-6e^3 + 51e^2 - 33e - 30 - 30(e - 1)\ln((e - 1)2^e)}{3(e + 1)} \approx 1.0394459}$

$\implies \int_{0}^{e - 2} m(x) - p(x) dx \approx \boxed{0.2606294}$ .