Area Bonanza 2

Let $|x| < 1$ .

$\sum_{n = 1}^{\infty} n x^n = \sum_{j = 1}^{\infty} \sum_{n = j}^{\infty} x^n = \dfrac{1}{1 - x}\sum_{j = 1}^{\infty} x^{j} = \dfrac{1}{1 - x}(x)\dfrac{1}{1 - x} = \dfrac{x}{(1 - x)^2}$

$(\dfrac{x}{x - 1})(\dfrac{1}{x})(\dfrac{x}{x - 1}) = \dfrac{x}{(x - 1)^2}.$

$g(x) = \sum_{n = 1}^{\infty} n^2 x^n = \sum_{j = 1}^{\infty} (\sum_{n = j}^{\infty} n x^n) =$ $\dfrac{1}{1 - x}\sum_{j = 1}^{\infty} (j x^j) + \dfrac{x}{(1 - x)^2}\sum_{j = 1}^{\infty} x^j =$

$(\dfrac{1}{1 - x})(\dfrac{x}{(1 - x)^2}) + (\dfrac{x}{(1 - x)^2})(\dfrac{x}{1 - x}) =$ $\dfrac{x^2 + x}{(1 - x)^3}$ .

$f(x) = \sum_{n = 1}^{\infty} n^3 x^n = \sum_{j = 1}^{\infty} \sum_{n = j}^{\infty} n^2 x^{n} =$ $\dfrac{1}{1 - x}\sum_{j = 1}^{\infty} j^2 x^j + \dfrac{2x}{(1 - x)^2}\sum_{j = 1}^{\infty} j x^j + \dfrac{x^2 + x}{(1 - x)^3}\sum_{j = 1}^{\infty} x^j = (\dfrac{1}{1 - x})(\dfrac{x^2 + x}{(1 - x)^3}) +$ $(\dfrac{2x}{(1 - x)^2})(\dfrac{x}{(1 - x)^2}) + (\dfrac{x^2 + x}{(1 - x)^3})(\dfrac{x}{1 - x}) =$ $\dfrac{x^3 + 4x^2 + x}{(1 - x)^4}$ .

Let $u = 1 -x \implies du = -dx \implies \int_{\frac{-1}{2}}^{0} f(x) dx = -\int_{1}^{\frac{3}{2}} \dfrac{u^3 - 7u^2 + 12u - 6}{u^4} du = -\int_{1}^{\frac{3}{2}} \dfrac{1}{u} - 7u^{-2} + 12u^{-3} - 6u^{-4} du =$

$-(\ln(u) + \dfrac{7}{u} - \dfrac{6}{u^2} + \dfrac{2}{u^3})|_{1}^{\frac{3}{2}} =$ $-(\ln(\dfrac{3}{2}) - \dfrac{11}{27}) = \dfrac{11}{27} - \ln(\dfrac{3}{2})$

and,

$\int_{\frac{-1}{2}}^{0} g(x) dx = \int_{1}^{\frac{3}{2}} \dfrac{u^2 - 3u + 2}{u^3} du =$ $\int_{1}^{\frac{3}{2}} \dfrac{1}{u} - 3u^{-2} + 2u^{-3} du =$ $(\ln(u) + \dfrac{3}{u} - \dfrac{1}{u^2})|_{1}^{\frac{3}{2}} = \ln(\dfrac{3}{2}) - \dfrac{4}{9} \implies$

$\int_{\frac{-1}{2}}^{0} f(x) - g(x) dx = \dfrac{23}{27} - \ln(\dfrac{9}{4}) =$ $\dfrac{23}{3^3} - \ln(\dfrac{3}{2})^2 = \dfrac{a}{b^b} - \ln(\dfrac{b}{c})^c \implies a + b + c = \boxed{28}$ .