Cubic function and Alternating Series

Let $(0 \leq x \leq 1)$ .

Let $f(x) = \sum_{n = 1}^{\infty} (-1)^{n + 1} \dfrac{x^n}{n}$ .

$\implies \dfrac{d}{dx}(f(x)) = \sum_{n = 1}^{\infty} (-1)^{n + 1} x^{n - 1} = \dfrac{1}{1 + x} \implies f(x) = \int_{0}^{x} \dfrac{1}{1 + x} dx = \ln(1 + x)$ .

$g(0) = f(0) = 0 \implies c = 0$ and $g(1) = f(1) = \ln(2) \implies \boxed{a + b = \ln(2) - 1}$

For $\int_{0}^{1} \ln(1 + x) dx$

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

$\implies 2\ln(2) - 1 = \int_{0}^{1} g(x) dx =\int_{0}^{1} x^3 + ax^2 + bx dx = (\dfrac{x^4}{4} + \dfrac{ax^3}{3} + \dfrac{bx^2}{2})|_{0}^{1} = \dfrac{1}{4} + \dfrac{a}{3} + \dfrac{b}{3} \implies$

$\boxed{4a + 6b = 24\ln(2) - 15}$

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

$\implies b = \dfrac{20\ln(2) - 11}{2}$ and $a = \dfrac{9}{2}(1 - 2\ln(2)) \implies$ $g(x) = x^3 + \dfrac{9}{2}(1 - 2\ln(2))x^2 + (\dfrac{20\ln(2) - 11}{2})x \implies$

$\int_{0}^{\frac{1}{2}} g(x) - f(x) dx = \dfrac{x}{4} + \dfrac{3}{2}(1 - 2\ln(2))x^3 + (\dfrac{20\ln(2) - 11}{4})x^2|_{0}^{\frac{1}{2}} - ((x + 1)\ln(x + 1) - x)|_{0}^{\frac{1}{2}}) =$

$\dfrac{56\ln(2) - 31}{64} + \dfrac{1}{2} - \dfrac{3}{2}\ln(\dfrac{3}{2}) \approx \boxed{0.01393112}$ .