Quartic functions 2

$f(0) = g(0) = 1 \implies c = 1$ and $f(\dfrac{\pi}{2}) = g(\dfrac{\pi}{2}) = 0 \implies \boxed{\pi^4a + 4\pi^2b = -16}$ and $\int_{0}^{\dfrac{\pi}{2}} g(x) dx = \int_{0}^{\dfrac{\pi}{2}} \cos(x) dx = 1$ $\implies 1 = \int_{0}^{\dfrac{\pi}{2}} (ax^4 + bx^2 + 1) dx =$ $\dfrac{ax^5}{5} + \dfrac{bx^3}{3} + x|_{0}^{\dfrac{\pi}{2}}$ $= \dfrac{\pi^5}{5 * 2^5}a + \dfrac{\pi^3}{3 * 2^3}b + \dfrac{\pi}{2} \implies$

$\boxed{3\pi^5a + 20\pi^3b = 480 - 240\pi}$

$\boxed{\pi^4a + 4\pi^2b = -16}$

Solving the system above we obtain: $a = \dfrac{80(\pi - 3)}{\pi^5}$ and $b = \dfrac{12(5 - 2\pi)}{\pi^3} \implies$

$\int_{0}^{\dfrac{\pi}{4}} f(x) dx = \int_{0}^{\dfrac{\pi}{4}} (\dfrac{80(\pi - 3)}{\pi^5}x^4 + \dfrac{12(5 - 2\pi)}{\pi^3}x^2 + 1) dx = \dfrac{16(\pi - 3)}{\pi^5}x^5 + \dfrac{4(5 - 2\pi)}{\pi^3}x^3 + x|_{0}^{\dfrac{\pi}{4}} =$

$\dfrac{\pi - 3}{4^3} + \dfrac{5 - 2\pi}{4^2} + \dfrac{\pi}{4} = \dfrac{\pi - 3 + 20 - 8\pi + 16\pi}{64} = \dfrac{9\pi + 17}{64}$

and,

$\int_{0}^{\dfrac{\pi}{4}} g(x) dx = \int_{0}^{\dfrac{\pi}{4}} \cos(x) dx = \dfrac{1}{\sqrt{2}} \implies$

$D = \int_{0}^{\dfrac{\pi}{4}} f(x) - g(x) dx = \dfrac{9\pi + 17}{64} - \dfrac{\sqrt{2}}{2} =$ $\dfrac{9\pi + 17 - 32\sqrt{2}}{64} = \dfrac{3^2\pi + 17 - 2^5\sqrt{2}}{2^{2 * 3}} = \dfrac{\alpha^{\beta}\pi + \lambda - \beta^{\gamma}\sqrt{\beta}}{\beta^{\alpha * \beta}} \implies$ $\alpha + \beta + \lambda + \gamma = \boxed{27}$ .

Note: I didn't show the graph in the problem, since the graph suggest that the areas are equal on $[0,\dfrac{\pi}{4}]$ which is not the case. As can be seen from the solution the difference is small ( $\dfrac{9\pi + 17 - 32\sqrt{2}}{64} \approx 0.0003046857$ ).