Why couldn't it be a 1?

$\large \displaystyle \int_{0}^{1} x^5 (2-x)^4 \, dx\\ \large \displaystyle = \int_0^1 x^5 (2^4 + x^4 - 4 \times 2^3 \times x - 4 \times 2 \times x^3 + 6 \times 2^2 \times x^2) \, dx\\ \large \displaystyle = \int_0^1 (16x^5 + x^9 - 32 \times x^6 - 8 \times x^8 + 24 \times x^7) \, dx\\ \large \displaystyle = \left[ \frac{16x^6}{6} + \frac{x^10}{10} - \frac{32x^7}{7} - \frac{8x^9}{9} + \frac{24x^8}{8} \right]_0^1\\ \large \displaystyle = \frac{16}{6} + \frac{1}{10} - \frac{32}{7} - \frac{8}{9} + \frac{24}{8} - 0\\ \large \displaystyle = \frac{80640 + 3024 - 138240 - 26880 + 90720}{30240} = \frac{9264}{30240} = \frac{193}{630}\\ \large \displaystyle \implies \frac{a}{b} = \frac{193}{630}\\ \large \displaystyle \therefore a + b = 193 + 630 = \color{#3D99F6}{\boxed{823}}.$

I used Beta function as mentioned by Hobart Pao .

$\begin{aligned} I & = \int_0^1 x^5(2-x)^4 \ dx \quad \quad \small \color{#3D99F6}{\text{Let }2-x=1+\sin \theta \implies x = 1-\sin \theta \implies dx = - \cos \theta \ d\theta} \\ & = \int_0^\frac{\pi}{2} (1-\sin \theta)^5(1+\sin \theta)^4 \cos \theta \ d\theta \\ & = \int_0^\frac{\pi}{2} (1-\sin \theta)(1-\sin^2 \theta)^4 \cos \theta \ d\theta \\ & = \int_0^\frac{\pi}{2} (1-\sin \theta) \cos^9 \theta \ d\theta \\ & = \int_0^\frac{\pi}{2} (\sin^0 \theta \cos^9 \theta -\sin \theta \cos^9 \theta) \ d\theta \\ & = \frac{1}{2}\left( B \left(\frac{1}{2}, 5 \right) - B \left(1, 5 \right) \right) \\ & = \frac{1}{2} \left(\frac{\Gamma \left(\frac{1}{2}\right) \Gamma (5)}{\Gamma \left(\frac{11}{2}\right)} - \frac{\Gamma (1) \Gamma (5)}{\Gamma (6)} \right) \\ & = \frac{4!}{2} \left(\frac{2^5 \sqrt{\pi}}{9!! \sqrt{\pi}} - \frac{1}{5!} \right) \\ & = \frac{256}{630} - \frac{1}{10} = \frac{193}{630} \end{aligned}$

$\implies a + b = 193 + 630 = \boxed{823}$

It will be better to inform that a/b is in the simplest form