If $I(m, n)=\int_0^1 x^{m-1}(1-x)^{n-1} d x, m, n>0$, then $I(9,14)+I(10,13)$ is

<p>$$\begin{aligned} & \mathrm{I}(\mathrm{~m}, \mathrm{~m})=\int_0^1 \mathrm{x}^{\mathrm{m}-1}(1-\mathrm{x})^{\mathrm{n}-1} \mathrm{dx} \\ & \text { Let } \mathrm{x}=\sin ^2 \theta \quad \mathrm{dx}=2 \sin \theta \cos \theta \mathrm{~d} \theta \\ & \mathrm{I}(\mathrm{~m}, \mathrm{n})=2 \int_0^{\pi / 2}(\sin \theta)^{2 \mathrm{~m}-1}(\cos \theta)^{2 \mathrm{n}-1} \mathrm{~d} \theta \\ & \mathrm{I}(9,14)+\mathrm{I}(10,13)=2 \int_0^{\pi / 2}(\sin \theta)^{17}(\cos \theta)^{27} \mathrm{~d} \theta \\ & +2 \int_0^{\pi / 2}(\sin \theta)^{19}(\cos \theta)^{25} \mathrm{~d} \theta \\ & =2 \int_0^{\pi / 2}(\sin \theta)^{17}(\cos \theta)^{25} \mathrm{~d} \theta \\ & =\mathrm{I}(9,13) \end{aligned}$$</p>