In $I(m, n) = \int_0^1 x^{m-1}(1-x)^{n-1} dx$,where $m, n > 0$,then $I(9, 14) + I(10, 13)$ is equal to:

$A$ polynomial function $f(x)$ satisfying the conditions $f(x) = [f'(x)]^2$ and $\int_{0}^{1} f(x) dx = \frac{19}{12}$ can be:

Let $f^{\prime}(x)=\frac{192 x^3}{2+\sin ^4 \pi x}$ for all $x \in R$ with $f\left(\frac{1}{2}\right)=0$. If $m \leq \int_{1 / 2}^1 f(x) d x \leq M$,then the possible values of $m$ and $M$ are

If $\int_{0}^{\pi} (\sin^{3} x) e^{-\sin^{2} x} dx = \alpha - \frac{\beta}{e} \int_{0}^{1} \sqrt{t} e^{t} dt$,then $\alpha + \beta$ is equal to $....$

If the abscissa of the vertex of the parabola $y = ax^2 + bx + c$ is $1$ $(a, b, c > 0)$ and $f(x) = \int_0^x (3at^2 + bt + c) dt$ is a strictly increasing function $\forall x \in R$,then the maximum possible value of $[\frac{a}{c}]$ is (where $[.]$ denotes the greatest integer function).

Let $\operatorname{Max} \limits _{0 \leq x \leq 2}\left\{\frac{9-x^{2}}{5-x}\right\}=\alpha$ and $\operatorname{Min} \limits _ {0 \leq x \leq 2}\left\{\frac{9-x^{2}}{5-x}\right\}=\beta$. If $\int\limits_{\beta-\frac{8}{3}}^{2 \alpha-1} \operatorname{Max}\left\{\frac{9- x ^{2}}{5- x }, x \right\} dx =\alpha_{1}+\alpha_{2} \log _{e}\left(\frac{8}{15}\right)$,then $\alpha_{1}+\alpha_{2}$ is equal to