The value of $\int_{3}^{4} \sqrt{(4-x)(x-3)} d x$ is

If $\int_{0}^{\sqrt{3}} \frac{15 x^{3}}{\sqrt{1+x^{2}+\sqrt{(1+x^{2})^{3}}}} dx = \alpha \sqrt{2} + \beta \sqrt{3}$,where $\alpha, \beta$ are integers,then $\alpha + \beta$ is equal to.

If $I = \int_{0}^{2} e^{x^{4}}(x - \alpha) dx = 0$, then $\alpha$ lies in the interval

Consider the matrices $A = \begin{bmatrix} 2 & -5 \\ 3 & m \end{bmatrix}$,$B = \begin{bmatrix} 20 \\ m \end{bmatrix}$,and $X = \begin{bmatrix} x \\ y \end{bmatrix}$. Let the set of all $m$,for which the system of equations $AX = B$ has a negative solution (i.e.,$x < 0$ and $y < 0$),be the interval $(a, b)$. Then $8 \int_a^b |A| dm$ is equal to.............

$A$ minimum value of $\int_0^x t e^{t^2} d t$ is

$\int\limits_{\frac{\pi }{6}}^{\frac{5\pi }{6}} {\left( {\frac{1}{2}{{(3\sin \theta )}^2} - \frac{1}{2}{{(1 + \sin \theta )}^2}} \right)\,d\theta } $