If $f(x) = \int_{x}^{x^2} (t - 1) dt$,$1 \le x \le 2$,then the global maximum value of $f(x)$ is

Let the slope of the line $45 x+5 y+3=0$ be $27 r_1+\frac{9 r_2}{2}$ for some $r_1, r_2 \in R$. Then $\lim_{x \rightarrow 3} \left( \int_3^x \frac{8 t^2}{\frac{3 r_2 x}{2}-r_2 x^2-r_1 x^3-3 x} dt \right)$ is equal to ...................

The points of extremum of $\int_0^{x^2} \frac{t^2 - 5t + 4}{2 + e^t} \,dt$ are

$\int_0^{\pi / 2} \sin ^m x \cos ^4 x \, dx = \frac{7 \pi}{2048} \Rightarrow m = ?$

Given that $\frac{d}{d x} \int_0^{\phi(x)} f(t) d t=f(\phi(x)) \phi^{\prime}(x)$. For all $x \in \left(0, \frac{\pi}{2}\right)$,if $\int_1^{\cos x} t^2 f(t) d t=\cos 2 x$,then $f\left(\frac{1}{\sqrt{2}}\right)=$

$\lim \limits_{x \rightarrow 1} \left( \frac{\int \limits_{0}^{(x-1)^{2}} t \cos(t^{2}) dt}{(x-1) \sin(x-1)} \right)$ is equal to