The true solution set of the inequality,$\sqrt{5x-6-x^2}+\left( \frac{\pi }{2}\int\limits_{0}^{x}{dz} \right)>x\int\limits_{0}^{\pi }{{{\sin }^{2}}xdx}$ is:

Let $f:[0,1] \rightarrow[0,1]$ be a continuous function such that $x^2+(f(x))^2 \leq 1$ for all $x \in[0,1]$ and $\int_0^1 f(x) d x=\frac{\pi}{4}$ Then, $\int_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}} \frac{f(x)}{1-x^2} d x$ equals

Let the function $f :[0,2] \rightarrow R$ be defined as

$f(x)=\left\{\begin{array}{cc}e^{\min \left[x^2, x-[x]\right\}}, & x \in[0,1) \\e^{\left[x-\log _e x\right]}, & x \in[1,2]\end{array}\right.$

where [t] denotes the greatest integer less than or equal to $t$. Then the value of the integral $\int \limits_0^2 x f(x) d x$ is

$I=\int \limits_{\pi / 4}^{\pi / 3}\left(\frac{8 \sin x-\sin 2 x}{x}\right) d x$. Then

Let $y=f(x)$ be a thrice differentiable function in $(-5,5)$. Let the tangents to the curve $y=f(x)$ at $(1, \mathrm{f}(1))$ and $(3, \mathrm{f}(3))$ make angles $\frac{\pi}{6}$ and $\frac{\pi}{4}$, respectively with positive $x$-axis. If $27 \int_1^3\left(\left(f^{\prime}(t)\right)^2+1\right) f^{\prime \prime}(t) d t=\alpha+\beta \sqrt{3} \quad$ where $\alpha, \quad \beta$ are integers, then the value of $\alpha+\beta$ equals

If $f(x)$ is a quadratic in $x$ , then $\int\limits_0^1 {f(x) dx}$ is