The value of $\int_1^3 {\sqrt {3 + {x^3}} \,dx} $ lies in the interval

The true solution set of the inequality,

$\sqrt {5\,x\,\, - \,\,6\,\, - \,\,{x^2}} \,\, + \,\,\frac{\pi }{2}\,\,\int\limits_0^x {} $$dz > x \int\limits_0^\pi  {} sin^2 x \,dx$ is :

The minimum value of the function $f(x)=\int \limits_0^2 e^{|x-t|} d t$ is

If $I$ is the greatest of the definite integrals

${I_1} = \int_0^1 {{e^{ - x}}{{\cos }^2}x\,dx} , \,\, {I_2} = \int_0^1 {{e^{ - {x^2}}}} {\cos ^2}x\,dx$

${I_3} = \int_0^1 {{e^{ - {x^2}}}dx} ,\,\,{I_4} = \int_0^1 {{e^{ - {x^2}/2}}dx} ,$ then

The points of intersection of
${F_1}(x) = \int_2^x {(2t - 5)\,dt} $ and ${F_2}(x) = \int_0^x {2t\,dt,} $ are

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