The number of continuous functions $f:[0,1] \rightarrow R$ that satisfy $\int \limits_0^1 x f(x) d x=\frac{1}{3}+\frac{1}{4} \int \limits_0^1(f(x))^2 d x$ is

If ${I_1} = \int_0^1 {{2^{{x^2}}}dx,\;} {I_2} = \int_0^1 {{2^{{x^3}}}dx} ,\;{I_3} = \int_1^2 {{2^{{x^2}}}dx} $,${I_4} = \int_1^2 {{2^{{x^3}}}dx} $, then

Let $f$ be a continuous function defined on $[0,1]$ such that $\int_0^1 f^2(x) d x=\left(\int_0^1 f(x) d x\right)^2$. Then, the range of $f$

The value of integral $\int_0^1 {\frac{{{x^b} - 1}}{{\log x}}} \,dx$ is

If $\frac{d}{{dx}}\,G\left( x \right) = \frac{{{e^{\tan \,x}}}}{x},\,x \in \left( {0,\pi /2} \right)$, then $\int\limits_{1/4}^{1/2} {\frac{2}{x}} .{e^{\tan \,\left( {\pi \,{x^2}} \right)}}dx$ is equal to

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