Let $\mathrm{f}$ be a non-negative function in $[0,1]$ and twice differentiable in $(0,1) .$ If $\int_{0}^{x} \sqrt{1-\left(f^{\prime}(t)\right)^{2}} \,d t=\int \limits_{0}^{x} f(t) \,d t$ $0 \leq x \leq 1$ and $f(0)=0$, then $\lim \limits _{x \rightarrow 0} \frac{1}{x^{2}} \int \limits_{0}^{x} f(t)\, d t:$

Let $I_1 = \int\limits_0^{\frac{\pi }{2}} {{e^{ - {x^2}}}\sin (x)dx} $ ; $I_2 = \int\limits_0^{\frac{\pi }{2}} {{e^{ - {x^2}}}dx} $ ; $I_3 = \int\limits_0^{\frac{\pi }{2}} {{e^{ - {x^2}}}(1 + x)\,dx} $

and consider the statements

$I\,:$ $I_1 < I_2$   

$II\,:$  $I_2 < I_3$ 

$III\,:$  $I_1 = I_3$

Which of the following is $(are)$ true?

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

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

Suppose $f(x)$ is a differentiable real function such that $f(x) + f'(x) \le 1$ for all $x$ and $f(0)=0$ . The largest possible value of $f(1)$ is