Let $f$ be a positive function. Let

${I_1} = \int_{1 - k}^k {x\,f\left\{ {x(1 - x)} \right\}} \,dx$, ${I_2} = \int_{1 - k}^k {\,f\left\{ {x(1 - x)} \right\}} \,dx$

when $2k - 1 > 0.$ Then ${I_1}/{I_2}$ is

The value of integral $\int_0^1 {{e^{{x^2}}}} dx$ lies in interval

Let $\frac{d}{{dx}}F(x) = \left( {\frac{{{e^{\sin x}}}}{x}} \right)\,;\,x > 0$. If $\int_{\,1}^{\,4} {\frac{3}{x}{e^{\sin {x^3}}}dx = F(k) - F(1)} $, then one of the possible value of $k$, is

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 $f$

Let $I_n=\int_0^{\pi / 2} x^n \cos x d x$, where $n$ is a non-negative integer. Then, $\sum \limits_{n=2}^{\infty}\left(\frac{I_n}{n !}+\frac{I_n-2}{(n-2) !}\right)$ equals

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