Let $J=\int_0^1 \frac{x}{1+x^8} d x$

Consider the following assertions:

$I$. $J>\frac{1}{4}$

$II$. $J<\frac{\pi}{8}$ Then,

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

Let $\mathrm{a}$ and $\mathrm{b}$ be real constants such that the function $f$ defined by $f(x)=\left\{\begin{array}{cc}x^2+3 x+a & x \leq 1 \\ b x+2, & x>1\end{array}\right.$ be differentiable on $R$. Then, the value of $\int_{-2}^2 f(x) d x$ equals

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 function $L(x) = \int_1^x {\frac{{dt}}{t}} $ satisfies the equation

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