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

$I=\int \limits_{\pi / 4}^{\pi / 3}\left(\frac{8 \sin x-\sin 2 x}{x}\right) d x$. Then

Let the function $f :[0,2] \rightarrow R$ be defined as

$f(x)=\left\{\begin{array}{cc}e^{\min \left[x^2, x-[x]\right\}}, & x \in[0,1) \\e^{\left[x-\log _e x\right]}, & x \in[1,2]\end{array}\right.$

where [t] denotes the greatest integer less than or equal to $t$. Then the value of the integral $\int \limits_0^2 x f(x) d x$ is

The numbers $P, Q$ and $R$ for which the function $f(x) = P{e^{2x}} + Q{e^x} + Rx$ satisfies the conditions $f(0) = - 1,$ $f'(\log 2) = 31$ and $\int_0^{\log 4} {[f(x) - Rx]\,dx = \frac{{39}}{2}} $ are given by

If $\int_{}^{} {f(x)\,dx} = x{e^{ - \log |x|}} + f(x),$ then $f(x)$ 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 $f$