$\int_{0}^{2} ( |2x^2 - 3x| + [x - \frac{1}{2}] ) dx$,where $[ \cdot ]$ is the greatest integer function,is equal to:

Let $f^{\prime}(x)=\frac{192 x^3}{2+\sin ^4 \pi x}$ for all $x \in R$ with $f\left(\frac{1}{2}\right)=0$. If $m \leq \int_{1 / 2}^1 f(x) d x \leq M$,then the possible values of $m$ and $M$ are

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

Let ${I_1} = \int\limits_0^1 {\frac{{{e^x}}}{{1 + x}}} \,dx$ and ${I_2} = \int\limits_0^1 {\frac{{{x^2}}}{{{e^{{x^3}}}\left( {2 - {x^3}} \right)}}} \,dx$,then the value of $\frac{{{I_1}}}{{{I_2}}}$ is equal to

If $b_{n} = \int_{0}^{\frac{\pi}{2}} \frac{\cos^{2} nx}{\sin x} dx$,$n \in N$,then

Let $f(\alpha) = \int_{0}^{\alpha} x^{2} \left(1 - \frac{x}{\alpha}\right)^{\alpha} dx$ (where $\alpha > 0$),then $\sum_{\alpha=1}^{5} \frac{f(\alpha)}{\alpha^{3}}$ is equal to-