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

Let $f(x)$ be a function satisfying $f'(x) = f(x)$ with $f(0) = 1$ and $g(x)$ be a function satisfying $f(x) + g(x) = x^2$. The value of the integral $\int_{0}^{1} f(x)g(x) \, dx$ is

Let $f: R \rightarrow R$ be a function defined as $f(x) = a \sin \left(\frac{\pi[x]}{2}\right) + [2-x]$,$a \in R$,where $[t]$ is the greatest integer less than or equal to $t$. If $\lim_{x \rightarrow -1} f(x)$ exists,then the value of $\int_{0}^{4} f(x) dx$ is equal to.

$A$ polynomial function $f(x)$ satisfying the conditions $f(x) = [f'(x)]^2$ and $\int_{0}^{1} f(x) dx = \frac{19}{12}$ can be:

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

The tangent to the graph of the function $y = f(x)$ at the point with abscissa $x = a$ forms with the $x$-axis an angle of $\pi/3$ and at the point with abscissa $x = b$ at an angle of $\pi/4$. Then the value of the integral $\int_{a}^{b} f(x) \cdot f''(x) \, dx$ is equal to (assume $f''(x)$ to be continuous).