$\int_0^1 \frac{x}{(1-x)^{3/4}} dx = $

The approximate value of $\int_1^3 \frac{dx}{2+3x}$ using Simpson's rule and dividing the interval $[1,3]$ into two equal parts is

If $f : R \to R$ is a continuous function such that $f(x) = \int\limits_1^x {tf(t)dt}$,then the correct statement is -

Let $f(x) = 2 + |x| - |x - 1| + |x + 1|$,$x \in R$. Consider:
$(S1): f^{\prime}\left(-\frac{3}{2}\right) + f^{\prime}\left(-\frac{1}{2}\right) + f^{\prime}\left(\frac{1}{2}\right) + f^{\prime}\left(\frac{3}{2}\right) = 4$
$(S2): \int_{-2}^{2} f(x) dx = 12$
Then,

If $\int_0^3 (3x^2 - 4x + 2) dx = k$, then a root of $3x^2 - 4x + 2 = \frac{3k}{5}$ that lies in the interval $[0, 3]$ is

Let the function $f :[0,2] \rightarrow R$ be defined as $f(x)=\begin{cases} e^{\min \{x^2, x-[x]\}}, & x \in[0,1) \\ e^{[x-\log_e x]}, & x \in[1,2] \end{cases}$ where $[t]$ denotes the greatest integer less than or equal to $t$. Then the value of the integral $\int_0^2 x f(x) dx$ is