Suppose we define the definite integral using the following formula $$\int_\limits{a}^{b} f(x) d x=\frac{b-a}{2}(f(a)+f(b))$$, for more accurate result for

$$c \in(a, b) \mathrm{F}(c)=\frac{c-a}{2}(f(a)+f(c))+\frac{b-c}{2}(f(b)+f(c))$$.

When $$c=\frac{a+b}{c}, \int_\limits{a}^{b} f(x) d x=\frac{b-a}{4}(f(a)+f(b)+2 f(c))$$

$$\int_\limits{0}^{\pi / 2} \sin x d x$$ is equal to:

Suppose we define the definite integral using the following formula $$\int_\limits{a}^{b} f(x) d x=\frac{b-a}{2}(f(a)+f(b))$$, for more accurate result for

$$c \in(a, b) \mathrm{F}(c)=\frac{c-a}{2}(f(a)+f(c))+\frac{b-c}{2}(f(b)+f(c))$$.

When $$c=\frac{a+b}{c}, \int_\limits{a}^{b} f(x) d x=\frac{b-a}{4}(f(a)+f(b)+2 f(c))$$

If $$\lim_\limits{t \rightarrow a} \frac{\int_{a}^{t} f(x) d x-\frac{(t-a)}{2}\{f(t)+f(a)\}}{(t-a)^{3}}=0$$ then the degree of polynomial function $$f(x)$$ almost is:

Suppose we define the definite integral using the following formula $$\int_\limits{a}^{b} f(x) d x=\frac{b-a}{2}(f(a)+f(b))$$, for more accurate result for

$$c \in(a, b) \mathrm{F}(c)=\frac{c-a}{2}(f(a)+f(c))+\frac{b-c}{2}(f(b)+f(c))$$.

When $$c=\frac{a+b}{c}, \int_\limits{a}^{b} f(x) d x=\frac{b-a}{4}(f(a)+f(b)+2 f(c))$$

$$f''(x) < 0 \forall x \in(a, b)$$ and $$c$$ is a point such that $$a < c < b$$, and $$(c, f(C))$$ is the point lying on the curve for which $$\mathrm{F}(C)$$ is maximum, then $$f'(C)$$ is equal to:

$$A=\left[\begin{array}{lll}1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1\end{array}\right]$$, if $$U_{1}, U_{2}$$ and $$U_{3}$$ are columns matrices satisfying. $$\mathrm{AU}_{1}=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right], \quad \mathrm{AU}_{2}=\left[\begin{array}{l}2 \\ 3 \\ 0\end{array}\right], \quad \mathrm{AU}_{3}=\left[\begin{array}{l}2 \\ 3 \\ 1\end{array}\right]$$ and $$\mathrm{U}$$ is $$3 \times 3$$ matrix whose columns are $$\mathrm{U}_{1}, \mathrm{U}_{2}, \mathrm{U}_{3}$$ then answer the following questions