Let the definite integral be defined by the formula
$$\int\limits_a^b {f\left( x \right)dx = {{b - a} \over 2}\left( {f\left( a \right) + f\left( b \right)} \right).} $$ For more accurate result for
$$c \in \left( {a,b} \right),$$ we can use $$\int\limits_a^b {f\left( x \right)dx = \int\limits_a^c {f\left( x \right)dx + \int\limits_c^b {f\left( x \right)dx = F\left( c \right)} } } $$ so
that for $$c = {{a + b} \over 2},$$ we get $$\int\limits_a^b {f\left( x \right)dx = {{b - a} \over 4}\left( {f\left( a \right) + f\left( b \right) + 2f\left( c \right)} \right).} $$

$$\int\limits_0^{\pi /2} {\sin x\,dx = } $$

$$\int\limits_{ - 2}^0 {\left\{ {{x^3} + 3{x^2} + 3x + 3 + \left( {x + 1} \right)\cos \left( {x + 1} \right)} \right\}\,\,dx} $$ is equal to

The value of the integral $$\int\limits_0^1 {\sqrt {{{1 - x} \over {1 + x}}} dx} $$ is

If $$f(x)$$ is differentiable and $$\int\limits_0^{{t^2}} {xf\left( x \right)dx = {2 \over 5}{t^5},} $$ then $$f\left( {{4 \over {25}}} \right)$$ equals