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

If $$f\left( x \right) = \int\limits_{{x^2}}^{{x^2} + 1} {{e^{ - {t^2}}}} dt,$$ then $$f(x)$$ increases in

If $$l\left( {m,n} \right) = \int\limits_0^1 {{t^m}{{\left( {1 + t} \right)}^n}dt,} $$ then the expression for $$l(m, n)$$ in terms of $$l(m+n, n-1)$$ is

Let $$T>0$$ be a fixed real number . Suppose $$f$$ is a continuous
function such that for all $$x \in R$$, $$f\left( {x + T} \right) = f\left( x \right)$$.

If $$I = \int\limits_0^T {f\left( x \right)dx} $$ then the value of $$\int\limits_3^{3 + 3T} {f\left( {2x} \right)dx} $$ is