Answer
$$\mathop {\lim }\limits_{b \to {0^ + }} \int_b^1 {\frac{1}{{\sqrt x }}dx} $$
Work Step by Step
$$\eqalign{
& \int_0^1 {{x^{ - 1/2}}dx} \cr
& {\text{The integrand }}{x^{ - 1/2}}{\text{ can be written as }}\frac{1}{{{x^{1/2}}}},{\text{ it is not defined for}} \cr
& x = 0,{\text{ then we must rewrite this value of the interval, note that}} \cr
& {\text{the upper limit is 1, the values are approaching to 0 to the right, }} \cr
& {\text{then }}\int_0^1 {{x^{ - 1/2}}dx} = \mathop {\lim }\limits_{b \to {0^ + }} \int_b^1 {\frac{1}{{{x^{1/2}}}}dx} \cr
& {\text{ }} = \mathop {\lim }\limits_{b \to {0^ + }} \int_b^1 {\frac{1}{{\sqrt x }}dx} \cr} $$
