Calculate the integral $$\displaystyle \int_0^1 \frac{1}{\sqrt {x}} \ dx$$
Development:
We identify the type of improper integral, and write it in the limit form.
$$\displaystyle \int_0^1 \frac{1}{\sqrt {x}} \ dx$$ is an improper integral of the second type, since it has a discontinuity in the integration interval. In particular, the discontinuity is at the end of the interval $$(x=0)$$.
$$\displaystyle \int_0^1 \frac{1}{\sqrt {x}} \ dx = \lim_{a \to 0}{\int_a^1 \frac{1}{\sqrt {x}} \ dx}$$
We calculate the integral according to parameter that we have introduced.
$$\displaystyle\int_a^1 \frac{1}{\sqrt {x}} \ dx=[2\sqrt{x}]_a^1=(2-2\sqrt{a})$$
We calculate the limit and, therefore, the result of the integral.
$$\lim_{a \to 0}{(2-2\sqrt{a})}=2$$
Solution:
$$\displaystyle \int_0^1 \frac{1}{\sqrt {x}} \ dx=2$$