Improper integrals

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.

$$\lim_{a \to 0}{(2-2\sqrt{a})}=2$$