Problems from Improper integrals

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 $(x=0)$.

${\displaystyle {\int }_0^1\frac{1}{\sqrt{x}}dx=\underset{a\to 0}{lim}{\int }_a^1\frac{1}{\sqrt{x}}dx}$

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.

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

Solution:

${\displaystyle {\int }_0^1\frac{1}{\sqrt{x}}dx=2}$

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