Introduction: In this solution, we have explained what is convergence of the integral example 1. We will test the convergence of the integral 0 to a 1/(sqrt(a-x)dx.
Qestion : Examine the convergence of the integral
\[\int_{0}^{a}\frac{dx}{\sqrt{a-x}}\]
Solution :
Let
\[I=\int_{0}^{a}\frac{dx}{\sqrt{a-x}}\]
Since integrand here is
\[\frac{1}{\sqrt{a-x}}\]
which is unbounded on Interval of integration [0, a] . So I is an improper integral of second kind.
\[I=limit_{\epsilon\rightarrow0+}\int_{0}^{a-\epsilon}\frac{dx}{\sqrt{a-x}}\]
\[=limit_{\epsilon\rightarrow0+}\int_{0}^{a-\epsilon}{\left(a-x\right)^{-\frac{1}{2}}dx}\]
\[=limit_{\epsilon\rightarrow0+}\left[-\frac{\left(a-x\right)^\frac{1}{2}}{\frac{1}{2}}\right]_0^{a-\epsilon}\]
\[=-2limit_{\epsilon\rightarrow0+}\left[\epsilon^\frac{1}{2}-a^\frac{1}{2}\right]\]
\[=-2\left(0^\frac{1}{2}-a^\frac{1}{2}\right)\]
\[=2a^\frac{1}{2}\] \[\int_{0}^{a}\frac{dx}{\sqrt{a-x}}\]
\[=2\sqrt{a}\]
thus we get
\[\int_{0}^{a}\frac{dx}{\sqrt{a-x}}=2\sqrt{a}\]
which is a finite number that is why given integral will be convergent.