compagnia04

Answered

2022-01-05

Solve the integral:

$\int}_{0}^{2}\frac{dx}{\sqrt{x}(x-1)$

Answer & Explanation

encolatgehu

Expert

2022-01-06Added 27 answers

You can break up the integral at any point or points you like. In this case, you could break it up into three integrals: pick a point c strictly between 0 and 1, and consider:

$\int}_{0}^{2}\frac{dx}{\sqrt{x}(x-1)}={\int}_{0}^{c}\frac{dx}{\sqrt{x}(x-1)}+{\int}_{c}^{1}\frac{dx}{\sqrt{x}(x-1)}+{\int}_{1}^{2}\frac{dx}{\sqrt{x}(x-1)$

$=\underset{a\to {0}^{+}}{lim}{\int}_{a}^{c}\frac{dx}{\sqrt{x}(x-1)}+\underset{b\to {1}^{-}}{lim}{\int}_{c}^{b}\frac{dx}{\sqrt{x}(x-1)}+\underset{d}{\overset{2}{lim}}\frac{dx}{\sqrt{x}(x-1)}$

The original improper integral exists if and only if each of the three improper integrals exist.

The original improper integral exists if and only if each of the three improper integrals exist.

Jenny Bolton

Expert

2022-01-07Added 32 answers

Consider the change of variables $t=\sqrt{x}$ . It transforms the interval $[0,2]$ into $[0,\sqrt{2}]$ and removes one of the problem zeroes. It is easy to see from the resulting expression that the integral diverges.

karton

Expert

2022-01-11Added 439 answers

Try to solve the integral. You'll find whether it converges. For example,

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