Examine the convergence of \(\Large\int\limits_{0}^{1}\frac{dx}{x^{2}}\)
Correct Answer: Description for Correct answer:
When \(x=0,\frac{1}{x^{2}}=\infty\)
\(\therefore\) The only point of infinite discontinuity is 0 in \([0,1]\)
Now \(\Large \int\limits_{0}^{1}\frac{dx}{x^{2}}dx\)
\(\Large =\lim\limits_{\lambda\rightarrow 0+}\int\limits_{\lambda}^{1}\frac{dx}{x^{2}}0<\lambda <1\)
\(\Large =\lim\limits_{\lambda\rightarrow 0+}\int\limits_{\lambda}^{1} \left(-\frac{1}{x}\right)_{\lambda}^{1} \)
\(\Large =\lim\limits_{\lambda\rightarrow 0+}\int\limits_{\lambda}^{1} \left(\frac{1}{\lambda}-1\right)=\infty\)
Thus the improper integral diverges.
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