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Examine the convergence \(\Large\int\limits_{0}^{2}\frac{dx}{2x^{2}-x^{2}}\)
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A) converges |
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B) diverges |
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C) converges to |
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D) none of these |
Correct Answer:
| B) diverges |
Description for Correct answer:
\(\Large \frac{1}{2x-x^{2}}=\frac{1}{x(2-x)}\)
0, 2 are the points of infinite discontituity.
Now \(\Large \int\limits_{0}^{2}\frac{dx}{2x-x^{2}}\)
\(\Large =\lim\limits_{\lambda\rightarrow 0+}\int\limits_{\lambda}^{1}\frac{dx}{x(2-x)}+\lim\limits_{\mu\rightarrow 0+}\int\limits_{1}^{2-\mu}\frac{2x}{2(2-x)}\)
\(\Large \frac{1}{2}\lim\limits_{\lambda\rightarrow 0+}\left[ log\frac{x}{2-x} \right]_{\lambda}^{1}+\frac{1}{2}\lim\limits_{\mu\rightarrow 0+}\left[ log\frac{x}{2-x} \right]_{\lambda}^{2-\mu}\)
\(\Large =-\frac{1}{2}\lim\limits_{\lambda\rightarrow 0+}\frac{\lambda}{2-\lambda}+\frac{1}{2}\lim\limits_{\mu\rightarrow 0+}\frac{2-\mu}{\mu}\)
\(\Large =\infty\)
\(\therefore\) The given integral diverges.
\(\Large \frac{1}{2x-x^{2}}=\frac{1}{x(2-x)}\)
0, 2 are the points of infinite discontituity.
Now \(\Large \int\limits_{0}^{2}\frac{dx}{2x-x^{2}}\)
\(\Large =\lim\limits_{\lambda\rightarrow 0+}\int\limits_{\lambda}^{1}\frac{dx}{x(2-x)}+\lim\limits_{\mu\rightarrow 0+}\int\limits_{1}^{2-\mu}\frac{2x}{2(2-x)}\)
\(\Large \frac{1}{2}\lim\limits_{\lambda\rightarrow 0+}\left[ log\frac{x}{2-x} \right]_{\lambda}^{1}+\frac{1}{2}\lim\limits_{\mu\rightarrow 0+}\left[ log\frac{x}{2-x} \right]_{\lambda}^{2-\mu}\)
\(\Large =-\frac{1}{2}\lim\limits_{\lambda\rightarrow 0+}\frac{\lambda}{2-\lambda}+\frac{1}{2}\lim\limits_{\mu\rightarrow 0+}\frac{2-\mu}{\mu}\)
\(\Large =\infty\)
\(\therefore\) The given integral diverges.
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