int_0^2 frac{2x-2}{2x-x^2} dx is equal to | vedclass

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(A) Let $I = \int_0^2 \frac{2x-2}{2x-x^2} dx$. Notice that the integrand is improper at the boundaries $x=0$ and $x=2$ because the denominator $2x-x^2

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$0$

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(A) Let $I = \int_0^2 \frac{2x-2}{2x-x^2} dx$. Notice that the integrand is improper at the boundaries $x=0$ and $x=2$ because the denominator $2x-x^2 = x(2-x)$ becomes zero at these points. We evaluate the improper integral as a limit: $I = \lim_{\epsilon 	o 0^+} \int_{\epsilon}^{2-\epsilon} \frac{2x-2}{2x-x^2} dx$. Let $f(x) = 2x-x^2$, then $f'(x) = 2-2x = -(2x-2)$. Thus, $\int \frac{2x-2}{2x-x^2} dx = -\int \frac{f'(x)}{f(x)} dx = -\ln|2x-x^2| + C$. Evaluating the definite integral: $I = \lim_{\epsilon 	o 0^+} [-\ln|2x-x^2|]_{\epsilon}^{2-\epsilon} = \lim_{\epsilon 	o 0^+} [-\ln|2(2-\epsilon)-(2-\epsilon)^2| - (-\ln|2\epsilon-\epsilon^2|)]$. $I = \lim_{\epsilon 	o 0^+} [\ln|2\epsilon-\epsilon^2| - \ln|4-2\epsilon-(4-4\epsilon+\epsilon^2)|] = \lim_{\epsilon 	o 0^+} [\ln|2\epsilon-\epsilon^2| - \ln|2\epsilon-\epsilon^2|] = 0$. Since the limits of the logarithmic terms cancel out symmetrically, the principal value of the integral is $0$.

$\int_0^2 \frac{2x-2}{2x-x^2} dx$ is equal to

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