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

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(A) Let $I = \int_0^2 \frac{2x-2}{2x-x^2} dx$. Note that the integrand $f(x) = \frac{2x-2}{2x-x^2} = \frac{-(2-2x)}{x(2-x)} = \frac{2(x-1)}{x(2-x)}$.

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$0$

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(A) Let $I = \int_0^2 \frac{2x-2}{2x-x^2} dx$. Note that the integrand $f(x) = \frac{2x-2}{2x-x^2} = \frac{-(2-2x)}{x(2-x)} = \frac{2(x-1)}{x(2-x)}$. The integral is improper because the denominator $2x-x^2 = x(2-x)$ becomes $0$ at $x=0$ and $x=2$. Let us evaluate the indefinite integral: $\int \frac{2x-2}{2x-x^2} dx = \int \frac{-(2-2x)}{2x-x^2} dx$. Let $u = 2x-x^2$,then $du = (2-2x) dx$,so $-(2-2x) dx = -du$. Thus,$\int \frac{-du}{u} = -\ln|u| + C = -\ln|2x-x^2| + C$. Evaluating the definite integral as a limit: $\lim_{\epsilon 	o 0^+} \int_{\epsilon}^{2-\epsilon} \frac{2x-2}{2x-x^2} dx = \lim_{\epsilon 	o 0^+} [-\ln|2x-x^2|]_{\epsilon}^{2-\epsilon}$. $= \lim_{\epsilon 	o 0^+} [-\ln|2(2-\epsilon)-(2-\epsilon)^2| - (-\ln|2\epsilon-\epsilon^2|)]$. $= \lim_{\epsilon 	o 0^+} [-\ln|4-2\epsilon-(4-4\epsilon+\epsilon^2)| + \ln|2\epsilon-\epsilon^2|]$. $= \lim_{\epsilon 	o 0^+} [-\ln|2\epsilon-\epsilon^2| + \ln|2\epsilon-\epsilon^2|] = 0$. However,since the integral diverges at both endpoints,the Cauchy Principal Value is $0$.

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

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