If int_3^b frac{x-1}{2x-x^2} dx = frac{1}{2}, | vedclass

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(C) Given the integral $\int_3^b \frac{x-1}{2x-x^2} dx = \frac{1}{2}$.We can rewrite the integrand as $-\frac{1}{2} \int_3^b \frac{-2x+2}{2x-x^2} dx =

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$1+\frac{3}{e}$

Vedclass · Answer

(C) Given the integral $\int_3^b \frac{x-1}{2x-x^2} dx = \frac{1}{2}$.We can rewrite the integrand as $-\frac{1}{2} \int_3^b \frac{-2x+2}{2x-x^2} dx = \frac{1}{2}$.Multiplying by $-2$,we get $\int_3^b \frac{2-2x}{2x-x^2} dx = -1$.Integrating,we have $\left[ \ln|2x-x^2| \right]_3^b = -1$.Substituting the limits: $\ln|2b-b^2| - \ln|6-9| = -1$.$\ln|2b-b^2| - \ln(3) = -1$.$\ln|2b-b^2| = \ln(3) - 1$.Taking the exponential of both sides: $|2b-b^2| = 3e^{-1} = \frac{3}{e}$.Since $2b-b^2 = -(b^2-2b+1-1) = -(b-1)^2+1$,we have $|1-(b-1)^2| = \frac{3}{e}$.Assuming $b$ is such that $2b-b^2  3$),we have $(b-1)^2 - 1 = \frac{3}{e}$.Therefore,$(b-1)^2 = 1 + \frac{3}{e}$.

If $\int_3^b \frac{x-1}{2x-x^2} dx = \frac{1}{2}$,then $(b-1)^2 =$

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