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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 =

Vedclass · Accepted Answer

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

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 $(b-1)^2 =$

Similar Questions

$\int_0^{\pi /4} {\frac{{4\sin 2\theta \,d\theta }}{{{{\sin }^4}\theta + {{\cos }^4}\theta }}} = $

$ \int_{-5}^{5} |x+2| \, dx $ is equal to

The value of the integral $\int_{1}^{3} [x^{2}-2x-2] dx$,where $[x]$ denotes the greatest integer function,is:

Choose the correct option regarding the following definite integrals:
$(i)$ $\int_0^{\pi / 2} \sin ^m(x) \cos (x) d x = \frac{1}{m+1}$
(ii) $\int_0^{\pi / 2} \sin (x) \cos ^n(x) d x = \frac{1}{n+1}$

What is the minimum value of $\int\limits_0^x {t{e^{ - {t^2}}}} dt$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module