Institution: বঙ্গবন্ধু শেখ মুজিবুর রহমান বিজ্ঞান ও প্রযুক্তি বিশ্ববিদ্যালয় (BSMRSTU) Exam: A ইউনিট || বঙ্গবন্ধু শেখ মুজিবুর রহমান বিজ্ঞান ও প্রযুক্তি বিশ্ববিদ্যালয় || 2011
Topic: Arithmetic Mathematics (পাটিগণিত), Subject: Mathematics (সাধারণ গণিত)

Best Answer Explained in Below-:

To evaluate the integral \( \int_0^1 \frac{-dx}{\sqrt{2x - x^2}} \), we first simplify the integrand. The expression under the square root, \( 2x - x^2 \), can be rewritten as \( - (x^2 - 2x) \) or \( - (x(x - 2)) \). Completing the square gives us \( 1 - (x-1)^2 \), allowing us to rewrite the integral:

\[
\sqrt{2x - x^2} = \sqrt{1 - (x-1)^2}
\]

This leads to the integral:

\[
\int_0^1 \frac{-dx}{\sqrt{1 - (x-1)^2}}
\]

Next, we can make a substitution. Let's set \( u = x - 1 \), which transforms our limits as \( x = 0 \) to \( u = -1 \) and \( x = 1 \) to \( u = 0 \). The integral becomes:

\[
\int_{-1}^0 \frac{-du}{\sqrt{1 - u^2}} = \int_0^1 \frac{du}{\sqrt{1 - u^2}}
\]

The integral \( \int_0^1 \frac{du}{\sqrt{1 - u^2}} \) is known to equal \( \frac{\pi}{2} \). Therefore, the value of the original integral \( \int_0^1 \frac{-dx}{\sqrt{2x - x^2}} \) is indeed \( \frac{\pi}{2} \), making option B the correct choice. This integral also represents the area of a quarter unit circle, reinforcing the result.