int_0^1dx/sqrt(1-x)=? 

প্রশ্ন: \( \int_{0}^{1} \frac{dx}{\sqrt{1-x}} = ? \)

সমাধান:

ধরি, \( u = 1 - x \)
সুতরাং, \( du = -dx \)
অতএব, \( dx = -du \)

সীমা পরিবর্তন:

• যখন \( x = 0 \), তখন \( u = 1 - 0 = 1 \)
• যখন \( x = 1 \), তখন \( u = 1 - 1 = 0 \)

এখন, ইন্টিগ্রালটি হবে: \( \int_{1}^{0} \frac{-du}{\sqrt{u}} = - \int_{1}^{0} u^{-\frac{1}{2}} du \)

আমরা জানি, \( \int x^n dx = \frac{x^{n+1}}{n+1} + C \)

সুতরাং, \( - \int_{1}^{0} u^{-\frac{1}{2}} du = - \left[ \frac{u^{\frac{1}{2}}}{\frac{1}{2}} \right]_{1}^{0} = - \left[ 2\sqrt{u} \right]_{1}^{0} \)

\( = - \left[ 2\sqrt{0} - 2\sqrt{1} \right] = - \left[ 0 - 2 \right] = 2 \) 🎉

অতএব, \( \int_{0}^{1} \frac{dx}{\sqrt{1-x}} = 2 \) 🥳