int_0^1(xdx)/(sqrt(9-x^2)) =কত? 

Explanation:

Another Explanation (5): প্রশ্ন: \( \int_{0}^{1} \frac{x \, dx}{\sqrt{9-x^2}} \) = কত? 🤔 সমাধান: ধরি, \( u = 9 - x^2 \) 🤓 তাহলে, \( du = -2x \, dx \) সুতরাং, \( x \, dx = -\frac{1}{2} du \) এখন, যখন \( x = 0 \), তখন \( u = 9 - 0^2 = 9 \) 🤩 এবং যখন \( x = 1 \), তখন \( u = 9 - 1^2 = 8 \) 😎 তাহলে, সমাকলনটি হবে: \( \int_{9}^{8} \frac{-\frac{1}{2} \, du}{\sqrt{u}} \) \( = -\frac{1}{2} \int_{9}^{8} u^{-\frac{1}{2}} \, du \) \( = -\frac{1}{2} \left[ \frac{u^{\frac{1}{2}}}{\frac{1}{2}} \right]_{9}^{8} \) \( = -\frac{1}{2} \cdot 2 \left[ \sqrt{u} \right]_{9}^{8} \) \( = - \left[ \sqrt{8} - \sqrt{9} \right] \) \( = - \left[ 2\sqrt{2} - 3 \right] \) \( = 3 - 2\sqrt{2} \) 🎉 অতএব, \( \int_{0}^{1} \frac{x \, dx}{\sqrt{9-x^2}} = 3 - 2\sqrt{2} \) 🥳