int_0^2x/(sqrt(9-2x^2))dx 

Explanation:

Another Explanation (5):

সমাধান

ধরি, \(I = \int_0^2 \frac{x}{\sqrt{9-2x^2}} dx\) এখন, \(9-2x^2 = z\) ধরি। সুতরাং, \(-4x dx = dz\) বা, \(x dx = -\frac{1}{4} dz\) সীমা পরিবর্তন করি: যখন \(x = 0\), তখন \(z = 9-2(0)^2 = 9\) যখন \(x = 2\), তখন \(z = 9-2(2)^2 = 9-8 = 1\) তাহলে, সমাকলনটি হবে: \(I = \int_9^1 \frac{-\frac{1}{4}}{\sqrt{z}} dz = -\frac{1}{4} \int_9^1 z^{-\frac{1}{2}} dz\) আমরা জ???নি, \(\int x^n dx = \frac{x^{n+1}}{n+1} + C\) সুতরাং, \(I = -\frac{1}{4} \left[ \frac{z^{\frac{1}{2}}}{\frac{1}{2}} \right]_9^1 = -\frac{1}{4} \cdot 2 \left[ \sqrt{z} \right]_9^1 = -\frac{1}{2} \left[ \sqrt{1} - \sqrt{9} \right]\) \(I = -\frac{1}{2} [1 - 3] = -\frac{1}{2} (-2) = 1\) অতএব, \(\int_0^2 \frac{x}{\sqrt{9-2x^2}} dx = 1\) 🥳🎉