X=int_0^4sqrt(2x+1)dx হলে, X-এর মান কত? 

Explanation:

Another Explanation (5): bài toán: \(X = \int_{0}^{4} \sqrt{2x+1} \, dx\) ধরি, \(u = 2x + 1\). তাহলে, \(\frac{du}{dx} = 2\) বা, \(dx = \frac{du}{2}\) সীমা পরিবর্তন করি: যখন \(x = 0\), তখন \(u = 2(0) + 1 = 1\) যখন \(x = 4\), তখন \(u = 2(4) + 1 = 9\) সুতরাং, \(X = \int_{1}^{9} \sqrt{u} \frac{du}{2}\) \(X = \frac{1}{2} \int_{1}^{9} u^{\frac{1}{2}} \, du\) \(X = \frac{1}{2} \left[ \frac{u^{\frac{3}{2}}}{\frac{3}{2}} \right]_{1}^{9}\) \(X = \frac{1}{2} \cdot \frac{2}{3} \left[ u^{\frac{3}{2}} \right]_{1}^{9}\) \(X = \frac{1}{3} \left[ 9^{\frac{3}{2}} - 1^{\frac{3}{2}} \right]\) \(X = \frac{1}{3} \left[ (3^2)^{\frac{3}{2}} - 1 \right]\) \(X = \frac{1}{3} \left[ 3^3 - 1 \right]\) \(X = \frac{1}{3} \left[ 27 - 1 \right]\) \(X = \frac{1}{3} \cdot 26\) \(X = \frac{26}{3}\) অতএব, \(X = \frac{26}{3}\) 🎉