int_0^1e^(sqrtx)/sqrtxdx = কত? 

Explanation:

Another Explanation (5): Solution: Let \(I = \int_{0}^{1} \frac{e^{\sqrt{x}}}{\sqrt{x}} dx\) Substitute \(u = \sqrt{x}\). Then \(x = u^2\) and \(dx = 2u du\). When \(x = 0\), \(u = \sqrt{0} = 0\). When \(x = 1\), \(u = \sqrt{1} = 1\). Therefore, \(I = \int_{0}^{1} \frac{e^{u}}{u} (2u) du\) \(I = 2 \int_{0}^{1} e^{u} du\) \(I = 2 [e^{u}]_{0}^{1}\) \(I = 2 (e^{1} - e^{0})\) \(I = 2 (e - 1)\) Therefore, \(\int_{0}^{1} \frac{e^{\sqrt{x}}}{\sqrt{x}} dx = 2(e-1)\).