int_0^1xe^(x²)dx =?

সমাধান:

ধরি, \(u = x^2\)
তাহলে, \(\frac{du}{dx} = 2x\)
সুতরাং, \(du = 2x dx\)
বা, \(x dx = \frac{1}{2} du\)

এখন, যখন \(x = 0\), তখন \(u = 0^2 = 0\)
এবং যখন \(x = 1\), তখন \(u = 1^2 = 1\)

অতএব, ∫₀¹ xe^x2 dx = ∫₀¹ e^u (\(\frac{1}{2}\)) du = \(\frac{1}{2}\) ∫₀¹ e^u du

আমরা জানি, ∫e^x dx = e^x + C
সুতরাং, \(\frac{1}{2}\) ∫₀¹ e^u du = \(\frac{1}{2}\) [e^u]₀¹

= \(\frac{1}{2}\) [e¹ - e⁰] = \(\frac{1}{2}\) [e - 1]

সুতরাং, ∫₀¹ xe^x2 dx = \(\frac{1}{2}\)(e - 1) 🥳