int_0^1 x^3 e^(x^2) dt =?  

Explanation:

Another Explanation (5): Solution: Let \(I = \int_0^1 x^3 e^{x^2} dx\). Let \(u = x^2\), then \(du = 2x dx\), so \(x dx = \frac{1}{2} du\). Also, \(x^2 = u\). When \(x = 0\), \(u = 0^2 = 0\). When \(x = 1\), \(u = 1^2 = 1\). Thus, \(I = \int_0^1 x^2 e^{x^2} x dx = \int_0^1 u e^u \frac{1}{2} du = \frac{1}{2} \int_0^1 u e^u du\). Now, we use integration by parts: \(\int v dw = vw - \int w dv\). Let \(v = u\) and \(dw = e^u du\). Then \(dv = du\) and \(w = e^u\). So, \(\int u e^u du = u e^u - \int e^u du = u e^u - e^u + C\). Therefore, \(I = \frac{1}{2} \left[ u e^u - e^u \right]_0^1 = \frac{1}{2} \left[ (1 \cdot e^1 - e^1) - (0 \cdot e^0 - e^0) \right] = \frac{1}{2} \left[ (e - e) - (0 - 1) \right] = \frac{1}{2} \left[ 0 - (-1) \right] = \frac{1}{2} \cdot 1 = \frac{1}{2}\). Hence, \(\int_0^1 x^3 e^{x^2} dx = \frac{1}{2}\). \(\boxed{1/2}\) 🎉