Discussion
Over the unit interval, the product \( x e^{x} \) integrates to
Answer: D
Take \( u = x \) and \( dv = e^{x}dx \), so that differentiating the polynomial factor removes it:
\[ \int_{0}^{1} x e^{x}dx \;=\; \left[x e^{x}\right]_{0}^{1} - \int_{0}^{1} e^{x}dx \;=\; e - \left[e^{x}\right]_{0}^{1} \;=\; e - (e - 1) \;=\; 1. \]
Choosing the factors the other way round would raise the power of \( x \) and make the integral harder rather than easier
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