Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.2 Exercises - Page 521: 39

$$\int_{0}^{3} x e^{x / 2} d x =2e^{3/2}+4$$

Since $$ \int_{0}^{3} x e^{x / 2} d x $$ Integrate by parts , let \begin{align*} u&=x\ \ \ \ \ \ \ dv= e^{x / 2} d x\\ du&=dx\ \ \ \ \ \ v=2 e^{x / 2} \end{align*} Then \begin{align*} \int_{0}^{3} x e^{x / 2} d x&=uv-\int vdu\\ &=2x e^{x / 2}\bigg|_{0}^{3} -\int_{0}^{3}2 e^{x / 2} dx\\ &=2x e^{x / 2}\bigg|_{0}^{3} -4e^{x / 2}\bigg|_{0}^{3} \\ &=2e^{3/2}+4 \end{align*}