Aloha :)
Das Integral kannst du mittels partieller Integration gut "knacken":$$\int\limits_0^3\underbrace{(2-x)}_{=u}\cdot\underbrace{e^x}_{=v'}dx=\left[\underbrace{(2-x)}_{=u}\cdot\underbrace{e^x}_{=v}\right]_0^3-\int\limits_0^3\underbrace{(-1)}_{=u'}\underbrace{e^x}_{=v}dx$$$$=(2-3)\cdot e^3-(2-0)e^0+\int\limits_0^3e^x\,dx=-e^3-2+\left[e^x\right]_0^3$$$$=-e^3-2+e^3-e^0=-3$$
