Question

Salut, am nevoie de ajutor la aceasta integrala

Answer 1

$f:[-1, 2]\to \mathbb{R},\,\,f(x) = \min(x,\, x^2)\\ \\ \min(x,\,x^2) = \begin{cases} x,\,\,\,x\in[-1,0]\\ x^2,\,\,x\in [0,1]\\ x,\,\,x\in [1,2]\end{cases}$

$\displaystyle \int_{-1}^2 e^x f(x)\, dx = \int_{-1}^0 e^x x\, dx +\int_{0}^1 e^x x^2\, dx+\int_{1}^2 e^x x\,dx\\ \\ = \int_{-1}^0 (e^x)'x\, dx+\int_{0}^1 (e^x)'x^2\, dx+\int_{1}^2(e^x)'x\, dx =\\ \\= (e^x x)\Big|_{-1}^0-e^x\Big|_{-1}^{0}+(e^xx^2)\Big|_{0}^1-2(e^xx)\Big|_{0}^1+2e^x\Big|_{0}^1+(e^x x)\Big|_{1}^2-e^x\Big|_{1}^2\\ \\ = (e^{-1})-(1-e^{-1})+(e^1)-(2e)+(2e-2)+(2e^2-e)-(e^2-e)\\ \\ =2e^{-1}-3+e+e^2\\ \\=\boxed{\dfrac{\left(e^{3}+e^{2}-3e+2\right)}{e}}$