Question

Cine ma ajuta la aceasta integrala

Answer 1

$f(x) = e^x\cdot \sqrt{x^2+1} \\ \\ \int_{-1}^1\sqrt{x^2+1}\cdot f(x)~dx= \int_{-1}^1 \sqrt{x^2+1}\cdot e^x\cdot \sqrt{x^2+1}~ dx = \\ \\ = \int_{-1}^1(x^2+1)\cdot e^x ~ dx = \int_{-1}^1 (x^2\cdot e^x+e^x)~ dx = \\ \\ = \int_{-1}^1 x^2\cdot e^x~dx +\int_{-1}^1 e^x~dx =\int_{-1}^1 x^2\cdot (e^x)'~dx+(e^x)\Big|_{-1}^1 = \\ \\ = (x^2\cdot e^x)\Big|_{-1}^1 - \int_{-1}^1(x^2)'\cdot e^x~dx+e^1-e^{-1} = \\ \\ = e^1 - e^{-1} - \int_{-1}^1 2x\cdot e^x~ dx+e-e^{-1} =$

$= -\int_{-1}^02x\cdot (e^x)'~dx +2e-2e^{-1} = \\ \\ = (-2x\cdot e^x)\Big|_{-1}^1+\int_{-1}^1 (2x)'\cdot e^x~ dx+2e-2e^{-1} = \\ \\ = -2e -2e^{-1}+\int _{-1}^02e^x~ dx+2e-2e^{-1} = \\ \\ = (2e^x)\Big|_{-1}^1 -4e^{-1} = \\ \\ = 2e-2e^{-1}-4e^{-1} = \\ \\ = 2e-6e^{-1}$