Respuesta :

konrad509
[tex]x^2=2+x\\ x^2-x-2=0\\ x^2+x-2x-2=0\\ x(x+1)-2(x+1)=0\\ (x-2)(x+1)=0\\ x=2 \vee =-1\\\\ \displaystyle A=\int \limits_{-1}^22+x-x^2\, dx\\ A=\left[2x+\dfrac{x^2}{2}-\dfrac{x^3}{3}\right]_{-1}^2\\ A=2\cdot2+\dfrac{2^2}{2}-\dfrac{2^3}{3}-\left(2\cdot(-1)+\dfrac{(-1)^2}{2}-\dfrac{(-1)^3}{3}\right)\\ A=4+2-\dfrac{8}{3}-\left(-2+\dfrac{1}{2}+\dfrac{1}{3}\right)\\ A=6-\dfrac{8}{3}+2-\dfrac{1}{2}-\dfrac{1}{3}\right)\\ A=8-\dfrac{9}{3}-\dfrac{1}{2}\\ A=8-3-\dfrac{1}{2}\\ A=5-\dfrac{1}{2}\\ A=4\dfrac{1}{2} [/tex]

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