Calcule a área da região limitada pela curva y=2+x-x^2 e pelo eixo x.
#CálculoII #Integração #SOS
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Primeira coisa que temos que ver eh quando essa curva intercepta o eixo x:
2+x-x²=0
x=-1 ou x=2
Agora temos os limites de integracao:
![\int\limits_{-1}^2 {2+x-x^2} \, dx =\\\\\\\left[2x+\dfrac{x^2}{2}-\dfrac{x^3}{3}\right]^2_{-1}=\\\\\\\left(2(2)+\dfrac{(2)^2}{2}-\dfrac{(2)^3}{3}\right)-\left(2(-1)+\dfrac{(-1)^2}{2}-\dfrac{(-1)^3}{3}\right)=\\\\\\\left(2(2)+\dfrac{(2)^2}{2}-\dfrac{(2)^3}{3}\right)-\left(2(-1)+\dfrac{(-1)^2}{2}-\dfrac{(-1)^3}{3}\right)=\dfrac92](https://tex.z-dn.net/?f=+%5Cint%5Climits_%7B-1%7D%5E2+%7B2%2Bx-x%5E2%7D+%5C%2C+dx+%3D%5C%5C%5C%5C%5C%5C%5Cleft%5B2x%2B%5Cdfrac%7Bx%5E2%7D%7B2%7D-%5Cdfrac%7Bx%5E3%7D%7B3%7D%5Cright%5D%5E2_%7B-1%7D%3D%5C%5C%5C%5C%5C%5C%5Cleft%282%282%29%2B%5Cdfrac%7B%282%29%5E2%7D%7B2%7D-%5Cdfrac%7B%282%29%5E3%7D%7B3%7D%5Cright%29-%5Cleft%282%28-1%29%2B%5Cdfrac%7B%28-1%29%5E2%7D%7B2%7D-%5Cdfrac%7B%28-1%29%5E3%7D%7B3%7D%5Cright%29%3D%5C%5C%5C%5C%5C%5C%5Cleft%282%282%29%2B%5Cdfrac%7B%282%29%5E2%7D%7B2%7D-%5Cdfrac%7B%282%29%5E3%7D%7B3%7D%5Cright%29-%5Cleft%282%28-1%29%2B%5Cdfrac%7B%28-1%29%5E2%7D%7B2%7D-%5Cdfrac%7B%28-1%29%5E3%7D%7B3%7D%5Cright%29%3D%5Cdfrac92 "\int\limits_{-1}^2 {2+x-x^2} \, dx =\\\\\\\left[2x+\dfrac{x^2}{2}-\dfrac{x^3}{3}\right]^2_{-1}=\\\\\\\left(2(2)+\dfrac{(2)^2}{2}-\dfrac{(2)^3}{3}\right)-\left(2(-1)+\dfrac{(-1)^2}{2}-\dfrac{(-1)^3}{3}\right)=\\\\\\\left(2(2)+\dfrac{(2)^2}{2}-\dfrac{(2)^3}{3}\right)-\left(2(-1)+\dfrac{(-1)^2}{2}-\dfrac{(-1)^3}{3}\right)=\dfrac92")
2+x-x²=0
x=-1 ou x=2
Agora temos os limites de integracao:
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