O valor da integral ![\displaystyle\int_0^1\left[\int_{x^2}^x dy\right]dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_0%5E1%5Cleft%5B%5Cint_%7Bx%5E2%7D%5Ex+dy%5Cright%5Ddx "\displaystyle\int_0^1\left[\int_{x^2}^x dy\right]dx")
é:
Escolha uma:
a. 1/5
b. 1/6
c. 1/7
d. 1/8
e. 1/4
Anexos:
Soluções para a tarefa
Respondido por
3
Resolve-se usando o Teorema de Fubini (integrais iteradas):
![\displaystyle\int_0^1\left[\int_{x^2}^x dy\right]dx\\\\\\ =\int_0^1y\big|_{x^2}^x\,dx\\\\\\ =\int_0^1(x-x^2)\,dx\\\\\\ =\left(\frac{x^2}{2}-\frac{x^3}{3}\right)\bigg|_0^1\\\\\\ =\frac{1^2}{2}-\frac{1^3}{3}\\\\\\ =\frac{1}{2}-\frac{1}{3}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_0%5E1%5Cleft%5B%5Cint_%7Bx%5E2%7D%5Ex+dy%5Cright%5Ddx%5C%5C%5C%5C%5C%5C+%3D%5Cint_0%5E1y%5Cbig%7C_%7Bx%5E2%7D%5Ex%5C%2Cdx%5C%5C%5C%5C%5C%5C+%3D%5Cint_0%5E1%28x-x%5E2%29%5C%2Cdx%5C%5C%5C%5C%5C%5C+%3D%5Cleft%28%5Cfrac%7Bx%5E2%7D%7B2%7D-%5Cfrac%7Bx%5E3%7D%7B3%7D%5Cright%29%5Cbigg%7C_0%5E1%5C%5C%5C%5C%5C%5C+%3D%5Cfrac%7B1%5E2%7D%7B2%7D-%5Cfrac%7B1%5E3%7D%7B3%7D%5C%5C%5C%5C%5C%5C+%3D%5Cfrac%7B1%7D%7B2%7D-%5Cfrac%7B1%7D%7B3%7D "\displaystyle\int_0^1\left[\int_{x^2}^x dy\right]dx\\\\\\ =\int_0^1y\big|_{x^2}^x\,dx\\\\\\ =\int_0^1(x-x^2)\,dx\\\\\\ =\left(\frac{x^2}{2}-\frac{x^3}{3}\right)\bigg|_0^1\\\\\\ =\frac{1^2}{2}-\frac{1^3}{3}\\\\\\ =\frac{1}{2}-\frac{1}{3}")
Resposta: alternativa b. 1/6.
Dúvidas? Comente.
Bons estudos! :-)
Resposta: alternativa b. 1/6.
Dúvidas? Comente.
Bons estudos! :-)
Lukyo:
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