A area limidada superiormente pelo grafico de X = F (X) é inferiormente por y = g ( x ) e pela integral
![A = \int\limits^b_a [ f( {x}) - g(x) ] \, dx A = \int\limits^b_a [ f( {x}) - g(x) ] \, dx](https://tex.z-dn.net/?f=+A+%3D+%5Cint%5Climits%5Eb_a+%5B+f%28+%7Bx%7D%29+-+g%28x%29+%5D+%5C%2C+dx+)
Calcule a area A da figura:
Figura em anexo
Alternativas:
a) A= 1
b) A= 2
c)
d)
Anexos:
Soluções para a tarefa
Respondido por
1
Pelo que me parece, deseja-se calcular a área entre os gráficos das funções

com
no intervalo ![[0,\,1]. [0,\,1].](https://tex.z-dn.net/?f=%5B0%2C%5C%2C1%5D.)
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(ATENÇÃO para o fato de
ser
no intervalo considerado!!!)
A área é dada por
![A=\displaystyle\int_0^1\big[f(x)-g(x)\big]dx\\\\\\ =\int_0^1\big[x-x^3\big]dx\\\\\\ =\left.\left(\dfrac{x^2}{2}-\dfrac{x^4}{4} \right )\right|_0^1\\\\\\ =\left(\dfrac{1^2}{2}-\dfrac{1^4}{4} \right )-\left(\dfrac{0^2}{2}-\dfrac{0^4}{4} \right )\\\\\\ =\dfrac{1}{2}-\dfrac{1}{4}\\\\\\ =\dfrac{2}{4}-\dfrac{1}{4}\\\\\\ =\dfrac{2-1}{4}\\\\\\ =\dfrac{1}{4}\mathrm{~u.a.} A=\displaystyle\int_0^1\big[f(x)-g(x)\big]dx\\\\\\ =\int_0^1\big[x-x^3\big]dx\\\\\\ =\left.\left(\dfrac{x^2}{2}-\dfrac{x^4}{4} \right )\right|_0^1\\\\\\ =\left(\dfrac{1^2}{2}-\dfrac{1^4}{4} \right )-\left(\dfrac{0^2}{2}-\dfrac{0^4}{4} \right )\\\\\\ =\dfrac{1}{2}-\dfrac{1}{4}\\\\\\ =\dfrac{2}{4}-\dfrac{1}{4}\\\\\\ =\dfrac{2-1}{4}\\\\\\ =\dfrac{1}{4}\mathrm{~u.a.}](https://tex.z-dn.net/?f=A%3D%5Cdisplaystyle%5Cint_0%5E1%5Cbig%5Bf%28x%29-g%28x%29%5Cbig%5Ddx%5C%5C%5C%5C%5C%5C+%3D%5Cint_0%5E1%5Cbig%5Bx-x%5E3%5Cbig%5Ddx%5C%5C%5C%5C%5C%5C+%3D%5Cleft.%5Cleft%28%5Cdfrac%7Bx%5E2%7D%7B2%7D-%5Cdfrac%7Bx%5E4%7D%7B4%7D+%5Cright+%29%5Cright%7C_0%5E1%5C%5C%5C%5C%5C%5C+%3D%5Cleft%28%5Cdfrac%7B1%5E2%7D%7B2%7D-%5Cdfrac%7B1%5E4%7D%7B4%7D+%5Cright+%29-%5Cleft%28%5Cdfrac%7B0%5E2%7D%7B2%7D-%5Cdfrac%7B0%5E4%7D%7B4%7D+%5Cright+%29%5C%5C%5C%5C%5C%5C+%3D%5Cdfrac%7B1%7D%7B2%7D-%5Cdfrac%7B1%7D%7B4%7D%5C%5C%5C%5C%5C%5C+%3D%5Cdfrac%7B2%7D%7B4%7D-%5Cdfrac%7B1%7D%7B4%7D%5C%5C%5C%5C%5C%5C+%3D%5Cdfrac%7B2-1%7D%7B4%7D%5C%5C%5C%5C%5C%5C+%3D%5Cdfrac%7B1%7D%7B4%7D%5Cmathrm%7B%7Eu.a.%7D)
Resposta: alternativa
com
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A área é dada por
Resposta: alternativa
Pirata2014:
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