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Calculate the area of the region bounded by the graphs of the functions y+x²=6 and y+2x-3=0

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Answer to a math question Calculate the area of the region bounded by the graphs of the functions y+x²=6 and y+2x-3=0

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Fred
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115 Answers
y + x^2 = 6
y + 2x - 3 = 0

[Solução]

A = \int_{-3}^1 (3 - 2x - (6 - x^2)) \, dx = \int_{-3}^1 (-x^2 - 2x - 3) \, dx = \left[ -\frac{x^3}{3} - x^2 - 3x \right]_{-3}^1 = \left( -\frac{1^3}{3} - 1^2 - 3\cdot1 \right) - \left( -\frac{(-3)^3}{3} - (-3)^2 - 3\cdot(-3) \right) = -\frac{1}{3} - 1 - 3 - \left( -\frac{-27}{3} - 9 + 9 \right) = \frac{32}{3}

[Passo a Passo]

1. Encontrar os pontos de interseção resolvendo o sistema:
y + x^2 = 6
y + 2x - 3 = 0
x^2 + 2x - 9 = 0
x = -3, \ x = 1

2. Determinar as funções que delimitam a região e a região de integração:
y = 3 - 2x
y = 6 - x^2
Região entre $x = -3$ e $x = 1$.

3. Calcular a área utilizando a integral:
A = \int_{-3}^1 (3 - 2x - (6 - x^2)) \, dx
A = \int_{-3}^1 (-x^2 - 2x - 3) \, dx

4. Resolver a integral:
\int (-x^2 - 2x - 3) \, dx = -\frac{x^3}{3} - x^2 - 3x

5. Avaliar nos limites:
\left[ -\frac{x^3}{3} - x^2 - 3x \right]_{-3}^1

6. Calcular cada termo:
\left( -\frac{1^3}{3} - 1^2 - 3\cdot1 \right) = -\frac{1}{3} - 1 - 3 = -\frac{10}{3}
\left( -\frac{(-3)^3}{3} - (-3)^2 - 3\cdot(-3) \right) = -\frac{-27}{3} - 9 + 9 = \frac{27}{3} = 9

7. Substituir nos limites de integração:
-\frac{10}{3} - 9 = \frac{32}{3}

Portanto, a área é:
\frac{32}{3}

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