What is area 66x, 2/3 x,40-x^2 ?

The answer to area 66x, 2/3 x,40-x^2 is 4092872/81

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area 66x, 2/3 x,40-x^2

area

66 x, \frac{2}{3} x, 40 - x^{2}

\frac{4092872}{81}

Solution steps

To find the intersection points solve

f_{i} (x) = f_{j} (x)

66 x = \frac{2}{3} x : x = 0

66 x = 40 - x^{2} : x = - 33 - 1129, x = 1129 - 33

\frac{2}{3} x = 40 - x^{2} : x = - \frac{20}{3}, x = 6

= \int_{- 33 - 1129}^{- \frac{20}{3}} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{- \frac{20}{3}}^{0} ∣ f_{2} (x) - f_{1} (x) ∣ d x + \int_{0}^{1129 - 33} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{1129 - 33}^{6} ∣ f_{2} (x) - f_{3} (x) ∣ d x

= \int_{- 33 - 1129}^{- \frac{20}{3}} 66 x - (40 - x^{2}) d x + \int_{- \frac{20}{3}}^{0} \frac{2}{3} x - 66 x d x + \int_{0}^{1129 - 33} 66 x - \frac{2}{3} x d x + \int_{1129 - 33}^{6} \frac{2}{3} x - (40 - x^{2}) d x

\int_{- 33 - 1129}^{- \frac{20}{3}} 66 x - (40 - x^{2}) d x = \frac{1915118 + 60966 1129}{81}

\int_{- \frac{20}{3}}^{0} \frac{2}{3} x - 66 x d x = \frac{39200}{27}

\int_{0}^{1129 - 33} 66 x - \frac{2}{3} x d x = 66 (1109 - 331129) - \frac{2 ( 1109 - 33 1129 )}{3}

\int_{1129 - 33}^{6} \frac{2}{3} x - (40 - x^{2}) d x = - 401129 + \frac{4330 1129 - 145526}{3} + 1488

= \frac{1915118 + 60966 1129}{81} + \frac{39200}{27} + 66 (1109 - 331129) - \frac{2 ( 1109 - 33 1129 )}{3} - 401129 + \frac{4330 1129 - 145526}{3} + 1488

Simplify

= \frac{4092872}{81}

f (x) = \frac{1}{x ^{3}}, at x = - 2

\int_{- 2}^{6} 4 y - (y^{2} - 12) d y

\frac{\partial}{\partial y} (2 x + y)

\frac{\partial}{\partial x} (40 x y^{- 4} z)

\int_{- 2}^{2} (4 - x^{2})^{2} d x

What is area 66x, 2/3 x,40-x^2 ?
The answer to area 66x, 2/3 x,40-x^2 is 4092872/81