What is area y=6x,y= 2/3 x,y=40-x^2 ?

The answer to area y=6x,y= 2/3 x,y=40-x^2 is 20192/81

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area y=6x,y= 2/3 x,y=40-x^2

area

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

\frac{20192}{81}

Decimal Notation

249.28395 \dots

Solution steps

To find the intersection points solve

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

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

6 x = 40 - x^{2} : x = - 10, x = 4

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

= \int_{- 10}^{- \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}^{4} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{4}^{6} ∣ f_{2} (x) - f_{3} (x) ∣ d x

= \int_{- 10}^{- \frac{20}{3}} 6 x - (40 - x^{2}) d x + \int_{- \frac{20}{3}}^{0} \frac{2}{3} x - 6 x d x + \int_{0}^{4} 6 x - \frac{2}{3} x d x + \int_{4}^{6} \frac{2}{3} x - (40 - x^{2}) d x

\int_{- 10}^{- \frac{20}{3}} 6 x - (40 - x^{2}) d x = \frac{5300}{81}

\int_{- \frac{20}{3}}^{0} \frac{2}{3} x - 6 x d x = \frac{3200}{27}

\int_{0}^{4} 6 x - \frac{2}{3} x d x = \frac{128}{3}

\int_{4}^{6} \frac{2}{3} x - (40 - x^{2}) d x = \frac{68}{3}

= \frac{5300}{81} + \frac{3200}{27} + \frac{128}{3} + \frac{68}{3}

Simplify

= \frac{20192}{81}

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y = 10 x^{2}, y^{2} = \frac{1}{10} x

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What is area y=6x,y= 2/3 x,y=40-x^2 ?
The answer to area y=6x,y= 2/3 x,y=40-x^2 is 20192/81