What is area y=9x,y= 1/3 x,y=10-x^2 ?

The answer to area y=9x,y= 1/3 x,y=10-x^2 is 16861/81

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area y=9x,y= 1/3 x,y=10-x^2

area

y = 9 x, y = \frac{1}{3} x, y = 10 - x^{2}

\frac{16861}{81}

Decimal Notation

208.16049 \dots

Solution steps

To find the intersection points solve

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

9 x = \frac{1}{3} x : x = 0

9 x = 10 - x^{2} : x = - 10, x = 1

\frac{1}{3} x = 10 - x^{2} : x = - \frac{10}{3}, x = 3

= \int_{- 10}^{- \frac{10}{3}} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{- \frac{10}{3}}^{0} ∣ f_{2} (x) - f_{1} (x) ∣ d x + \int_{0}^{1} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{1}^{3} ∣ f_{2} (x) - f_{3} (x) ∣ d x

= \int_{- 10}^{- \frac{10}{3}} 9 x - (10 - x^{2}) d x + \int_{- \frac{10}{3}}^{0} \frac{1}{3} x - 9 x d x + \int_{0}^{1} 9 x - \frac{1}{3} x d x + \int_{1}^{3} \frac{1}{3} x - (10 - x^{2}) d x

\int_{- 10}^{- \frac{10}{3}} 9 x - (10 - x^{2}) d x = \frac{11800}{81}

\int_{- \frac{10}{3}}^{0} \frac{1}{3} x - 9 x d x = \frac{1300}{27}

\int_{0}^{1} 9 x - \frac{1}{3} x d x = \frac{13}{3}

\int_{1}^{3} \frac{1}{3} x - (10 - x^{2}) d x = 10

= \frac{11800}{81} + \frac{1300}{27} + \frac{13}{3} + 10

Simplify

= \frac{16861}{81}

tan (x y)

1 - 8 u^{2}

\int_{0}^{1} 2 π x (e^{- x^{2}}) d x

\frac{\partial}{\partial x} (- x e^{- 2 x y})

\frac{d ^{2}}{d x ^{2}} (θ cos (θ))

What is area y=9x,y= 1/3 x,y=10-x^2 ?
The answer to area y=9x,y= 1/3 x,y=10-x^2 is 16861/81