What is area y=9x,y= 8/3 x,y=52-x^2 ?

The answer to area y=9x,y= 8/3 x,y=52-x^2 is 72523/162

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area y=9x,y= 8/3 x,y=52-x^2

area

y = 9 x, y = \frac{8}{3} x, y = 52 - x^{2}

\frac{72523}{162}

Decimal Notation

447.67283 \dots

Solution steps

To find the intersection points solve

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

9 x = \frac{8}{3} x : x = 0

9 x = 52 - x^{2} : x = - 13, x = 4

\frac{8}{3} x = 52 - x^{2} : x = - \frac{26}{3}, x = 6

= \int_{- 13}^{- \frac{26}{3}} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{- \frac{26}{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_{- 13}^{- \frac{26}{3}} 9 x - (52 - x^{2}) d x + \int_{- \frac{26}{3}}^{0} \frac{8}{3} x - 9 x d x + \int_{0}^{4} 9 x - \frac{8}{3} x d x + \int_{4}^{6} \frac{8}{3} x - (52 - x^{2}) d x

\int_{- 13}^{- \frac{26}{3}} 9 x - (52 - x^{2}) d x = \frac{21463}{162}

\int_{- \frac{26}{3}}^{0} \frac{8}{3} x - 9 x d x = \frac{6422}{27}

\int_{0}^{4} 9 x - \frac{8}{3} x d x = \frac{152}{3}

\int_{4}^{6} \frac{8}{3} x - (52 - x^{2}) d x = \frac{80}{3}

= \frac{21463}{162} + \frac{6422}{27} + \frac{152}{3} + \frac{80}{3}

Simplify

= \frac{72523}{162}

x^{2} y^{^{'}} = y - x y, y (- 1) = - 1

\int x (x - 1)^{2} d x

\frac{d y}{d x} = e^{3 x + 4 y}

x \to 0 lim (7 x + 3)

y 3 - x y 2 + x 3 = 5

What is area y=9x,y= 8/3 x,y=52-x^2 ?
The answer to area y=9x,y= 8/3 x,y=52-x^2 is 72523/162