What is area x=((y-3)^2)/3 ,y=9-x ?

The answer to area x=((y-3)^2)/3 ,y=9-x is 81/2

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area x=((y-3)^2)/3 ,y=9-x

area

x = \frac{( y - 3 ) ^{2}}{3}, y = 9 - x

\frac{81}{2}

Decimal Notation

40.5

Solution steps

Isolate

y

for

x = \frac{( y - 3 ) ^{2}}{3} : y = 3 x + 3, y = - 3 x + 3

f_{1} (x) = 3 x + 3

f_{2} (x) = - 3 x + 3

To find the intersection points solve

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

3 x + 3 = - 3 x + 3 : x = 0

3 x + 3 = 9 - x : x = 3

- 3 x + 3 = 9 - x : x = 12

= \int_{0}^{3} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{3}^{12} ∣ f_{2} (x) - f_{3} (x) ∣ d x

= \int_{0}^{3} 3 x + 3 - (- 3 x + 3) d x + \int_{3}^{12} - 3 x + 3 - (9 - x) d x

\int_{0}^{3} 3 x + 3 - (- 3 x + 3) d x = 12

\int_{3}^{12} - 3 x + 3 - (9 - x) d x = \frac{57}{2}

= 12 + \frac{57}{2}

Simplify

= \frac{81}{2}

7 x y \frac{d y}{d x} = 2

\frac{d}{d x} (1 + 5 x)

\frac{d}{d x} (\frac{x - 2}{x ^{2} + 4})

12 = \frac{( 33000 ) d y}{d t} + \frac{1}{22 \cdot 1 0 ^{- 6}} y

h \to 0 lim (\frac{3}{3 h + 1 + 1})

What is area x=((y-3)^2)/3 ,y=9-x ?
The answer to area x=((y-3)^2)/3 ,y=9-x is 81/2