What is area x=((y-2)^2)/3 ,y=8-x ?

The answer to area x=((y-2)^2)/3 ,y=8-x is 81/2

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area x=((y-2)^2)/3 ,y=8-x

area

x = \frac{( y - 2 ) ^{2}}{3}, y = 8 - x

\frac{81}{2}

Decimal

40.5

Solution steps

Isolate

y

for

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

f_{1} (x) = 3 x + 2

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

To find the intersection points solve

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

3 x + 2 = - 3 x + 2 : x = 0

3 x + 2 = 8 - x : x = 3

- 3 x + 2 = 8 - 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 + 2 - (- 3 x + 2) d x + \int_{3}^{12} - 3 x + 2 - (8 - x) d x

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

\int_{3}^{12} - 3 x + 2 - (8 - x) d x = \frac{57}{2}

= 12 + \frac{57}{2}

Simplify

= \frac{81}{2}

What is area x=((y-2)^2)/3 ,y=8-x ?
The answer to area x=((y-2)^2)/3 ,y=8-x is 81/2