What is area x=((y-2)^2)/2 ,y=6-x ?

The answer to area x=((y-2)^2)/2 ,y=6-x is 18

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area x=((y-2)^2)/2 ,y=6-x

area

x = \frac{( y - 2 ) ^{2}}{2}, y = 6 - x

18

Solution steps

Isolate

y

for

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

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

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

To find the intersection points solve

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

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

2 x + 2 = 6 - x : x = 2

- 2 x + 2 = 6 - x : x = 8

= \int_{0}^{2} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{2}^{8} ∣ f_{2} (x) - f_{3} (x) ∣ d x

= \int_{0}^{2} 2 x + 2 - (- 2 x + 2) d x + \int_{2}^{8} - 2 x + 2 - (6 - x) d x

\int_{0}^{2} 2 x + 2 - (- 2 x + 2) d x = \frac{16}{3}

\int_{2}^{8} - 2 x + 2 - (6 - x) d x = \frac{38}{3}

= \frac{16}{3} + \frac{38}{3}

Simplify

= 18

f (x) = \frac{x}{( 3 x - 2 ) ^{6}}, at x = 1

f (x) = (x - 1) x^{2} - 2 x + 2

\int x^{5} e^{x^{3} - 7} d x

n = 1 \sum \infty \frac{3}{4 n ^{2} - 1}

1 + cos (2 x)

What is area x=((y-2)^2)/2 ,y=6-x ?
The answer to area x=((y-2)^2)/2 ,y=6-x is 18