What is area x=((y-4)^2)/4 ,y=7-x ?

The answer to area x=((y-4)^2)/4 ,y=7-x is 64/3

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area x=((y-4)^2)/4 ,y=7-x

area

x = \frac{( y - 4 ) ^{2}}{4}, y = 7 - x

\frac{64}{3}

Decimal Notation

21.33333 \dots

Solution steps

Isolate

y

for

x = \frac{( y - 4 ) ^{2}}{4} : y = 2 x + 4, y = - 2 x + 4

f_{1} (x) = 2 x + 4

f_{2} (x) = - 2 x + 4

To find the intersection points solve

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

2 x + 4 = - 2 x + 4 : x = 0

2 x + 4 = 7 - x : x = 1

- 2 x + 4 = 7 - x : x = 9

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

= \int_{0}^{1} 2 x + 4 - (- 2 x + 4) d x + \int_{1}^{9} - 2 x + 4 - (7 - x) d x

\int_{0}^{1} 2 x + 4 - (- 2 x + 4) d x = \frac{8}{3}

\int_{1}^{9} - 2 x + 4 - (7 - x) d x = \frac{56}{3}

= \frac{8}{3} + \frac{56}{3}

Simplify

= \frac{64}{3}

\frac{d}{d x} (- 243 sin (3 x))

ln (e^{3 x})

(t^{2} - 3)^{2}

\frac{d}{d x} (\frac{x + 3}{x - 2})

\frac{\partial}{\partial x} (x y^{3} + 2 x y)

What is area x=((y-4)^2)/4 ,y=7-x ?
The answer to area x=((y-4)^2)/4 ,y=7-x is 64/3