What is area x-2=-(1/2)(y-4)^2,x+y=6 ?

The answer to area x-2=-(1/2)(y-4)^2,x+y=6 is No Solution

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area x-2=-(1/2)(y-4)^2,x+y=6

area

x - 2 = - (\frac{1}{2}) (y - 4)^{2}, x + y = 6

No Solution

Solution steps

Isolate

y

for

x - 2 = - (\frac{1}{2}) (y - 4)^{2} : y = - 2 x + 4 + 4, y = - - 2 x + 4 + 4

Isolate

y

for

x + y = 6 : y = 6 - x

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

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

To find the intersection points solve

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

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

- 2 x + 4 + 4 = 6 - x : x = 2, x = 0

- - 2 x + 4 + 4 = 6 - x : x = 2

Curves not bounded

= No Solution

\int t^{12} d t

\frac{d}{d x} (\frac{e ^{- x}}{( 1 + e ^{- x} ) ^{2}})

\int u d u

\frac{1}{t}

\frac{d}{d x} ((x + y) e^{y})

What is area x-2=-(1/2)(y-4)^2,x+y=6 ?
The answer to area x-2=-(1/2)(y-4)^2,x+y=6 is No Solution