What is area 2x=y^2,x+y=12 ?

The answer to area 2x=y^2,x+y=12 is 250/3

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area 2x=y^2,x+y=12

area

2 x = y^{2}, x + y = 12

\frac{250}{3}

Decimal Notation

83.33333 \dots

Solution steps

Isolate

y

for

2 x = y^{2} : y = 2 x, y = - 2 x

Isolate

y

for

x + y = 12 : y = 12 - x

f_{1} (x) = 2 x

f_{2} (x) = - 2 x

To find the intersection points solve

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

2 x = - 2 x : x = 0

2 x = 12 - x : x = 8

- 2 x = 12 - x : x = 18

= \int_{0}^{8} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{8}^{18} ∣ f_{2} (x) - f_{3} (x) ∣ d x

= \int_{0}^{8} 2 x - (- 2 x) d x + \int_{8}^{18} - 2 x - (12 - x) d x

\int_{0}^{8} 2 x - (- 2 x) d x = \frac{128}{3}

\int_{8}^{18} - 2 x - (12 - x) d x = \frac{122}{3}

= \frac{128}{3} + \frac{122}{3}

Simplify

= \frac{250}{3}

x \to 0 lim ((1 - 4 x)^{\frac{3}{x}})

y^{^{'}} (x) = (x - y + 2)^{2}, y (0) = 2

x \to 8 + lim (8 x - 8)

\frac{d}{d t} (\frac{cos ( t )}{- sin ( t )})

\frac{d}{d x} (- e^{- x})