What is area x=y^2,x=2-y ?

The answer to area x=y^2,x=2-y is 9/2

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area x=y^2,x=2-y

area

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

\frac{9}{2}

Solution steps

Isolate

y

for

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

Isolate

y

for

x = 2 - y : y = 2 - x

f_{1} (x) = x

f_{2} (x) = - x

To find the intersection points solve

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

x = - x : x = 0

x = 2 - x : x = 1

- x = 2 - x : x = 4

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

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

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

\int_{1}^{4} - x - (2 - x) d x = \frac{19}{6}

= \frac{4}{3} + \frac{19}{6}

Simplify

= \frac{9}{2}

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

cos (θ^{2})

\int x (x^{2} + 4)^{5} d x

\frac{\partial}{\partial x} (e^{y x})

\int \frac{e ^{2 y}}{e ^{y} - 1} d y