What is area x=y^2-4y,x=2y-y^2 ?

The answer to area x=y^2-4y,x=2y-y^2 is 9

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area x=y^2-4y,x=2y-y^2

area

x = y^{2} - 4 y, x = 2 y - y^{2}

9

Solution steps

Isolate

y

for

x = y^{2} - 4 y : y = 2 + x + 4, y = 2 - x + 4

Isolate

y

for

x = 2 y - y^{2} : y = 1 - - x + 1, y = - x + 1 + 1

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 = - 4

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

No Solution for

x \in R

2 + x + 4 = - x + 1 + 1 : x = - 3

2 - x + 4 = 1 - - x + 1 : x = 0

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

No Solution for

x \in R

1 - - x + 1 = - x + 1 + 1 : x = 1

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

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

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

\int_{- 3}^{0} 2 - x + 4 - (- x + 1 + 1) d x = \frac{19}{3}

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

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

Simplify

= 9

\int \frac{1}{t ^{2} 400 - t ^{2}} d t

\frac{d}{d x} (cos (arctan (3 x)))

\int_{0}^{\frac{1}{10}} x ln (10 x) d x

x \to 0 lim (2)

f (x) = 3 tan (x), at x = \frac{π}{3}