What is area x=y^2-2y,x-y-4=0 ?

The answer to area x=y^2-2y,x-y-4=0 is 125/6

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area x=y^2-2y,x-y-4=0

area

x = y^{2} - 2 y, x - y - 4 = 0

\frac{125}{6}

Decimal Notation

20.83333 \dots

Solution steps

Isolate

y

for

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

Isolate

y

for

x - y - 4 = 0 : y = x - 4

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

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

To find the intersection points solve

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

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

1 + x + 1 = x - 4 : x = 8

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

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

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

\int_{- 1}^{3} 1 + x + 1 - (1 - x + 1) d x = \frac{32}{3}

\int_{3}^{8} 1 + x + 1 - (x - 4) d x = \frac{61}{6}

= \frac{32}{3} + \frac{61}{6}

Simplify

= \frac{125}{6}

h^{^{'}}

\int cos^{3} (θ) d θ

4 y^{^{''}} + 4 y^{^{'}} + y = 0

\frac{d}{d x} (\frac{x ^{6}}{x ^{3}})

(3 x + 1)^{7}