What is area x=y^2-8,x=2y ?

The answer to area x=y^2-8,x=2y is 36

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area x=y^2-8,x=2y

area

x = y^{2} - 8, x = 2 y

36

Solution steps

Isolate

y

for

x = y^{2} - 8 : y = x + 8, y = - x + 8

Isolate

y

for

x = 2 y : y = \frac{x}{2}

f_{1} (x) = x + 8

f_{2} (x) = - x + 8

To find the intersection points solve

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

x + 8 = - x + 8 : x = - 8

x + 8 = \frac{x}{2} : x = 8

- x + 8 = \frac{x}{2} : x = - 4

= \int_{- 8}^{- 4} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{- 4}^{8} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{- 8}^{- 4} x + 8 - (- x + 8) d x + \int_{- 4}^{8} x + 8 - \frac{x}{2} d x

\int_{- 8}^{- 4} x + 8 - (- x + 8) d x = \frac{32}{3}

\int_{- 4}^{8} x + 8 - \frac{x}{2} d x = \frac{76}{3}

= \frac{32}{3} + \frac{76}{3}

Simplify

= 36

\frac{d}{d x} (\frac{x - 3}{x} + ln (\frac{x}{3}))

x \to \infty lim (3 ln (x))

k

\int_{2}^{2} \frac{u}{u ^{2} - 1} d u

\frac{\partial}{\partial y} (e^{x + y + 1})