What is area y=x-6,y^2=2x+108 ?

The answer to area y=x-6,y^2=2x+108 is 2662/3

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area y=x-6,y^2=2x+108

area

y = x - 6, y^{2} = 2 x + 108

\frac{2662}{3}

Solution steps

Isolate

y

for

y^{2} = 2 x + 108 : y = 2 x + 108, y = - 2 x + 108

f_{1} (x) = x - 6

f_{2} (x) = 2 x + 108

To find the intersection points solve

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

x - 6 = 2 x + 108 : x = 18

x - 6 = - 2 x + 108 : x = - 4

2 x + 108 = - 2 x + 108 : x = - 54

= \int_{- 54}^{- 4} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{- 4}^{18} ∣ f_{1} (x) - f_{2} (x) ∣ d x

= \int_{- 54}^{- 4} 2 x + 108 - (- 2 x + 108) d x + \int_{- 4}^{18} x - 6 - 2 x + 108 d x

\int_{- 54}^{- 4} 2 x + 108 - (- 2 x + 108) d x = \frac{2000}{3}

\int_{- 4}^{18} x - 6 - 2 x + 108 d x = \frac{662}{3}

= \frac{2000}{3} + \frac{662}{3}

Simplify

= \frac{2662}{3}

x \to 3 lim (\frac{13 sin ( 6 x )}{x})

\frac{d}{d x} (2 x - 15)

f (x) = ln (c^{3} - 1)

n = 4 \sum \infty e^{7 - 6 n}

\frac{d x}{d t} + 2 t^{7} x^{9} + \frac{x}{t} = 0