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

The answer to area x=y^2,x=y+6 is 125/6

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

area

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

\frac{125}{6}

Solution steps

Isolate

y

for

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

Isolate

y

for

x = y + 6 : y = x - 6

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 = x - 6 : x = 9

- x = x - 6 : x = 4

= \int_{0}^{4} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{4}^{9} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{0}^{4} x - (- x) d x + \int_{4}^{9} x - (x - 6) d x

\int_{0}^{4} x - (- x) d x = \frac{32}{3}

\int_{4}^{9} x - (x - 6) d x = \frac{61}{6}

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

Simplify

= \frac{125}{6}

\frac{\partial}{\partial s} (s cos (t - \frac{π}{2}))

\frac{d}{d x} (\frac{- 3}{x ^{2} + 9})

\int \frac{x ^{2}}{a - x ^{2}} d x

\frac{d}{d x} (csc (x) cot (x))

f (t) = 80 e^{- \frac{t}{10}} - 40 t