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Popular Calculus >

area y^2=6-x,3y=x+12

Solution

area

y^{2} = 6 - x, 3 y = x + 12

Solution

\frac{243}{2}

+1

Decimal

121.5

Solution steps

Isolate

y

for

y^{2} = 6 - x : y = 6 - x, y = - 6 - x

Isolate

y

for

3 y = x + 12 : y = \frac{x + 12}{3}

f_{1} (x) = 6 - x

f_{2} (x) = - 6 - x

To find the intersection points solve

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

6 - x = - 6 - x : x = 6

6 - x = \frac{x + 12}{3} : x = - 3

- 6 - x = \frac{x + 12}{3} : x = - 30

= \int_{- 30}^{- 3} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{- 3}^{6} ∣ f_{1} (x) - f_{2} (x) ∣ d x

= \int_{- 30}^{- 3} - 6 - x - \frac{x + 12}{3} d x + \int_{- 3}^{6} 6 - x - (- 6 - x) d x

\int_{- 30}^{- 3} - 6 - x - \frac{x + 12}{3} d x = \frac{171}{2}

\int_{- 3}^{6} 6 - x - (- 6 - x) d x = 36

= \frac{171}{2} + 36

Simplify

= \frac{243}{2}

Graph

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Frequently Asked Questions (FAQ)

What is area y^2=6-x,3y=x+12 ?
The answer to area y^2=6-x,3y=x+12 is 243/2