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Popular Calculus >

area x+y^2=12,x+y=0

Solution

area

x + y^{2} = 12, x + y = 0

Solution

\frac{343}{6}

+1

Decimal

57.16666 \dots

Solution steps

Isolate

y

for

x + y^{2} = 12 : y = 12 - x, y = - 12 - x

Isolate

y

for

x + y = 0 : y = - x

f_{1} (x) = 12 - x

f_{2} (x) = - 12 - x

To find the intersection points solve

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

12 - x = - 12 - x : x = 12

12 - x = - x : x = - 4

- 12 - x = - x : x = 3

= \int_{- 4}^{3} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{3}^{12} ∣ f_{2} (x) - f_{1} (x) ∣ d x

= \int_{- 4}^{3} 12 - x - (- x) d x + \int_{3}^{12} - 12 - x - 12 - x d x

\int_{- 4}^{3} 12 - x - (- x) d x = \frac{127}{6}

\int_{3}^{12} - 12 - x - 12 - x d x = 36

= \frac{127}{6} + 36

Simplify

= \frac{343}{6}

Graph

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Frequently Asked Questions (FAQ)

What is area x+y^2=12,x+y=0 ?
The answer to area x+y^2=12,x+y=0 is 343/6