What is area x^3,x+6=y,(-x)/2 =y ?

The answer to area x^3,x+6=y,(-x)/2 =y is 22

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area x^3,x+6=y,(-x)/2 =y

area

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

22

Solution steps

Isolate

y

for

x + 6 = y : y = x + 6

Isolate

y

for

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

f_{1} (x) = x + 6

f_{2} (x) = - \frac{x}{2}

To find the intersection points solve

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

x + 6 = - \frac{x}{2} : x = - 4

x + 6 = x^{3} : x = 2

- \frac{x}{2} = x^{3} : x = 0

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

= \int_{- 4}^{0} x + 6 - (- \frac{x}{2}) d x + \int_{0}^{2} x + 6 - x^{3} d x

\int_{- 4}^{0} x + 6 - (- \frac{x}{2}) d x = 12

\int_{0}^{2} x + 6 - x^{3} d x = 10

= 12 + 10

Simplify

= 22

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