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

The answer to area y=x+2,y=-3x+6,y=(2-x)/3 is 4

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

area

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

4

Solution steps

To find the intersection points solve

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

x + 2 = - 3 x + 6 : x = 1

x + 2 = \frac{2 - x}{3} : x = - 1

- 3 x + 6 = \frac{2 - x}{3} : x = 2

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

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

\int_{- 1}^{1} x + 2 - \frac{2 - x}{3} d x = \frac{8}{3}

\int_{1}^{2} \frac{2 - x}{3} - (- 3 x + 6) d x = \frac{4}{3}

= \frac{8}{3} + \frac{4}{3}

Simplify

= 4

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What is area y=x+2,y=-3x+6,y=(2-x)/3 ?
The answer to area y=x+2,y=-3x+6,y=(2-x)/3 is 4