What is area y=x^2,y=2x^2-4x,y=0 ?

The answer to area y=x^2,y=2x^2-4x,y=0 is 40/3

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area y=x^2,y=2x^2-4x,y=0

area

y = x^{2}, y = 2 x^{2} - 4 x, y = 0

\frac{40}{3}

Decimal

13.33333 \dots

Solution steps

To find the intersection points solve

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

x^{2} = 2 x^{2} - 4 x : x = 4, x = 0

x^{2} = 0 : x = 0

2 x^{2} - 4 x = 0 : x = 2, x = 0

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

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

\int_{0}^{4} x^{2} - (2 x^{2} - 4 x) d x = \frac{32}{3}

\int_{0}^{2} 2 x^{2} - 4 x - 0 d x = \frac{8}{3}

= \frac{8}{3} + \frac{32}{3}

Simplify

= \frac{40}{3}

What is area y=x^2,y=2x^2-4x,y=0 ?
The answer to area y=x^2,y=2x^2-4x,y=0 is 40/3