What is area y=4x^2,y=6x^2,5x+y=6>= 0 ?

The answer to area y=4x^2,y=6x^2,5x+y=6>= 0 is 2/3

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area y=4x^2,y=6x^2,5x+y=6>= 0

area

y = 4 x^{2}, y = 6 x^{2}, 5 x + y = 6 \geq 0

\frac{2}{3}

Decimal Notation

0.66666 \dots

Solution steps

Apply the area formula:

\int_{0}^{1} 2 ∣ x ∣^{2} d x

Solve

\int_{0}^{1} 2 ∣ x ∣^{2} d x : \frac{2}{3}

The area is:

= \frac{2}{3}

\int e^{3 x + 8} d x

\frac{d}{d x} (\frac{cot ( x )}{1 + csc ( x )})

\frac{\partial}{\partial y} (x^{2} + 2 y + 2 z)

c^{^{'}}

What is area y=4x^2,y=6x^2,5x+y=6>= 0 ?
The answer to area y=4x^2,y=6x^2,5x+y=6>= 0 is 2/3