What is area y=3x^2,y^2= 1/3 x ?

The answer to area y=3x^2,y^2= 1/3 x is 1/27

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area y=3x^2,y^2= 1/3 x

area

y = 3 x^{2}, y^{2} = \frac{1}{3} x

\frac{1}{27}

Decimal

0.03703 \dots

Solution steps

Isolate

y

for

y^{2} = \frac{1}{3} x : y = \frac{x}{3}, y = - \frac{x}{3}

f_{1} (x) = 3 x^{2}

f_{2} (x) = \frac{x}{3}

To find the intersection points solve

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

3 x^{2} = \frac{x}{3} : x = 0, x = \frac{1}{3}

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

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

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

= \int_{0}^{\frac{1}{3}} 3 x^{2} - \frac{x}{3} d x

\int_{0}^{\frac{1}{3}} 3 x^{2} - \frac{x}{3} d x = \frac{1}{27}

= \frac{1}{27}

\int \frac{ln ^{2} ( x ^{2} )}{x} d x

\frac{\partial}{\partial x} (y e^{y} x)

\frac{d}{d x} ((3 - 2 x)^{3})

f (X) = 7 (X - 5)^{\frac{2}{3}}

y = ln (9 x)