What is area y=4x^2,y=x^2+2 ?

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

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area y=4x^2,y=x^2+2

area

y = 4 x^{2}, y = x^{2} + 2

- \frac{4 2}{3 3} + 4 \frac{2}{3}

Decimal

2.17732 \dots

Solution steps

Apply the area formula:

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

Solve

\int_{- \frac{2}{3}}^{\frac{2}{3}} 4 x^{2} - (x^{2} + 2) d x : - \frac{4 2}{3 3} + 4 \frac{2}{3}

The area is:

= - \frac{4 2}{3 3} + 4 \frac{2}{3}

x \to 0 lim ((1 - 4 x)^{\frac{1}{x}})

x y^{^{'}} + x y = 1 - y, y (1) = 0

\frac{\partial}{\partial x} (x^{2} y^{2} (2 z + 1)^{2})

\int \frac{x ^{2}}{a ^{2} + b ^{2}} d x

\int \frac{x ^{2} - x}{x ^{3} - 6 x ^{2} + 11 x - 6} d x

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