What is area y=3x^2,y=2,x=0,x>= 0 ?

The answer to area y=3x^2,y=2,x=0,x>= 0 is -(4sqrt(2))/(3sqrt(3))+4sqrt(2/3)

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area y=3x^2,y=2,x=0,x>= 0

area

y = 3 x^{2}, y = 2, x = 0, x \geq 0

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

Decimal Notation

2.17732 \dots

Solution steps

Apply the area formula:

\int_{- \frac{2}{3}}^{\frac{2}{3}} 3 x^{2} - 2 d x

Solve

\int_{- \frac{2}{3}}^{\frac{2}{3}} 3 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}

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

\frac{x}{2 + x ^{2}}

(3 x + 7) x^{2}

\frac{d}{d x} (- \frac{32}{x ^{3}} + \frac{4}{x ^{2}})

\int_{2}^{4} \frac{x ^{2} + 1}{x ^{2} - 5} d x

What is area y=3x^2,y=2,x=0,x>= 0 ?
The answer to area y=3x^2,y=2,x=0,x>= 0 is -(4sqrt(2))/(3sqrt(3))+4sqrt(2/3)