What is area y=10x^2,y=x^2+12 ?

The answer to area y=10x^2,y=x^2+12 is -(16)/(sqrt(3))+16sqrt(3)

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area y=10x^2,y=x^2+12

area

y = 10 x^{2}, y = x^{2} + 12

- \frac{16}{3} + 163

Decimal

18.47520 \dots

Solution steps

Apply the area formula:

\int_{- \frac{2}{3}}^{\frac{2}{3}} 10 x^{2} - (x^{2} + 12) d x

Solve

\int_{- \frac{2}{3}}^{\frac{2}{3}} 10 x^{2} - (x^{2} + 12) d x : - \frac{16}{3} + 163

The area is:

= - \frac{16}{3} + 163

\int e^{\frac{- 4 x}{3}} d x

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

n = 0 \sum \infty (\frac{2}{5})^{n}

\frac{d x}{d t} = \frac{t}{x e ^{t + 2 x}}

y^{^{'}} + sin (x) y = sin (x), y (0) = π

What is area y=10x^2,y=x^2+12 ?
The answer to area y=10x^2,y=x^2+12 is -(16)/(sqrt(3))+16sqrt(3)