What is area f(x)=-4x^2-4,g(x)=2x^2-9 ?

The answer to area f(x)=-4x^2-4,g(x)=2x^2-9 is -(5sqrt(10))/(3sqrt(3))+10sqrt(5/6)

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area f(x)=-4x^2-4,g(x)=2x^2-9

area

f (x) = - 4 x^{2} - 4, g (x) = 2 x^{2} - 9

- \frac{5 10}{3 3} + 10 \frac{5}{6}

Decimal Notation

6.08580 \dots

Solution steps

Apply the area formula:

\int_{- \frac{5}{6}}^{\frac{5}{6}} 6 x^{2} - 5 d x

Solve

\int_{- \frac{5}{6}}^{\frac{5}{6}} 6 x^{2} - 5 d x : - \frac{5 10}{3 3} + 10 \frac{5}{6}

The area is:

= - \frac{5 10}{3 3} + 10 \frac{5}{6}

\frac{d}{d x} (- 3 ln (cos (\frac{x}{3})))

x \to 0 lim (\frac{ln ( x )}{3})

\frac{\partial}{\partial y} (\frac{y + x}{y - x})

x \to 4 - lim (\frac{1}{x ^{2} - 3 x - 4})

\int \frac{( x - 1 )}{x ^{2}} e^{x} d x

What is area f(x)=-4x^2-4,g(x)=2x^2-9 ?
The answer to area f(x)=-4x^2-4,g(x)=2x^2-9 is -(5sqrt(10))/(3sqrt(3))+10sqrt(5/6)