What is area 3x^2,x^2+4,-2,2 ?

The answer to area 3x^2,x^2+4,-2,2 is (32-16sqrt(2))/3+16sqrt(2)-16

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area 3x^2,x^2+4,-2,2

area

3 x^{2}, x^{2} + 4, - 2, 2

\frac{32 - 16 2}{3} + 162 - 16

Decimal Notation

9.75161 \dots

Solution steps

Apply the area formula:

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

Solve

\int_{- 2}^{2} 3 x^{2} - (x^{2} + 4) d x : \frac{32 - 16 2}{3} + 162 - 16

The area is:

= \frac{32 - 16 2}{3} + 162 - 16

ln (\frac{x}{x ^{2} + 6})

\int \frac{- 1}{x} d x

y (x + 7) + y^{^{'}} = 0

\frac{d}{d x} (tan (5 x + 2))

\frac{d}{d x} (x^{3} \cdot e^{- x})

What is area 3x^2,x^2+4,-2,2 ?
The answer to area 3x^2,x^2+4,-2,2 is (32-16sqrt(2))/3+16sqrt(2)-16