What is the integral from-1 to 1 of pi(2-2x^2)^2 ?

The integral from-1 to 1 of pi(2-2x^2)^2 is pi 64/15

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integral from-1 to 1 of pi(2-2x^2)^2

\int_{- 1}^{1} π (2 - 2 x^{2})^{2} d x

π \frac{64}{15}

Decimal Notation

13.40412 \dots

Solution steps

\int_{- 1}^{1} π (2 - 2 x^{2})^{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= π \cdot \int_{- 1}^{1} (2 - 2 x^{2})^{2} d x

Expand

(2 - 2 x^{2})^{2} : 4 - 8 x^{2} + 4 x^{4}

= π \cdot \int_{- 1}^{1} 4 - 8 x^{2} + 4 x^{4} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= π (\int_{- 1}^{1} 4 d x - \int_{- 1}^{1} 8 x^{2} d x + \int_{- 1}^{1} 4 x^{4} d x)

\int_{- 1}^{1} 4 d x = 8

\int_{- 1}^{1} 8 x^{2} d x = \frac{16}{3}

\int_{- 1}^{1} 4 x^{4} d x = \frac{8}{5}

= π (8 - \frac{16}{3} + \frac{8}{5})

Join

8 - \frac{16}{3} + \frac{8}{5} : \frac{64}{15}

= π \frac{64}{15}

y^{^{''}} + 2 y^{^{'}} - 15 y = 4 e^{- t}

\frac{\partial}{\partial y} (x^{2} - y)

2 y^{^{''}} + 5 y^{^{'}} - 3 y = 0

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

\frac{\partial}{\partial x} (3 x ln (y) + y ln (z) - 2 x)

What is the integral from-1 to 1 of pi(2-2x^2)^2 ?
The integral from-1 to 1 of pi(2-2x^2)^2 is pi 64/15