What is the integral from-1 to 1 of pi(8-8x^2)^2 ?

The integral from-1 to 1 of pi(8-8x^2)^2 is pi 1024/15

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integral from-1 to 1 of pi(8-8x^2)^2

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

π \frac{1024}{15}

Decimal Notation

214.46605 \dots

Solution steps

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

Expand

(8 - 8 x^{2})^{2} : 64 - 128 x^{2} + 64 x^{4}

= π \cdot \int_{- 1}^{1} 64 - 128 x^{2} + 64 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} 64 d x - \int_{- 1}^{1} 128 x^{2} d x + \int_{- 1}^{1} 64 x^{4} d x)

\int_{- 1}^{1} 64 d x = 128

\int_{- 1}^{1} 128 x^{2} d x = \frac{256}{3}

\int_{- 1}^{1} 64 x^{4} d x = \frac{128}{5}

= π (128 - \frac{256}{3} + \frac{128}{5})

Join

128 - \frac{256}{3} + \frac{128}{5} : \frac{1024}{15}

= π \frac{1024}{15}

f (x) = x - (x + 1)^{4}

y^{^{'}} = x - 7

e^{x} (4 - x)

\int \frac{9 r ^{2}}{1 - r ^{3}} d r

n = 1 \sum \infty n^{3} e^{- n^{2}}

What is the integral from-1 to 1 of pi(8-8x^2)^2 ?
The integral from-1 to 1 of pi(8-8x^2)^2 is pi 1024/15