What is the integral from-1 to 1 of 1/2 (4-4x^2)^2 ?

The integral from-1 to 1 of 1/2 (4-4x^2)^2 is 128/15

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integral from-1 to 1 of 1/2 (4-4x^2)^2

\int_{- 1}^{1} \frac{1}{2} (4 - 4 x^{2})^{2} d x

\frac{128}{15}

Decimal

8.53333 \dots

Solution steps

\int_{- 1}^{1} \frac{1}{2} (4 - 4 x^{2})^{2} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int_{- 1}^{1} (4 - 4 x^{2})^{2} d x

Expand

(4 - 4 x^{2})^{2} : 16 - 32 x^{2} + 16 x^{4}

= \frac{1}{2} \cdot \int_{- 1}^{1} 16 - 32 x^{2} + 16 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

= \frac{1}{2} (\int_{- 1}^{1} 16 d x - \int_{- 1}^{1} 32 x^{2} d x + \int_{- 1}^{1} 16 x^{4} d x)

\int_{- 1}^{1} 16 d x = 32

\int_{- 1}^{1} 32 x^{2} d x = \frac{64}{3}

\int_{- 1}^{1} 16 x^{4} d x = \frac{32}{5}

= \frac{1}{2} (32 - \frac{64}{3} + \frac{32}{5})

Simplify

\frac{1}{2} (32 - \frac{64}{3} + \frac{32}{5}) : \frac{128}{15}

= \frac{128}{15}

\frac{\partial}{\partial t} (\frac{s - 3 r}{t})

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

\frac{d}{d x} (\frac{7}{3 x ^{4}})

x \to 8 lim (\frac{x - 8}{x})

x \to \infty lim (\frac{6 x ^{2}}{e ^{5 x}})

What is the integral from-1 to 1 of 1/2 (4-4x^2)^2 ?
The integral from-1 to 1 of 1/2 (4-4x^2)^2 is 128/15