What is the integral from-1 to 1 of 1/2 (6-6x^2)^2 ?

The integral from-1 to 1 of 1/2 (6-6x^2)^2 is 96/5

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integral from-1 to 1 of 1/2 (6-6x^2)^2

\int_{- 1}^{1} \frac{1}{2} (6 - 6 x^{2})^{2} d x

\frac{96}{5}

Decimal

19.2

Solution steps

\int_{- 1}^{1} \frac{1}{2} (6 - 6 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} (6 - 6 x^{2})^{2} d x

Expand

(6 - 6 x^{2})^{2} : 36 - 72 x^{2} + 36 x^{4}

= \frac{1}{2} \cdot \int_{- 1}^{1} 36 - 72 x^{2} + 36 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} 36 d x - \int_{- 1}^{1} 72 x^{2} d x + \int_{- 1}^{1} 36 x^{4} d x)

\int_{- 1}^{1} 36 d x = 72

\int_{- 1}^{1} 72 x^{2} d x = 48

\int_{- 1}^{1} 36 x^{4} d x = \frac{72}{5}

= \frac{1}{2} (72 - 48 + \frac{72}{5})

Simplify

\frac{1}{2} (72 - 48 + \frac{72}{5}) : \frac{96}{5}

= \frac{96}{5}

What is the integral from-1 to 1 of 1/2 (6-6x^2)^2 ?
The integral from-1 to 1 of 1/2 (6-6x^2)^2 is 96/5