What is the integral from-1 to 1 of 1/2 (9-9x^2)^2 ?

The integral from-1 to 1 of 1/2 (9-9x^2)^2 is 216/5

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integral from-1 to 1 of 1/2 (9-9x^2)^2

\int_{- 1}^{1} \frac{1}{2} (9 - 9 x^{2})^{2} d x

\frac{216}{5}

Decimal Notation

43.2

Solution steps

\int_{- 1}^{1} \frac{1}{2} (9 - 9 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} (9 - 9 x^{2})^{2} d x

Expand

(9 - 9 x^{2})^{2} : 81 - 162 x^{2} + 81 x^{4}

= \frac{1}{2} \cdot \int_{- 1}^{1} 81 - 162 x^{2} + 81 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} 81 d x - \int_{- 1}^{1} 162 x^{2} d x + \int_{- 1}^{1} 81 x^{4} d x)

\int_{- 1}^{1} 81 d x = 162

\int_{- 1}^{1} 162 x^{2} d x = 108

\int_{- 1}^{1} 81 x^{4} d x = \frac{162}{5}

= \frac{1}{2} (162 - 108 + \frac{162}{5})

Simplify

\frac{1}{2} (162 - 108 + \frac{162}{5}) : \frac{216}{5}

= \frac{216}{5}

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What is the integral from-1 to 1 of 1/2 (9-9x^2)^2 ?
The integral from-1 to 1 of 1/2 (9-9x^2)^2 is 216/5