What is the integral from-1 to 1 of 1/2 (3-3x^2)^2 ?

The integral from-1 to 1 of 1/2 (3-3x^2)^2 is 24/5

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integral from-1 to 1 of 1/2 (3-3x^2)^2

\int_{- 1}^{1} \frac{1}{2} (3 - 3 x^{2})^{2} d x

\frac{24}{5}

Decimal

4.8

Solution steps

\int_{- 1}^{1} \frac{1}{2} (3 - 3 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} (3 - 3 x^{2})^{2} d x

Expand

(3 - 3 x^{2})^{2} : 9 - 18 x^{2} + 9 x^{4}

= \frac{1}{2} \cdot \int_{- 1}^{1} 9 - 18 x^{2} + 9 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} 9 d x - \int_{- 1}^{1} 18 x^{2} d x + \int_{- 1}^{1} 9 x^{4} d x)

\int_{- 1}^{1} 9 d x = 18

\int_{- 1}^{1} 18 x^{2} d x = 12

\int_{- 1}^{1} 9 x^{4} d x = \frac{18}{5}

= \frac{1}{2} (18 - 12 + \frac{18}{5})

Simplify

\frac{1}{2} (18 - 12 + \frac{18}{5}) : \frac{24}{5}

= \frac{24}{5}

f (x) = x^{3} cos (x)

\frac{\partial}{\partial x} (\frac{4 x + 3}{4 y + 1})

\frac{d}{d x} (\frac{ln ( x )}{x ^{\frac{5}{4}}})

\int y cos (y) sin^{2} (y) d y

\frac{1}{x}, at x = 2

What is the integral from-1 to 1 of 1/2 (3-3x^2)^2 ?
The integral from-1 to 1 of 1/2 (3-3x^2)^2 is 24/5