What is the integral from-1 to 0 of (x^3-2)^2*3x^2 ?

The integral from-1 to 0 of (x^3-2)^2*3x^2 is 19/3

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integral from-1 to 0 of (x^3-2)^2*3x^2

\int_{- 1}^{0} (x^{3} - 2)^{2} \cdot 3 x^{2} d x

\frac{19}{3}

Decimal

6.33333 \dots

Solution steps

\int_{- 1}^{0} (x^{3} - 2)^{2} \cdot 3 x^{2} d x

Take the constant out:

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

= 3 \cdot \int_{- 1}^{0} (x^{3} - 2)^{2} x^{2} d x

Expand

(x^{3} - 2)^{2} x^{2} : x^{8} - 4 x^{5} + 4 x^{2}

= 3 \cdot \int_{- 1}^{0} x^{8} - 4 x^{5} + 4 x^{2} 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

= 3 (\int_{- 1}^{0} x^{8} d x - \int_{- 1}^{0} 4 x^{5} d x + \int_{- 1}^{0} 4 x^{2} d x)

\int_{- 1}^{0} x^{8} d x = \frac{1}{9}

\int_{- 1}^{0} 4 x^{5} d x = - \frac{2}{3}

\int_{- 1}^{0} 4 x^{2} d x = \frac{4}{3}

= 3 (\frac{1}{9} - (- \frac{2}{3}) + \frac{4}{3})

Simplify

3 (\frac{1}{9} - (- \frac{2}{3}) + \frac{4}{3}) : \frac{19}{3}

= \frac{19}{3}

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What is the integral from-1 to 0 of (x^3-2)^2*3x^2 ?
The integral from-1 to 0 of (x^3-2)^2*3x^2 is 19/3