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

The integral from-1 to 0 of 2x^2(2x^3-1)^3 is -20/3

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

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

- \frac{20}{3}

Decimal

- 6.66666 \dots

Solution steps

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

Take the constant out:

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

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

Expand

x^{2} (2 x^{3} - 1)^{3} : 8 x^{11} - 12 x^{8} + 6 x^{5} - x^{2}

= 2 \cdot \int_{- 1}^{0} 8 x^{11} - 12 x^{8} + 6 x^{5} - 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

= 2 (\int_{- 1}^{0} 8 x^{11} d x - \int_{- 1}^{0} 12 x^{8} d x + \int_{- 1}^{0} 6 x^{5} d x - \int_{- 1}^{0} x^{2} d x)

\int_{- 1}^{0} 8 x^{11} d x = - \frac{2}{3}

\int_{- 1}^{0} 12 x^{8} d x = \frac{4}{3}

\int_{- 1}^{0} 6 x^{5} d x = - 1

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

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

Simplify

2 (- \frac{2}{3} - \frac{4}{3} - 1 - \frac{1}{3}) : - \frac{20}{3}

= - \frac{20}{3}

What is the integral from-1 to 0 of 2x^2(2x^3-1)^3 ?
The integral from-1 to 0 of 2x^2(2x^3-1)^3 is -20/3