What is the integral from-1 to 0 of-2x^3(-x^4+2)^4 ?

The integral from-1 to 0 of-2x^3(-x^4+2)^4 is 31/10

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integral from-1 to 0 of-2x^3(-x^4+2)^4

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

\frac{31}{10}

Decimal

3.1

Solution steps

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

Expand

x^{3} (- x^{4} + 2)^{4} : x^{19} - 8 x^{15} + 24 x^{11} - 32 x^{7} + 16 x^{3}

= - 2 \cdot \int_{- 1}^{0} x^{19} - 8 x^{15} + 24 x^{11} - 32 x^{7} + 16 x^{3} 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} x^{19} d x - \int_{- 1}^{0} 8 x^{15} d x + \int_{- 1}^{0} 24 x^{11} d x - \int_{- 1}^{0} 32 x^{7} d x + \int_{- 1}^{0} 16 x^{3} d x)

\int_{- 1}^{0} x^{19} d x = - \frac{1}{20}

\int_{- 1}^{0} 8 x^{15} d x = - \frac{1}{2}

\int_{- 1}^{0} 24 x^{11} d x = - 2

\int_{- 1}^{0} 32 x^{7} d x = - 4

\int_{- 1}^{0} 16 x^{3} d x = - 4

= - 2 (- \frac{1}{20} - (- \frac{1}{2}) - 2 - (- 4) - 4)

Simplify

- 2 (- \frac{1}{20} - (- \frac{1}{2}) - 2 - (- 4) - 4) : \frac{31}{10}

= \frac{31}{10}

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What is the integral from-1 to 0 of-2x^3(-x^4+2)^4 ?
The integral from-1 to 0 of-2x^3(-x^4+2)^4 is 31/10