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

The integral from-1 to 0 of x^3(4x^4-3)^4 is -61/20

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

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

- \frac{61}{20}

Decimal

- 3.05

Solution steps

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

Expand

x^{3} (4 x^{4} - 3)^{4} : 256 x^{19} - 768 x^{15} + 864 x^{11} - 432 x^{7} + 81 x^{3}

= \int_{- 1}^{0} 256 x^{19} - 768 x^{15} + 864 x^{11} - 432 x^{7} + 81 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

= \int_{- 1}^{0} 256 x^{19} d x - \int_{- 1}^{0} 768 x^{15} d x + \int_{- 1}^{0} 864 x^{11} d x - \int_{- 1}^{0} 432 x^{7} d x + \int_{- 1}^{0} 81 x^{3} d x

\int_{- 1}^{0} 256 x^{19} d x = - \frac{64}{5}

\int_{- 1}^{0} 768 x^{15} d x = - 48

\int_{- 1}^{0} 864 x^{11} d x = - 72

\int_{- 1}^{0} 432 x^{7} d x = - 54

\int_{- 1}^{0} 81 x^{3} d x = - \frac{81}{4}

= - \frac{64}{5} - (- 48) - 72 - (- 54) - \frac{81}{4}

Simplify

- \frac{64}{5} - (- 48) - 72 - (- 54) - \frac{81}{4} : - \frac{61}{20}

= - \frac{61}{20}

What is the integral from-1 to 0 of x^3(4x^4-3)^4 ?
The integral from-1 to 0 of x^3(4x^4-3)^4 is -61/20