What is the integral from-1 to 0 of (x^5-4x^3+4x+4) ?

The integral from-1 to 0 of (x^5-4x^3+4x+4) is 17/6

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integral from-1 to 0 of (x^5-4x^3+4x+4)

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

\frac{17}{6}

Decimal

2.83333 \dots

Solution steps

\int_{- 1}^{0} x^{5} - 4 x^{3} + 4 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

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

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

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

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

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

= - \frac{1}{6} - (- 1) - 2 + 4

Simplify

- \frac{1}{6} - (- 1) - 2 + 4 : \frac{17}{6}

= \frac{17}{6}

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What is the integral from-1 to 0 of (x^5-4x^3+4x+4) ?
The integral from-1 to 0 of (x^5-4x^3+4x+4) is 17/6