What is the integral from-1 to 1 of (x^5-4x^2+x-1) ?

The integral from-1 to 1 of (x^5-4x^2+x-1) is -14/3

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integral from-1 to 1 of (x^5-4x^2+x-1)

\int_{- 1}^{1} (x^{5} - 4 x^{2} + x - 1) d x

- \frac{14}{3}

Decimal

- 4.66666 \dots

Solution steps

\int_{- 1}^{1} x^{5} - 4 x^{2} + x - 1 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}^{1} x^{5} d x - \int_{- 1}^{1} 4 x^{2} d x + \int_{- 1}^{1} x d x - \int_{- 1}^{1} 1 d x

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

\int_{- 1}^{1} 4 x^{2} d x = \frac{8}{3}

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

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

= 0 - \frac{8}{3} + 0 - 2

Simplify

0 - \frac{8}{3} + 0 - 2 : - \frac{14}{3}

= - \frac{14}{3}

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What is the integral from-1 to 1 of (x^5-4x^2+x-1) ?
The integral from-1 to 1 of (x^5-4x^2+x-1) is -14/3