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

The integral from-1 to 1 of 3x^4-x^2+x-1 is -22/15

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

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

- \frac{22}{15}

Decimal

- 1.46666 \dots

Solution steps

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

\int_{- 1}^{1} 3 x^{4} d x = \frac{6}{5}

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

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

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

= \frac{6}{5} - \frac{2}{3} + 0 - 2

Simplify

\frac{6}{5} - \frac{2}{3} + 0 - 2 : - \frac{22}{15}

= - \frac{22}{15}

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What is the integral from-1 to 1 of 3x^4-x^2+x-1 ?
The integral from-1 to 1 of 3x^4-x^2+x-1 is -22/15