What is the integral from-1 to 1 of (8x^3-6x+1) ?

The integral from-1 to 1 of (8x^3-6x+1) is 2

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integral from-1 to 1 of (8x^3-6x+1)

\int_{- 1}^{1} (8 x^{3} - 6 x + 1) d x

2

Solution steps

\int_{- 1}^{1} 8 x^{3} - 6 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} 8 x^{3} d x - \int_{- 1}^{1} 6 x d x + \int_{- 1}^{1} 1 d x

\int_{- 1}^{1} 8 x^{3} d x = 0

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

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

= 0 - 0 + 2

Simplify

= 2

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What is the integral from-1 to 1 of (8x^3-6x+1) ?
The integral from-1 to 1 of (8x^3-6x+1) is 2