What is the integral from-1 to 3 of (6x^3+3x^2-7x) ?

The integral from-1 to 3 of (6x^3+3x^2-7x) is 120

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integral from-1 to 3 of (6x^3+3x^2-7x)

\int_{- 1}^{3} (6 x^{3} + 3 x^{2} - 7 x) d x

120

Solution steps

\int_{- 1}^{3} 6 x^{3} + 3 x^{2} - 7 x 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}^{3} 6 x^{3} d x + \int_{- 1}^{3} 3 x^{2} d x - \int_{- 1}^{3} 7 x d x

\int_{- 1}^{3} 6 x^{3} d x = 120

\int_{- 1}^{3} 3 x^{2} d x = 28

\int_{- 1}^{3} 7 x d x = 28

= 120 + 28 - 28

Simplify

= 120

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What is the integral from-1 to 3 of (6x^3+3x^2-7x) ?
The integral from-1 to 3 of (6x^3+3x^2-7x) is 120