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

The integral from-1 to 3 of 6x+x^2-x^3 is 40/3

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

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

\frac{40}{3}

Decimal Notation

13.33333 \dots

Solution steps

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

\int_{- 1}^{3} 6 x d x = 24

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

\int_{- 1}^{3} x^{3} d x = 20

= 24 + \frac{28}{3} - 20

Simplify

24 + \frac{28}{3} - 20 : \frac{40}{3}

= \frac{40}{3}

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What is the integral from-1 to 3 of 6x+x^2-x^3 ?
The integral from-1 to 3 of 6x+x^2-x^3 is 40/3