What is the integral from-2 to 0 of (x^3-x^2-6x) ?

The integral from-2 to 0 of (x^3-x^2-6x) is 16/3

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integral from-2 to 0 of (x^3-x^2-6x)

\int_{- 2}^{0} (x^{3} - x^{2} - 6 x) d x

\frac{16}{3}

Decimal

5.33333 \dots

Solution steps

\int_{- 2}^{0} x^{3} - x^{2} - 6 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_{- 2}^{0} x^{3} d x - \int_{- 2}^{0} x^{2} d x - \int_{- 2}^{0} 6 x d x

\int_{- 2}^{0} x^{3} d x = - 4

\int_{- 2}^{0} x^{2} d x = \frac{8}{3}

\int_{- 2}^{0} 6 x d x = - 12

= - 4 - \frac{8}{3} - (- 12)

Simplify

- 4 - \frac{8}{3} - (- 12) : \frac{16}{3}

= \frac{16}{3}

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What is the integral from-2 to 0 of (x^3-x^2-6x) ?
The integral from-2 to 0 of (x^3-x^2-6x) is 16/3