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

The integral from 0 to 3 of (x^3-x^2-6x) is -63/4

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

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

- \frac{63}{4}

Decimal Notation

- 15.75

Solution steps

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

\int_{0}^{3} x^{3} d x = \frac{81}{4}

\int_{0}^{3} x^{2} d x = 9

\int_{0}^{3} 6 x d x = 27

= \frac{81}{4} - 9 - 27

Simplify

\frac{81}{4} - 9 - 27 : - \frac{63}{4}

= - \frac{63}{4}

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What is the integral from 0 to 3 of (x^3-x^2-6x) ?
The integral from 0 to 3 of (x^3-x^2-6x) is -63/4