What is the integral from-3 to 0 of x(x+3)(x+1) ?

The integral from-3 to 0 of x(x+3)(x+1) is 9/4

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integral from-3 to 0 of x(x+3)(x+1)

\int_{- 3}^{0} x (x + 3) (x + 1) d x

\frac{9}{4}

Solution steps

\int_{- 3}^{0} x (x + 3) (x + 1) d x

Expand

x (x + 3) (x + 1) : x^{3} + 4 x^{2} + 3 x

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

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

\int_{- 3}^{0} 4 x^{2} d x = 36

\int_{- 3}^{0} 3 x d x = - \frac{27}{2}

= - \frac{81}{4} + 36 - \frac{27}{2}

Simplify

- \frac{81}{4} + 36 - \frac{27}{2} : \frac{9}{4}

= \frac{9}{4}

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What is the integral from-3 to 0 of x(x+3)(x+1) ?
The integral from-3 to 0 of x(x+3)(x+1) is 9/4