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

The integral from-1 to 1 of (x+1)(x+2)(x+3) is 16

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

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

16

Solution steps

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

Expand

(x + 1) (x + 2) (x + 3) : x^{3} + 6 x^{2} + 11 x + 6

= \int_{- 1}^{1} x^{3} + 6 x^{2} + 11 x + 6 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}^{1} x^{3} d x + \int_{- 1}^{1} 6 x^{2} d x + \int_{- 1}^{1} 11 x d x + \int_{- 1}^{1} 6 d x

\int_{- 1}^{1} x^{3} d x = 0

\int_{- 1}^{1} 6 x^{2} d x = 4

\int_{- 1}^{1} 11 x d x = 0

\int_{- 1}^{1} 6 d x = 12

= 0 + 4 + 0 + 12

Simplify

= 16

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What is the integral from-1 to 1 of (x+1)(x+2)(x+3) ?
The integral from-1 to 1 of (x+1)(x+2)(x+3) is 16