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

The integral from-1 to 2 of (x-1)(x+1) is 0

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

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

0

Solution steps

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

Expand

(x - 1) (x + 1) : x^{2} - 1

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

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

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

= 3 - 3

Simplify

= 0

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What is the integral from-1 to 2 of (x-1)(x+1) ?
The integral from-1 to 2 of (x-1)(x+1) is 0