What is the integral from-1 to 2 of (4-x^2)-(x^2-2x) ?

The integral from-1 to 2 of (4-x^2)-(x^2-2x) is 9

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integral from-1 to 2 of (4-x^2)-(x^2-2x)

\int_{- 1}^{2} (4 - x^{2}) - (x^{2} - 2 x) d x

9

Solution steps

\int_{- 1}^{2} 4 - x^{2} - (x^{2} - 2 x) d x

Simplify

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

\int_{- 1}^{2} 4 d x = 12

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

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

= 12 - 6 + 3

Simplify

= 9

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What is the integral from-1 to 2 of (4-x^2)-(x^2-2x) ?
The integral from-1 to 2 of (4-x^2)-(x^2-2x) is 9