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

The integral from-1 to 2 of [(x^2)-(x+2)] is -9/2

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integral from-1 to 2 of [(x^2)-(x+2)]

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

- \frac{9}{2}

Decimal

- 4.5

Solution steps

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

Simplify

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

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

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

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

= 3 - \frac{3}{2} - 6

Simplify

3 - \frac{3}{2} - 6 : - \frac{9}{2}

= - \frac{9}{2}

What is the integral from-1 to 2 of [(x^2)-(x+2)] ?
The integral from-1 to 2 of [(x^2)-(x+2)] is -9/2