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

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

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

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

\frac{9}{2}

Decimal

4.5

Solution steps

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

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

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

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

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

Simplify

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

= \frac{9}{2}

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