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

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

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

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

\frac{9}{2}

Decimal Notation

4.5

Solution steps

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

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

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

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

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

Simplify

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

= \frac{9}{2}

\int_{0}^{3} \frac{x ^{2}}{9} d x

\int_{- π}^{0} - 1 d x

\int_{- 8}^{0} 64 - x^{2} d x

\int_{0}^{2 π} x^{2} + 1 d x

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

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