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

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

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

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

\frac{9}{2}

Decimal Notation

4.5

Solution steps

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

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

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

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

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

Simplify

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

= \frac{9}{2}

x \to \frac{1}{3} lim (\frac{x + 1}{x + 2})

x \to 3 lim (1)

\frac{d}{d x} (8 x^{2} + 1)

\frac{d}{d x} ((sin (2 x)) (cos (2 x)))

\frac{\partial}{\partial x} (x^{0.9})

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