What is the integral from-2 to 0 of (x-2)(x+1) ?

The integral from-2 to 0 of (x-2)(x+1) is 2/3

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integral from-2 to 0 of (x-2)(x+1)

\int_{- 2}^{0} (x - 2) (x + 1) d x

\frac{2}{3}

Decimal Notation

0.66666 \dots

Solution steps

\int_{- 2}^{0} (x - 2) (x + 1) d x

Expand

(x - 2) (x + 1) : x^{2} - x - 2

= \int_{- 2}^{0} 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_{- 2}^{0} x^{2} d x - \int_{- 2}^{0} x d x - \int_{- 2}^{0} 2 d x

\int_{- 2}^{0} x^{2} d x = \frac{8}{3}

\int_{- 2}^{0} x d x = - 2

\int_{- 2}^{0} 2 d x = 4

= \frac{8}{3} - (- 2) - 4

Simplify

\frac{8}{3} - (- 2) - 4 : \frac{2}{3}

= \frac{2}{3}

y = 8 x, y = \frac{5}{2} x, y = 9 - x^{2}

x \to 4 - lim (4 - x)

\frac{d y}{d x} = \frac{4}{x e ^{y}}

\int_{1}^{3} 11 r^{3} \cdot ln (r) d r

6 x^{2} \cdot e^{\frac{x ^{3}}{3}}

What is the integral from-2 to 0 of (x-2)(x+1) ?
The integral from-2 to 0 of (x-2)(x+1) is 2/3