What is the integral from-2 to 4 of (x^2-2x-8) ?

The integral from-2 to 4 of (x^2-2x-8) is -36

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integral from-2 to 4 of (x^2-2x-8)

\int_{- 2}^{4} (x^{2} - 2 x - 8) d x

- 36

Solution steps

\int_{- 2}^{4} x^{2} - 2 x - 8 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}^{4} x^{2} d x - \int_{- 2}^{4} 2 x d x - \int_{- 2}^{4} 8 d x

\int_{- 2}^{4} x^{2} d x = 24

\int_{- 2}^{4} 2 x d x = 12

\int_{- 2}^{4} 8 d x = 48

= 24 - 12 - 48

Simplify

= - 36

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\frac{d}{d x} (\frac{x cos ( x )}{1 + e ^{x}})

\frac{\partial}{\partial x} (y^{5} sin (2 x))

4 - x^{2}, x^{2} - 2 x

What is the integral from-2 to 4 of (x^2-2x-8) ?
The integral from-2 to 4 of (x^2-2x-8) is -36