What is the integral from-4 to 2 of (12-x-x^2)-(x+4) ?

The integral from-4 to 2 of (12-x-x^2)-(x+4) is 36

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integral from-4 to 2 of (12-x-x^2)-(x+4)

\int_{- 4}^{2} (12 - x - x^{2}) - (x + 4) d x

36

Solution steps

\int_{- 4}^{2} 12 - x - x^{2} - (x + 4) d x

Simplify

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

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

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

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

= - 24 - (- 12) + 48

Simplify

= 36

\frac{\partial}{\partial x} (x^{5} + 7 x^{5} y)

x \to - 4 lim (\frac{x ^{2} + 3}{x - 1})

e^{y} y^{^{'}} + \frac{t}{y + 1} = \frac{2}{y + 1}

x \to \infty lim (\frac{1}{6 ^{x + 1}})

(6, 2), (- 3, 5)

What is the integral from-4 to 2 of (12-x-x^2)-(x+4) ?
The integral from-4 to 2 of (12-x-x^2)-(x+4) is 36