What is the integral from-2 to 6 of 4x-x^2+12 ?

The integral from-2 to 6 of 4x-x^2+12 is 256/3

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integral from-2 to 6 of 4x-x^2+12

\int_{- 2}^{6} 4 x - x^{2} + 12 d x

\frac{256}{3}

Decimal Notation

85.33333 \dots

Solution steps

\int_{- 2}^{6} 4 x - x^{2} + 12 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}^{6} 4 x d x - \int_{- 2}^{6} x^{2} d x + \int_{- 2}^{6} 12 d x

\int_{- 2}^{6} 4 x d x = 64

\int_{- 2}^{6} x^{2} d x = \frac{224}{3}

\int_{- 2}^{6} 12 d x = 96

= 64 - \frac{224}{3} + 96

Simplify

64 - \frac{224}{3} + 96 : \frac{256}{3}

= \frac{256}{3}

\int_{5}^{7} [\frac{k}{2} x^{2}] d x

y^{^{''}} + 5 y^{^{'}} = e^{- t}, y (0) = 0, y^{^{'}} (0) = 1

\frac{\partial}{\partial y} (e^{x y} ln (z))

\int \frac{( ln ( 2 x ) )}{x} d x

x \to 4 lim (f (x))

What is the integral from-2 to 6 of 4x-x^2+12 ?
The integral from-2 to 6 of 4x-x^2+12 is 256/3