What is the integral from-2 to 6 of 4y-(y^2-12) ?

The integral from-2 to 6 of 4y-(y^2-12) is 256/3

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integral from-2 to 6 of 4y-(y^2-12)

\int_{- 2}^{6} 4 y - (y^{2} - 12) d y

\frac{256}{3}

Solution steps

\int_{- 2}^{6} 4 y - (y^{2} - 12) d y

Simplify

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

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 y d y - \int_{- 2}^{6} y^{2} d y + \int_{- 2}^{6} 12 d y

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

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

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

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

Simplify

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

= \frac{256}{3}

\frac{\partial}{\partial y} (2 x + y)

\frac{\partial}{\partial x} (40 x y^{- 4} z)

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

\frac{d}{d x} ((5 x - 3)^{4} (2 x - 7)^{- 3})

(\frac{1}{2 x})^{^{'}}

What is the integral from-2 to 6 of 4y-(y^2-12) ?
The integral from-2 to 6 of 4y-(y^2-12) is 256/3