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

Simplify

= \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}

(y^{2} + y \cdot x) d x - x^{2} d y = 0

\frac{\partial}{\partial x} (\frac{x ^{3} \cdot y}{x ^{2} + y ^{2}})

2^{n}

\int \frac{cos ^{4} ( x )}{sin ( x )} d x

y^{^{'}} = x e^{x^{2} - 4}

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