What is the integral from 0 to 2 of (x^3-6x^2+8x) ?

The integral from 0 to 2 of (x^3-6x^2+8x) is 4

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integral from 0 to 2 of (x^3-6x^2+8x)

\int_{0}^{2} (x^{3} - 6 x^{2} + 8 x) d x

4

Solution steps

\int_{0}^{2} x^{3} - 6 x^{2} + 8 x 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_{0}^{2} x^{3} d x - \int_{0}^{2} 6 x^{2} d x + \int_{0}^{2} 8 x d x

\int_{0}^{2} x^{3} d x = 4

\int_{0}^{2} 6 x^{2} d x = 16

\int_{0}^{2} 8 x d x = 16

= 4 - 16 + 16

Simplify

= 4

\int_{0}^{1} 6 cos (x^{2}) d x

\frac{\partial}{\partial x} (4 x^{3} - 4 y)

\frac{d y}{d x} + \frac{3 y}{x} = 6 x^{2}

sin (2 t)

\frac{\partial}{\partial x} (x^{0.75})

What is the integral from 0 to 2 of (x^3-6x^2+8x) ?
The integral from 0 to 2 of (x^3-6x^2+8x) is 4