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

The integral from 0 to 3 of x^3-6x^2+8x is 9/4

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

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

\frac{9}{4}

Solution steps

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

\int_{0}^{3} x^{3} d x = \frac{81}{4}

\int_{0}^{3} 6 x^{2} d x = 54

\int_{0}^{3} 8 x d x = 36

= \frac{81}{4} - 54 + 36

Simplify

\frac{81}{4} - 54 + 36 : \frac{9}{4}

= \frac{9}{4}

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What is the integral from 0 to 3 of x^3-6x^2+8x ?
The integral from 0 to 3 of x^3-6x^2+8x is 9/4