What is the integral from 0 to 2 of (x^3-5x^2+6x) ?

The integral from 0 to 2 of (x^3-5x^2+6x) is 8/3

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integral from 0 to 2 of (x^3-5x^2+6x)

\int_{0}^{2} (x^{3} - 5 x^{2} + 6 x) d x

\frac{8}{3}

Decimal

2.66666 \dots

Solution steps

\int_{0}^{2} x^{3} - 5 x^{2} + 6 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} 5 x^{2} d x + \int_{0}^{2} 6 x d x

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

\int_{0}^{2} 5 x^{2} d x = \frac{40}{3}

\int_{0}^{2} 6 x d x = 12

= 4 - \frac{40}{3} + 12

Simplify

4 - \frac{40}{3} + 12 : \frac{8}{3}

= \frac{8}{3}

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What is the integral from 0 to 2 of (x^3-5x^2+6x) ?
The integral from 0 to 2 of (x^3-5x^2+6x) is 8/3