What is the integral from 0 to 1 of (x^3-3x^2+2x) ?

The integral from 0 to 1 of (x^3-3x^2+2x) is 1/4

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integral from 0 to 1 of (x^3-3x^2+2x)

\int_{0}^{1} (x^{3} - 3 x^{2} + 2 x) d x

\frac{1}{4}

Decimal

0.25

Solution steps

\int_{0}^{1} x^{3} - 3 x^{2} + 2 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}^{1} x^{3} d x - \int_{0}^{1} 3 x^{2} d x + \int_{0}^{1} 2 x d x

\int_{0}^{1} x^{3} d x = \frac{1}{4}

\int_{0}^{1} 3 x^{2} d x = 1

\int_{0}^{1} 2 x d x = 1

= \frac{1}{4} - 1 + 1

Simplify

= \frac{1}{4}

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What is the integral from 0 to 1 of (x^3-3x^2+2x) ?
The integral from 0 to 1 of (x^3-3x^2+2x) is 1/4