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

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

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

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

- \frac{1}{4}

Decimal

- 0.25

Solution steps

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

\int_{1}^{2} x^{3} d x = \frac{15}{4}

\int_{1}^{2} 3 x^{2} d x = 7

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

= \frac{15}{4} - 7 + 3

Simplify

\frac{15}{4} - 7 + 3 : - \frac{1}{4}

= - \frac{1}{4}

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