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

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

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

\int_{- 1}^{1} (x^{3} - 2 x^{2} + 4 x) d x

- \frac{4}{3}

Decimal

- 1.33333 \dots

Solution steps

\int_{- 1}^{1} x^{3} - 2 x^{2} + 4 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}^{1} x^{3} d x - \int_{- 1}^{1} 2 x^{2} d x + \int_{- 1}^{1} 4 x d x

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

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

\int_{- 1}^{1} 4 x d x = 0

= 0 - \frac{4}{3} + 0

Simplify

= - \frac{4}{3}

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