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

The integral from 0 to 1 of (2x^3-2x^2-4x) is -13/6

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integral from 0 to 1 of (2x^3-2x^2-4x)

\int_{0}^{1} (2 x^{3} - 2 x^{2} - 4 x) d x

- \frac{13}{6}

Decimal

- 2.16666 \dots

Solution steps

\int_{0}^{1} 2 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_{0}^{1} 2 x^{3} d x - \int_{0}^{1} 2 x^{2} d x - \int_{0}^{1} 4 x d x

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

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

\int_{0}^{1} 4 x d x = 2

= \frac{1}{2} - \frac{2}{3} - 2

Simplify

\frac{1}{2} - \frac{2}{3} - 2 : - \frac{13}{6}

= - \frac{13}{6}

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What is the integral from 0 to 1 of (2x^3-2x^2-4x) ?
The integral from 0 to 1 of (2x^3-2x^2-4x) is -13/6