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

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

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

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

\frac{4}{3}

Decimal Notation

1.33333 \dots

Solution steps

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

\int_{0}^{2} 2 x^{2} d x = \frac{16}{3}

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

= \frac{16}{3} - 4

Simplify

\frac{16}{3} - 4 : \frac{4}{3}

= \frac{4}{3}

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