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

The integral from 0 to 2 of (-x^3+2x^2+x) is 10/3

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

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

\frac{10}{3}

Decimal

3.33333 \dots

Solution steps

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

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

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

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

= - 4 + \frac{16}{3} + 2

Simplify

- 4 + \frac{16}{3} + 2 : \frac{10}{3}

= \frac{10}{3}

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What is the integral from 0 to 2 of (-x^3+2x^2+x) ?
The integral from 0 to 2 of (-x^3+2x^2+x) is 10/3