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

The integral from 0 to 3 of (2x-x^2+x) is 9/2

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

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

\frac{9}{2}

Decimal Notation

4.5

Solution steps

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

Simplify

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

\int_{0}^{3} x^{2} d x = 9

\int_{0}^{3} 3 x d x = \frac{27}{2}

= - 9 + \frac{27}{2}

Simplify

- 9 + \frac{27}{2} : \frac{9}{2}

= \frac{9}{2}

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What is the integral from 0 to 3 of (2x-x^2+x) ?
The integral from 0 to 3 of (2x-x^2+x) is 9/2