What is the integral from 0 to 1 of (-5x^3-4x^2+9x) ?

The integral from 0 to 1 of (-5x^3-4x^2+9x) is 23/12

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integral from 0 to 1 of (-5x^3-4x^2+9x)

\int_{0}^{1} (- 5 x^{3} - 4 x^{2} + 9 x) d x

\frac{23}{12}

Decimal

1.91666 \dots

Solution steps

\int_{0}^{1} - 5 x^{3} - 4 x^{2} + 9 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} 5 x^{3} d x - \int_{0}^{1} 4 x^{2} d x + \int_{0}^{1} 9 x d x

\int_{0}^{1} 5 x^{3} d x = \frac{5}{4}

\int_{0}^{1} 4 x^{2} d x = \frac{4}{3}

\int_{0}^{1} 9 x d x = \frac{9}{2}

= - \frac{5}{4} - \frac{4}{3} + \frac{9}{2}

Simplify

- \frac{5}{4} - \frac{4}{3} + \frac{9}{2} : \frac{23}{12}

= \frac{23}{12}

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What is the integral from 0 to 1 of (-5x^3-4x^2+9x) ?
The integral from 0 to 1 of (-5x^3-4x^2+9x) is 23/12