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

The integral from 0 to 1 of (2x^3-5x+9) is 7

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

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

7

Solution steps

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

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

\int_{0}^{1} 5 x d x = \frac{5}{2}

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

= \frac{1}{2} - \frac{5}{2} + 9

Simplify

\frac{1}{2} - \frac{5}{2} + 9 : 7

= 7

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What is the integral from 0 to 1 of (2x^3-5x+9) ?
The integral from 0 to 1 of (2x^3-5x+9) is 7