What is the integral from 0 to 1 of 9x^8(9-x^9) ?

The integral from 0 to 1 of 9x^8(9-x^9) is 17/2

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integral from 0 to 1 of 9x^8(9-x^9)

\int_{0}^{1} 9 x^{8} (9 - x^{9}) d x

\frac{17}{2}

Decimal

8.5

Solution steps

\int_{0}^{1} 9 x^{8} (9 - x^{9}) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 9 \cdot \int_{0}^{1} x^{8} (9 - x^{9}) d x

Expand

x^{8} (9 - x^{9}) : 9 x^{8} - x^{17}

= 9 \cdot \int_{0}^{1} 9 x^{8} - x^{17} 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

= 9 (\int_{0}^{1} 9 x^{8} d x - \int_{0}^{1} x^{17} d x)

\int_{0}^{1} 9 x^{8} d x = 1

\int_{0}^{1} x^{17} d x = \frac{1}{18}

= 9 (1 - \frac{1}{18})

Simplify

9 (1 - \frac{1}{18}) : \frac{17}{2}

= \frac{17}{2}

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What is the integral from 0 to 1 of 9x^8(9-x^9) ?
The integral from 0 to 1 of 9x^8(9-x^9) is 17/2