What is the integral from 0 to 1 of 6x^5(9-x^6) ?

The integral from 0 to 1 of 6x^5(9-x^6) is 17/2

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integral from 0 to 1 of 6x^5(9-x^6)

\int_{0}^{1} 6 x^{5} (9 - x^{6}) d x

\frac{17}{2}

Decimal

8.5

Solution steps

\int_{0}^{1} 6 x^{5} (9 - x^{6}) d x

Take the constant out:

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

= 6 \cdot \int_{0}^{1} x^{5} (9 - x^{6}) d x

Expand

x^{5} (9 - x^{6}) : 9 x^{5} - x^{11}

= 6 \cdot \int_{0}^{1} 9 x^{5} - x^{11} 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

= 6 (\int_{0}^{1} 9 x^{5} d x - \int_{0}^{1} x^{11} d x)

\int_{0}^{1} 9 x^{5} d x = \frac{3}{2}

\int_{0}^{1} x^{11} d x = \frac{1}{12}

= 6 (\frac{3}{2} - \frac{1}{12})

Simplify

6 (\frac{3}{2} - \frac{1}{12}) : \frac{17}{2}

= \frac{17}{2}

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What is the integral from 0 to 1 of 6x^5(9-x^6) ?
The integral from 0 to 1 of 6x^5(9-x^6) is 17/2