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

The integral from 0 to 1 of 5x^4(9-x^5) is 17/2

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

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

\frac{17}{2}

Decimal

8.5

Solution steps

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

Take the constant out:

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

= 5 \cdot \int_{0}^{1} x^{4} (9 - x^{5}) d x

Expand

x^{4} (9 - x^{5}) : 9 x^{4} - x^{9}

= 5 \cdot \int_{0}^{1} 9 x^{4} - 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

= 5 (\int_{0}^{1} 9 x^{4} d x - \int_{0}^{1} x^{9} d x)

\int_{0}^{1} 9 x^{4} d x = \frac{9}{5}

\int_{0}^{1} x^{9} d x = \frac{1}{10}

= 5 (\frac{9}{5} - \frac{1}{10})

Simplify

5 (\frac{9}{5} - \frac{1}{10}) : \frac{17}{2}

= \frac{17}{2}

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What is the integral from 0 to 1 of 5x^4(9-x^5) ?
The integral from 0 to 1 of 5x^4(9-x^5) is 17/2