What is the integral from 0 to 1 of 72x^8(1-x) ?

The integral from 0 to 1 of 72x^8(1-x) is 4/5

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integral from 0 to 1 of 72x^8(1-x)

\int_{0}^{1} 72 x^{8} (1 - x) d x

\frac{4}{5}

Solution steps

\int_{0}^{1} 72 x^{8} (1 - x) d x

Take the constant out:

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

= 72 \cdot \int_{0}^{1} x^{8} (1 - x) d x

Apply Integration By Parts

: [\frac{1}{9} x^{9} (1 - x) - \int - \frac{x ^{9}}{9} d x]_{0}^{1}

= 72 [\frac{1}{9} x^{9} (1 - x) - \int - \frac{x ^{9}}{9} d x]_{0}^{1}

\int - \frac{x ^{9}}{9} d x = - \frac{x ^{10}}{90}

= 72 [\frac{1}{9} x^{9} (1 - x) - (- \frac{x ^{10}}{90})]_{0}^{1}

Simplify

72 [\frac{1}{9} x^{9} (1 - x) - (- \frac{x ^{10}}{90})]_{0}^{1} : 72 [\frac{1}{90} (10 x^{9} - 9 x^{10})]_{0}^{1}

= 72 [\frac{1}{90} (10 x^{9} - 9 x^{10})]_{0}^{1}

Compute the boundaries:

\frac{1}{90}

= 72 \cdot \frac{1}{90}

Simplify

= \frac{4}{5}

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What is the integral from 0 to 1 of 72x^8(1-x) ?
The integral from 0 to 1 of 72x^8(1-x) is 4/5