What is the integral from 0 to 6 of 8/729 (x+3)^3 ?

The integral from 0 to 6 of 8/729 (x+3)^3 is 160/9

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integral from 0 to 6 of 8/729 (x+3)^3

\int_{0}^{6} \frac{8}{729} (x + 3)^{3} d x

\frac{160}{9}

Solution steps

\int_{0}^{6} \frac{8}{729} (x + 3)^{3} d x

Take the constant out:

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

= \frac{8}{729} \cdot \int_{0}^{6} (x + 3)^{3} d x

Apply u-substitution

: \int_{3}^{9} u^{3} d u

= \frac{8}{729} \cdot \int_{3}^{9} u^{3} d u

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{3}^{9}

= \frac{8}{729} [\frac{u ^{4}}{4}]_{3}^{9}

Compute the boundaries:

1620

= \frac{8}{729} \cdot 1620

Simplify

= \frac{160}{9}

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What is the integral from 0 to 6 of 8/729 (x+3)^3 ?
The integral from 0 to 6 of 8/729 (x+3)^3 is 160/9