What is the integral from 0 to 3 of (x^4)/(81) ?

The integral from 0 to 3 of (x^4)/(81) is 3/5

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integral from 0 to 3 of (x^4)/(81)

\int_{0}^{3} \frac{x ^{4}}{81} d x

\frac{3}{5}

Decimal Notation

0.6

Solution steps

\int_{0}^{3} \frac{x ^{4}}{81} d x

Take the constant out:

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

= \frac{1}{81} \cdot \int_{0}^{3} x^{4} d x

Apply the Power Rule

: [\frac{x ^{5}}{5}]_{0}^{3}

= \frac{1}{81} [\frac{x ^{5}}{5}]_{0}^{3}

Compute the boundaries:

\frac{243}{5}

= \frac{1}{81} \cdot \frac{243}{5}

Simplify

= \frac{3}{5}

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What is the integral from 0 to 3 of (x^4)/(81) ?
The integral from 0 to 3 of (x^4)/(81) is 3/5