What is the integral from 0 to 3 of (x^3)/9 ?

The integral from 0 to 3 of (x^3)/9 is 9/4

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integral from 0 to 3 of (x^3)/9

\int_{0}^{3} \frac{x ^{3}}{9} d x

\frac{9}{4}

Decimal Notation

2.25

Solution steps

\int_{0}^{3} \frac{x ^{3}}{9} d x

Take the constant out:

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

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

Apply the Power Rule

: [\frac{x ^{4}}{4}]_{0}^{3}

= \frac{1}{9} [\frac{x ^{4}}{4}]_{0}^{3}

Compute the boundaries:

\frac{81}{4}

= \frac{1}{9} \cdot \frac{81}{4}

Simplify

= \frac{9}{4}

\int_{0}^{2} (2 - x^{2}) d x

\int_{1}^{2} x (x^{2} - 1)^{3} d x

\int_{0}^{1} (3 - 3 x^{2}) d x

\int_{1}^{4} 3 x^{2} - 2 x d x

\int_{- 3}^{4} 5 d x

What is the integral from 0 to 3 of (x^3)/9 ?
The integral from 0 to 3 of (x^3)/9 is 9/4