What is the integral from 0 to 1 of (x^3+1)/9 ?

The integral from 0 to 1 of (x^3+1)/9 is 5/36

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integral from 0 to 1 of (x^3+1)/9

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

\frac{5}{36}

Decimal Notation

0.13888 \dots

Solution steps

\int_{0}^{1} \frac{x ^{3} + 1}{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}^{1} x^{3} + 1 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

= \frac{1}{9} (\int_{0}^{1} x^{3} d x + \int_{0}^{1} 1 d x)

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

\int_{0}^{1} 1 d x = 1

= \frac{1}{9} (\frac{1}{4} + 1)

Simplify

\frac{1}{9} (\frac{1}{4} + 1) : \frac{5}{36}

= \frac{5}{36}

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What is the integral from 0 to 1 of (x^3+1)/9 ?
The integral from 0 to 1 of (x^3+1)/9 is 5/36