What is the integral from 1 to n of (x^3)/(n^3) ?

The integral from 1 to n of (x^3)/(n^3) is (n^4-1)/(4n^3)

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integral from 1 to n of (x^3)/(n^3)

\int_{1}^{n} \frac{x ^{3}}{n ^{3}} d x

\frac{n ^{4} - 1}{4 n ^{3}}

Solution steps

\int_{1}^{n} \frac{x ^{3}}{n ^{3}} d x

Take the constant out:

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

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

Apply the Power Rule

: [\frac{x ^{4}}{4}]_{1}^{n}

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

Simplify

\frac{1}{n ^{3}} [\frac{x ^{4}}{4}]_{1}^{n} : \frac{[ \frac{x ^{4}}{4} ] _{1}^{n}}{n ^{3}}

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

Compute the boundaries:

\frac{n ^{4}}{4} - \frac{1}{4}

= \frac{\frac{n ^{4}}{4} - \frac{1}{4}}{n ^{3}}

Simplify

= \frac{n ^{4} - 1}{4 n ^{3}}

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What is the integral from 1 to n of (x^3)/(n^3) ?
The integral from 1 to n of (x^3)/(n^3) is (n^4-1)/(4n^3)