What is the integral from 1 to 2^n of x^{n-1} ?

The integral from 1 to 2^n of x^{n-1} is (2^{n^2}-1)/n

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integral from 1 to 2^n of x^{n-1}

\int_{1}^{2^{n}} x^{n - 1} d x

\frac{2 ^{n^{2}} - 1}{n}

Solution steps

\int_{1}^{2^{n}} x^{n - 1} d x

Apply the Power Rule

: [\frac{x ^{n}}{n}]_{1}^{2^{n}}

= [\frac{x ^{n}}{n}]_{1}^{2^{n}}

Compute the boundaries:

\frac{2 ^{n^{2}}}{n} - \frac{1}{n}

= \frac{2 ^{n^{2}}}{n} - \frac{1}{n}

Simplify

= \frac{2 ^{n^{2}} - 1}{n}

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What is the integral from 1 to 2^n of x^{n-1} ?
The integral from 1 to 2^n of x^{n-1} is (2^{n^2}-1)/n