What is the integral from 0 to 1 of y^x ?

The integral from 0 to 1 of y^x is (y-1)/(ln(y))

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integral from 0 to 1 of y^x

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

\frac{y - 1}{ln ( y )}

Solution steps

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

\int a^{x} d x = \frac{a ^{x}}{ln a}

= [\frac{y ^{x}}{ln ( y )}]_{0}^{1}

Compute the boundaries:

\frac{y}{ln ( y )} - \frac{1}{ln ( y )}

= \frac{y}{ln ( y )} - \frac{1}{ln ( y )}

Simplify

= \frac{y - 1}{ln ( y )}

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What is the integral from 0 to 1 of y^x ?
The integral from 0 to 1 of y^x is (y-1)/(ln(y))