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

The integral from 0 to 1 of a^xe^x is (ea-1)/(ln(ea))

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integral from 0 to 1 of a^xe^x

\int_{0}^{1} a^{x} e^{x} d x

\frac{e a - 1}{ln ( e a )}

Solution steps

\int_{0}^{1} a^{x} e^{x} d x

Apply exponent rule:

a^{n} \cdot b^{n} = (a \cdot b)^{n}

a^{x} e^{x} = (a e)^{x}

= \int_{0}^{1} (a e)^{x} d x

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

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

Compute the boundaries:

\frac{e a}{ln ( e a )} - \frac{1}{ln ( e a )}

= \frac{e a}{ln ( e a )} - \frac{1}{ln ( e a )}

Simplify

= \frac{e a - 1}{ln ( e a )}

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