What is the integral from 0 to 1 of 3^xe^x ?

The integral from 0 to 1 of 3^xe^x is (3e-1)/(ln(3e))

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integral from 0 to 1 of 3^xe^x

\int_{0}^{1} 3^{x} e^{x} d x

\frac{3 e - 1}{ln ( 3 e )}

Decimal Notation

3.40932 \dots

Solution steps

\int_{0}^{1} 3^{x} e^{x} d x

Apply exponent rule:

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

3^{x} e^{x} = (3 e)^{x}

= \int_{0}^{1} (3 e)^{x} d x

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

= [\frac{( 3 e ) ^{x}}{ln ( 3 e )}]_{0}^{1}

Compute the boundaries:

\frac{3 e - 1}{ln ( 3 e )}

= \frac{3 e - 1}{ln ( 3 e )}

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What is the integral from 0 to 1 of 3^xe^x ?
The integral from 0 to 1 of 3^xe^x is (3e-1)/(ln(3e))