What is the integral from 0 to 1 of 2^xe^x ?

The integral from 0 to 1 of 2^xe^x is (2e-1)/(ln(2e))

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integral from 0 to 1 of 2^xe^x

\int_{0}^{1} 2^{x} e^{x} d x

\frac{2 e - 1}{ln ( 2 e )}

Decimal

2.62030 \dots

Solution steps

\int_{0}^{1} 2^{x} e^{x} d x

Apply exponent rule:

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

2^{x} e^{x} = (2 e)^{x}

= \int_{0}^{1} (2 e)^{x} d x

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

= [\frac{( 2 e ) ^{x}}{ln ( 2 e )}]_{0}^{1}

Compute the boundaries:

\frac{2 e - 1}{ln ( 2 e )}

= \frac{2 e - 1}{ln ( 2 e )}

\int_{- 1}^{0} 2 x - 1 d x

\int_{1}^{2} \frac{2 x}{( x ^{2} + 2 ) ^{3}} d x

\int_{2}^{5} 4^{2 x} d x

\int_{0}^{12} 2000 - 12 x^{2} d x

\int_{3}^{4} 4 x^{5} d x

What is the integral from 0 to 1 of 2^xe^x ?
The integral from 0 to 1 of 2^xe^x is (2e-1)/(ln(2e))