What is the integral from 0 to 1 of 2^{x+1} ?

The integral from 0 to 1 of 2^{x+1} is 2/(ln(2))

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integral from 0 to 1 of 2^{x+1}

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

\frac{2}{ln ( 2 )}

Decimal Notation

2.88539 \dots

Solution steps

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

2^{x + 1} = 2^{1} \cdot 2^{x}

= \int_{0}^{1} 2^{1} \cdot 2^{x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2^{1} \cdot \int_{0}^{1} 2^{x} d x

Simplify

= 2 \cdot \int_{0}^{1} 2^{x} d x

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

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

Compute the boundaries:

\frac{1}{ln ( 2 )}

= 2 \cdot \frac{1}{ln ( 2 )}

Simplify

= \frac{2}{ln ( 2 )}

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What is the integral from 0 to 1 of 2^{x+1} ?
The integral from 0 to 1 of 2^{x+1} is 2/(ln(2))