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

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

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integral from 0 to 1 of (2^x)/(ln(2))

\int_{0}^{1} \frac{2 ^{x}}{ln ( 2 )} d x

\frac{1}{ln ^{2} ( 2 )}

Decimal

2.08136 \dots

Solution steps

\int_{0}^{1} \frac{2 ^{x}}{ln ( 2 )} d x

Take the constant out:

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

= \frac{1}{ln ( 2 )} \cdot \int_{0}^{1} 2^{x} d x

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

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

Compute the boundaries:

\frac{1}{ln ( 2 )}

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

Simplify

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

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