What is the integral from 0 to 1 of 4^x ?

The integral from 0 to 1 of 4^x is 3/(2ln(2))

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integral from 0 to 1 of 4^x

\int_{0}^{1} 4^{x} d x

\frac{3}{2 ln ( 2 )}

Decimal

2.16404 \dots

Solution steps

\int_{0}^{1} 4^{x} d x

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

= [\frac{4 ^{x}}{ln ( 4 )}]_{0}^{1}

\frac{4 ^{x}}{ln ( 4 )} = \frac{2 ^{2 x - 1}}{ln ( 2 )}

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

Compute the boundaries:

\frac{3}{2 ln ( 2 )}

= \frac{3}{2 ln ( 2 )}

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What is the integral from 0 to 1 of 4^x ?
The integral from 0 to 1 of 4^x is 3/(2ln(2))