What is the integral from 0 to 1 of 4^{2x} ?

The integral from 0 to 1 of 4^{2x} is (15)/(4ln(2))

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integral from 0 to 1 of 4^{2x}

\int_{0}^{1} 4^{2 x} d x

\frac{15}{4 ln ( 2 )}

Decimal

5.41010 \dots

Solution steps

\int_{0}^{1} 4^{2 x} d x

Simplify

4^{2 x} : 1 6^{x}

= \int_{0}^{1} 1 6^{x} d x

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

= [\frac{1 6 ^{x}}{ln ( 16 )}]_{0}^{1}

\frac{1 6 ^{x}}{ln ( 16 )} = \frac{2 ^{4 x - 2}}{ln ( 2 )}

= [\frac{2 ^{4 x - 2}}{ln ( 2 )}]_{0}^{1}

Compute the boundaries:

\frac{15}{4 ln ( 2 )}

= \frac{15}{4 ln ( 2 )}

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What is the integral from 0 to 1 of 4^{2x} ?
The integral from 0 to 1 of 4^{2x} is (15)/(4ln(2))