What is the integral from 1 to 3 of 4^{4x} ?

The integral from 1 to 3 of 4^{4x} is (2097120)/(ln(2))

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integral from 1 to 3 of 4^{4x}

\int_{1}^{3} 4^{4 x} d x

\frac{2097120}{ln ( 2 )}

Decimal

3025504.62414 \dots

Solution steps

\int_{1}^{3} 4^{4 x} d x

Simplify

4^{4 x} : 25 6^{x}

= \int_{1}^{3} 25 6^{x} d x

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

= [\frac{25 6 ^{x}}{ln ( 256 )}]_{1}^{3}

\frac{25 6 ^{x}}{ln ( 256 )} = \frac{2 ^{8 x - 3}}{ln ( 2 )}

= [\frac{2 ^{8 x - 3}}{ln ( 2 )}]_{1}^{3}

Compute the boundaries:

\frac{2097120}{ln ( 2 )}

= \frac{2097120}{ln ( 2 )}

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What is the integral from 1 to 3 of 4^{4x} ?
The integral from 1 to 3 of 4^{4x} is (2097120)/(ln(2))