What is the integral from 0 to 1 of 6^{4x} ?

The integral from 0 to 1 of 6^{4x} is (1295)/(4ln(6))

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integral from 0 to 1 of 6^{4x}

\int_{0}^{1} 6^{4 x} d x

\frac{1295}{4 ln ( 6 )}

Decimal Notation

180.68831 \dots

Solution steps

\int_{0}^{1} 6^{4 x} d x

Simplify

6^{4 x} : 129 6^{x}

= \int_{0}^{1} 129 6^{x} d x

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

= [\frac{129 6 ^{x}}{ln ( 1296 )}]_{0}^{1}

\frac{129 6 ^{x}}{ln ( 1296 )} = \frac{8 1 ^{x} \cdot 2 ^{4 x - 2}}{ln ( 6 )}

= [\frac{8 1 ^{x} \cdot 2 ^{4 x - 2}}{ln ( 6 )}]_{0}^{1}

Compute the boundaries:

\frac{1295}{4 ln ( 6 )}

= \frac{1295}{4 ln ( 6 )}

y^{^{''}} - 3 y^{^{'}} = 18 t

\frac{d}{d t} (A e^{- 2 t} cos (t) + B e^{- 2 t} sin (t))

\frac{4 x}{x ^{2} + 2 x - 3}

2 e^{(5 t - 2)}

\frac{d}{d x} (\frac{1}{5 - 11 e ^{x}})

What is the integral from 0 to 1 of 6^{4x} ?
The integral from 0 to 1 of 6^{4x} is (1295)/(4ln(6))