What is the integral from 0 to 2 of 4^{3x} ?

The integral from 0 to 2 of 4^{3x} is (1365)/(2ln(2))

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integral from 0 to 2 of 4^{3x}

\int_{0}^{2} 4^{3 x} d x

\frac{1365}{2 ln ( 2 )}

Decimal Notation

984.63936 \dots

Solution steps

\int_{0}^{2} 4^{3 x} d x

Simplify

4^{3 x} : 6 4^{x}

= \int_{0}^{2} 6 4^{x} d x

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

= [\frac{6 4 ^{x}}{ln ( 64 )}]_{0}^{2}

\frac{6 4 ^{x}}{ln ( 64 )} = \frac{2 ^{6 x - 1}}{3 ln ( 2 )}

= [\frac{2 ^{6 x - 1}}{3 ln ( 2 )}]_{0}^{2}

Compute the boundaries:

\frac{1365}{2 ln ( 2 )}

= \frac{1365}{2 ln ( 2 )}

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

\int_{0}^{1} \frac{1}{x ^{q}} d x

\int_{0}^{2} 1 + (- 2 x)^{2} d x

\int_{0}^{2} (e^{x} - 1) d x

What is the integral from 0 to 2 of 4^{3x} ?
The integral from 0 to 2 of 4^{3x} is (1365)/(2ln(2))