What is the integral from 0 to 1 of 4^{3x} ?

The integral from 0 to 1 of 4^{3x} is (21)/(2ln(2))

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integral from 0 to 1 of 4^{3x}

\int_{0}^{1} 4^{3 x} d x

\frac{21}{2 ln ( 2 )}

Decimal Notation

15.14829 \dots

Solution steps

\int_{0}^{1} 4^{3 x} d x

Simplify

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

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

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

= [\frac{6 4 ^{x}}{ln ( 64 )}]_{0}^{1}

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

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

Compute the boundaries:

\frac{21}{2 ln ( 2 )}

= \frac{21}{2 ln ( 2 )}

\int_{- 1}^{x} 1 - ∣ x ∣ d x

\int_{- 3}^{- 1} \frac{x ^{5} - 2 x}{x ^{3}} d x

\int_{0}^{2} x (\frac{3}{4} x (2 - x)) d x

\int_{y}^{y + 1} (x + y) d x

\int_{0}^{3} (x^{3}) d x

What is the integral from 0 to 1 of 4^{3x} ?
The integral from 0 to 1 of 4^{3x} is (21)/(2ln(2))