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

The integral from 0 to 1 of 4^{4x} is (255)/(8ln(2))

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

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

\frac{255}{8 ln ( 2 )}

Decimal Notation

45.98590 \dots

Solution steps

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

Simplify

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

= \int_{0}^{1} 25 6^{x} d x

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

= [\frac{25 6 ^{x}}{ln ( 256 )}]_{0}^{1}

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

= [\frac{2 ^{8 x - 3}}{ln ( 2 )}]_{0}^{1}

Compute the boundaries:

\frac{255}{8 ln ( 2 )}

= \frac{255}{8 ln ( 2 )}

\int \frac{cos ( x )}{sin ^{2} ( x ) - 64} d x

\int cos (2 π) d x

\frac{d}{d x} (\frac{cos ( 6 x )}{e ^{x}})

y = arctan (x)

x^{2} y^{^{'}} = 2 x^{2} + x y + 2 y^{2}

What is the integral from 0 to 1 of 4^{4x} ?
The integral from 0 to 1 of 4^{4x} is (255)/(8ln(2))