What is the integral from 0 to 1 of 8^{2x} ?

The integral from 0 to 1 of 8^{2x} is (21)/(2ln(2))

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integral from 0 to 1 of 8^{2x}

\int_{0}^{1} 8^{2 x} d x

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

Decimal Notation

15.14829 \dots

Solution steps

\int_{0}^{1} 8^{2 x} d x

Simplify

8^{2 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 )}

\frac{\partial}{\partial y} (x \cdot sin (x + y))

(x^{2} + 6) y^{^{'}} = x y

- 4 y^{^{''}} + 16 y = 0

\frac{d}{d x} (ln (x \cdot y))

\int x^{4} e^{x^{5}} d x

What is the integral from 0 to 1 of 8^{2x} ?
The integral from 0 to 1 of 8^{2x} is (21)/(2ln(2))