What is the integral from 0 to 1 of 6^{2x} ?

The integral from 0 to 1 of 6^{2x} is (35)/(2ln(6))

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integral from 0 to 1 of 6^{2x}

\int_{0}^{1} 6^{2 x} d x

\frac{35}{2 ln ( 6 )}

Decimal Notation

9.76693 \dots

Solution steps

\int_{0}^{1} 6^{2 x} d x

Simplify

6^{2 x} : 3 6^{x}

= \int_{0}^{1} 3 6^{x} d x

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

= [\frac{3 6 ^{x}}{ln ( 36 )}]_{0}^{1}

\frac{3 6 ^{x}}{ln ( 36 )} = \frac{9 ^{x} \cdot 2 ^{2 x - 1}}{ln ( 6 )}

= [\frac{9 ^{x} \cdot 2 ^{2 x - 1}}{ln ( 6 )}]_{0}^{1}

Compute the boundaries:

\frac{35}{2 ln ( 6 )}

= \frac{35}{2 ln ( 6 )}

\frac{\partial}{\partial x} (12 x e^{x - 3 y})

x \to 0 lim ((x + 1) sin (\frac{1}{x}))

\frac{d}{d x} (cos^{12} (x))

5 t + \frac{7}{t}

\int \frac{1}{3 ^{- x}} d x

What is the integral from 0 to 1 of 6^{2x} ?
The integral from 0 to 1 of 6^{2x} is (35)/(2ln(6))