What is the integral from 5 to 6 of f(4^x)4^x ?

The integral from 5 to 6 of f(4^x)4^x is (3932160f)/(ln(2))

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integral from 5 to 6 of f(4^x)4^x

\int_{5}^{6} f (4^{x}) 4^{x} d x

\frac{3932160 f}{ln ( 2 )}

Solution steps

\int_{5}^{6} f \cdot 4^{x} \cdot 4^{x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= f \cdot \int_{5}^{6} 4^{x} \cdot 4^{x} d x

Simplify

4^{x} \cdot 4^{x} : 1 6^{x}

= f \cdot \int_{5}^{6} 1 6^{x} d x

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

= f [\frac{1 6 ^{x}}{ln ( 16 )}]_{5}^{6}

\frac{1 6 ^{x}}{ln ( 16 )} = \frac{2 ^{4 x - 2}}{ln ( 2 )}

= f [\frac{2 ^{4 x - 2}}{ln ( 2 )}]_{5}^{6}

Compute the boundaries:

\frac{3932160}{ln ( 2 )}

= f \frac{3932160}{ln ( 2 )}

Simplify

= \frac{3932160 f}{ln ( 2 )}

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What is the integral from 5 to 6 of f(4^x)4^x ?
The integral from 5 to 6 of f(4^x)4^x is (3932160f)/(ln(2))