What is the integral from 3 to 4 of f(2^x)2^x ?

The integral from 3 to 4 of f(2^x)2^x is (96f)/(ln(2))

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integral from 3 to 4 of f(2^x)2^x

\int_{3}^{4} f (2^{x}) 2^{x} d x

\frac{96 f}{ln ( 2 )}

Solution steps

\int_{3}^{4} f \cdot 2^{x} \cdot 2^{x} d x

Take the constant out:

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

= f \cdot \int_{3}^{4} 2^{x} \cdot 2^{x} d x

Simplify

2^{x} \cdot 2^{x} : 4^{x}

= f \cdot \int_{3}^{4} 4^{x} d x

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

= f [\frac{4 ^{x}}{ln ( 4 )}]_{3}^{4}

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

= f [\frac{2 ^{2 x - 1}}{ln ( 2 )}]_{3}^{4}

Compute the boundaries:

\frac{96}{ln ( 2 )}

= f \frac{96}{ln ( 2 )}

Simplify

= \frac{96 f}{ln ( 2 )}

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What is the integral from 3 to 4 of f(2^x)2^x ?
The integral from 3 to 4 of f(2^x)2^x is (96f)/(ln(2))