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

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

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

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

\frac{6}{ln ( 2 )}

Decimal

8.65617 \dots

Solution steps

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

Simplify

2^{2 x} : 4^{x}

= \int_{1}^{2} 4^{x} d x

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

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

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

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

Compute the boundaries:

\frac{6}{ln ( 2 )}

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

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