What is the integral from 1 to 3 of (4^{x+1}+2^x) ?

The integral from 1 to 3 of (4^{x+1}+2^x) is (126)/(ln(2))

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integral from 1 to 3 of (4^{x+1}+2^x)

\int_{1}^{3} (4^{x + 1} + 2^{x}) d x

\frac{126}{ln ( 2 )}

Decimal Notation

181.77957 \dots

Solution steps

\int_{1}^{3} 4^{x + 1} + 2^{x} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{3} 4^{x + 1} d x + \int_{1}^{3} 2^{x} d x

\int_{1}^{3} 4^{x + 1} d x = \frac{120}{ln ( 2 )}

\int_{1}^{3} 2^{x} d x = \frac{6}{ln ( 2 )}

= \frac{120}{ln ( 2 )} + \frac{6}{ln ( 2 )}

Simplify

\frac{120}{ln ( 2 )} + \frac{6}{ln ( 2 )} : \frac{126}{ln ( 2 )}

= \frac{126}{ln ( 2 )}

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What is the integral from 1 to 3 of (4^{x+1}+2^x) ?
The integral from 1 to 3 of (4^{x+1}+2^x) is (126)/(ln(2))