What is the integral from 1 to 12 of (0.5)^x ?

The integral from 1 to 12 of (0.5)^x is 0.72099…

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integral from 1 to 12 of (0.5)^x

\int_{1}^{12} (0.5)^{x} d x

0.72099 \dots

Solution steps

\int_{1}^{12} 0. 5^{x} d x

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

= [\frac{0. 5 ^{x}}{ln ( 0.5 )}]_{1}^{12}

\frac{0. 5 ^{x}}{ln ( 0.5 )} = - 1.44269 \dots \cdot 0. 5^{x}

= [- 1.44269 \dots \cdot 0. 5^{x}]_{1}^{12}

Compute the boundaries:

0.72099 \dots

= 0.72099 \dots

\frac{d}{d x} (\frac{1}{1 - cos ( x )})

\int \frac{2 x}{x ^{2} - 4 x + 13} d x

\frac{d}{d x} (x e^{x} cot (x))

\frac{d}{d x} (- 10 x + 3 cos (x))

\frac{d ^{2}}{d x ^{2}} ((2 x + 5)^{7})

What is the integral from 1 to 12 of (0.5)^x ?
The integral from 1 to 12 of (0.5)^x is 0.72099…