What is the integral from-1 to 1 of 8^x ?

The integral from-1 to 1 of 8^x is (21)/(8ln(2))

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integral from-1 to 1 of 8^x

\int_{- 1}^{1} 8^{x} d x

\frac{21}{8 ln ( 2 )}

Decimal Notation

3.78707 \dots

Solution steps

\int_{- 1}^{1} 8^{x} d x

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

= [\frac{8 ^{x}}{ln ( 8 )}]_{- 1}^{1}

Simplify

ln (8) : 3 ln (2)

= [\frac{8 ^{x}}{3 ln ( 2 )}]_{- 1}^{1}

Compute the boundaries:

\frac{21}{8 ln ( 2 )}

= \frac{21}{8 ln ( 2 )}

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What is the integral from-1 to 1 of 8^x ?
The integral from-1 to 1 of 8^x is (21)/(8ln(2))