What is the integral from 1 to 4 of 3^{x-1} ?

The integral from 1 to 4 of 3^{x-1} is (26)/(ln(3))

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integral from 1 to 4 of 3^{x-1}

\int_{1}^{4} 3^{x - 1} d x

\frac{26}{ln ( 3 )}

Decimal Notation

23.66621 \dots

Solution steps

\int_{1}^{4} 3^{x - 1} d x

Simplify

3^{x - 1} : 3^{x} \cdot 3^{- 1}

= \int_{1}^{4} 3^{x} \cdot 3^{- 1} d x

Take the constant out:

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

= 3^{- 1} \cdot \int_{1}^{4} 3^{x} d x

Apply exponent rule:

a^{- 1} = \frac{1}{a}

= \frac{1}{3} \cdot \int_{1}^{4} 3^{x} d x

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

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

Compute the boundaries:

\frac{78}{ln ( 3 )}

= \frac{1}{3} \cdot \frac{78}{ln ( 3 )}

Simplify

= \frac{26}{ln ( 3 )}

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What is the integral from 1 to 4 of 3^{x-1} ?
The integral from 1 to 4 of 3^{x-1} is (26)/(ln(3))