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

The integral from 1 to 3 of 3(2^{x+1}) is (36)/(ln(2))

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

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

\frac{36}{ln ( 2 )}

Decimal

51.93702 \dots

Solution steps

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

Take the constant out:

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

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

2^{x + 1} = 2^{1} \cdot 2^{x}

= 3 \cdot \int_{1}^{3} 2^{1} \cdot 2^{x} d x

Take the constant out:

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

= 3 \cdot 2^{1} \cdot \int_{1}^{3} 2^{x} d x

Simplify

= 3 \cdot 2 \cdot \int_{1}^{3} 2^{x} d x

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

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

Simplify

= 6 [\frac{2 ^{x}}{ln ( 2 )}]_{1}^{3}

Compute the boundaries:

\frac{6}{ln ( 2 )}

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

Simplify

= \frac{36}{ln ( 2 )}

What is the integral from 1 to 3 of 3(2^{x+1}) ?
The integral from 1 to 3 of 3(2^{x+1}) is (36)/(ln(2))