What is the integral from 0 to 1 of 2^{2x+1} ?

The integral from 0 to 1 of 2^{2x+1} is 3/(ln(2))

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integral from 0 to 1 of 2^{2x+1}

\int_{0}^{1} 2^{2 x + 1} d x

\frac{3}{ln ( 2 )}

Decimal

4.32808 \dots

Solution steps

\int_{0}^{1} 2^{2 x + 1} d x

Apply u-substitution

: \int_{1}^{3} 2^{u - 1} d u

= \int_{1}^{3} 2^{u - 1} d u

Simplify

2^{u - 1} : 2^{u} \cdot 2^{- 1}

= \int_{1}^{3} 2^{u} \cdot 2^{- 1} d u

Take the constant out:

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

= 2^{- 1} \cdot \int_{1}^{3} 2^{u} d u

Apply exponent rule:

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

= \frac{1}{2} \cdot \int_{1}^{3} 2^{u} d u

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

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

Compute the boundaries:

\frac{6}{ln ( 2 )}

= \frac{1}{2} \cdot \frac{6}{ln ( 2 )}

Simplify

= \frac{3}{ln ( 2 )}

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What is the integral from 0 to 1 of 2^{2x+1} ?
The integral from 0 to 1 of 2^{2x+1} is 3/(ln(2))