What is the integral from 0 to 1 of 2^u2u ?

The integral from 0 to 1 of 2^u2u is (2(2ln(2)-1))/(ln^2(2))

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integral from 0 to 1 of 2^u2u

\int_{0}^{1} 2^{u} 2 u d u

\frac{2 ( 2 ln ( 2 ) - 1 )}{ln ^{2} ( 2 )}

Decimal Notation

1.60804 \dots

Solution steps

\int_{0}^{1} 2^{u} \cdot 2 u d u

Take the constant out:

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

= 2 \cdot \int_{0}^{1} 2^{u} u d u

Apply Integration By Parts

: [\frac{2 ^{u} u}{ln ( 2 )} - \int \frac{2 ^{u}}{ln ( 2 )} d u]_{0}^{1}

= 2 [\frac{2 ^{u} u}{ln ( 2 )} - \int \frac{2 ^{u}}{ln ( 2 )} d u]_{0}^{1}

\int \frac{2 ^{u}}{ln ( 2 )} d u = \frac{2 ^{u}}{ln ^{2} ( 2 )}

= 2 [\frac{2 ^{u} u}{ln ( 2 )} - \frac{2 ^{u}}{ln ^{2} ( 2 )}]_{0}^{1}

Compute the boundaries:

\frac{2 ln ( 2 ) - 1}{ln ^{2} ( 2 )}

= 2 \cdot \frac{2 ln ( 2 ) - 1}{ln ^{2} ( 2 )}

Simplify

= \frac{2 ( 2 ln ( 2 ) - 1 )}{ln ^{2} ( 2 )}

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What is the integral from 0 to 1 of 2^u2u ?
The integral from 0 to 1 of 2^u2u is (2(2ln(2)-1))/(ln^2(2))