What is the integral from 0 to 1 of x^22^x ?

The integral from 0 to 1 of x^22^x is (2ln^2(2)-4ln(2)+2)/(ln^3(2))

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integral from 0 to 1 of x^22^x

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

\frac{2 ln ^{2} ( 2 ) - 4 ln ( 2 ) + 2}{ln ^{3} ( 2 )}

Decimal Notation

0.56547 \dots

Solution steps

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

Apply Integration By Parts

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

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

\int \frac{2 ^{x + 1} x}{ln ( 2 )} d x = \frac{1}{ln ( 2 )} (\frac{2 ^{x + 1} x}{ln ( 2 )} - \frac{1}{ln ^{2} ( 2 )} \cdot 2^{1 + x})

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

Compute the boundaries:

\frac{2 ln ^{2} ( 2 ) - 4 ln ( 2 ) + 2}{ln ^{3} ( 2 )}

= \frac{2 ln ^{2} ( 2 ) - 4 ln ( 2 ) + 2}{ln ^{3} ( 2 )}

\int_{0}^{1} x^{2} (x^{\frac{2}{3}}) d x

\int_{1}^{2} x^{2} - x d x

\int_{0}^{\frac{π}{6}} 2 cos (x) d x

\int_{0}^{4} 4 x + 5 d x

\int_{0}^{π} cos (6 x) d x

What is the integral from 0 to 1 of x^22^x ?
The integral from 0 to 1 of x^22^x is (2ln^2(2)-4ln(2)+2)/(ln^3(2))