What is the integral from 0 to 1 of x2^{-x} ?

The integral from 0 to 1 of x2^{-x} is (-ln(2)+1)/(2ln^2(2))

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integral from 0 to 1 of x2^{-x}

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Compute the boundaries:

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

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

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\int_{- \frac{π}{2}}^{\frac{π}{2}} 2 t + cos (t) d t

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\int_{0}^{2} 2 t sin (4 π t) d t

What is the integral from 0 to 1 of x2^{-x} ?
The integral from 0 to 1 of x2^{-x} is (-ln(2)+1)/(2ln^2(2))