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

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

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integral from 0 to 1 of x2^x

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

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

Decimal Notation

0.80402 \dots

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{2 ^{x}}{ln ^{2} ( 2 )}

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

Compute the boundaries:

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

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

\int_{0}^{l n (2)} 6 e^{2 x} d x

\int_{1.5}^{2} \frac{1}{2} d x

\int_{0}^{0.5} x d x

\int_{0}^{1} e^{y} d y

\int_{- 1}^{1} - 2 cos (\frac{π}{3} x) d x

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