What is the integral from 0 to 1 of x4^x ?

The integral from 0 to 1 of x4^x is (8ln(2)-3)/(4ln^2(2))

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integral from 0 to 1 of x4^x

\int_{0}^{1} x 4^{x} d x

\frac{8 ln ( 2 ) - 3}{4 ln ^{2} ( 2 )}

Decimal Notation

1.32436 \dots

Solution steps

\int_{0}^{1} x \cdot 4^{x} d x

Apply Integration By Parts

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

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

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

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

Compute the boundaries:

\frac{8 ln ( 2 ) - 3}{4 ln ^{2} ( 2 )}

= \frac{8 ln ( 2 ) - 3}{4 ln ^{2} ( 2 )}

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

\int_{3}^{12} x - 3 d x

\int_{0}^{2} \frac{24}{x ^{2} + 4} d x

\int_{0}^{\infty} 25 e^{- 5 x} d x

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

What is the integral from 0 to 1 of x4^x ?
The integral from 0 to 1 of x4^x is (8ln(2)-3)/(4ln^2(2))