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

The integral from 0 to 1 of x^2ln(x) is -1/9

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integral from 0 to 1 of x^2ln(x)

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

- \frac{1}{9}

Decimal Notation

- 0.11111 \dots

Solution steps

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

Apply Integration By Parts

: [\frac{1}{3} x^{3} ln (x) - \int \frac{x ^{2}}{3} d x]_{0}^{1}

= [\frac{1}{3} x^{3} ln (x) - \int \frac{x ^{2}}{3} d x]_{0}^{1}

\int \frac{x ^{2}}{3} d x = \frac{x ^{3}}{9}

= [\frac{1}{3} x^{3} ln (x) - \frac{x ^{3}}{9}]_{0}^{1}

Compute the boundaries:

- \frac{1}{9}

= - \frac{1}{9}

\frac{d}{d x} (\frac{1}{( 1 - x ^{6} ) ^{\frac{3}{2}}})

\int_{0}^{N} e^{- s t} e^{- 3 t} d t

y^{^{'}} - (4 x - y + 1)^{2} = 0

\frac{d}{d x} (\frac{1}{0})

\int sin (2) d x

What is the integral from 0 to 1 of x^2ln(x) ?
The integral from 0 to 1 of x^2ln(x) is -1/9