What is the integral from 0 to 2 of 38z^2ln(z) ?

The integral from 0 to 2 of 38z^2ln(z) is 38(8/3 ln(2)-8/9)

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integral from 0 to 2 of 38z^2ln(z)

\int_{0}^{2} 38 z^{2} ln (z) d z

38 (\frac{8}{3} ln (2) - \frac{8}{9})

Decimal Notation

36.46113 \dots

Solution steps

\int_{0}^{2} 38 z^{2} ln (z) d z

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 38 \cdot \int_{0}^{2} z^{2} ln (z) d z

Apply Integration By Parts

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

= 38 [\frac{1}{3} z^{3} ln (z) - \int \frac{z ^{2}}{3} d z]_{0}^{2}

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

= 38 [\frac{1}{3} z^{3} ln (z) - \frac{z ^{3}}{9}]_{0}^{2}

Compute the boundaries:

\frac{8}{3} ln (2) - \frac{8}{9}

= 38 (\frac{8}{3} ln (2) - \frac{8}{9})

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What is the integral from 0 to 2 of 38z^2ln(z) ?
The integral from 0 to 2 of 38z^2ln(z) is 38(8/3 ln(2)-8/9)