What is the integral from 1 to 9 of y^{1/2}ln(y) ?

The integral from 1 to 9 of y^{1/2}ln(y) is 36ln(3)-104/9

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integral from 1 to 9 of y^{1/2}ln(y)

\int_{1}^{9} y^{\frac{1}{2}} ln (y) d y

36 ln (3) - \frac{104}{9}

Decimal

27.99448 \dots

Solution steps

\int_{1}^{9} y^{\frac{1}{2}} ln (y) d y

Apply Integration By Parts

: [\frac{2}{3} y^{\frac{3}{2}} ln (y) - \int \frac{2}{3} y^{\frac{1}{2}} d y]_{1}^{9}

= [\frac{2}{3} y^{\frac{3}{2}} ln (y) - \int \frac{2}{3} y^{\frac{1}{2}} d y]_{1}^{9}

\int \frac{2}{3} y^{\frac{1}{2}} d y = \frac{4}{9} y^{\frac{3}{2}}

= [\frac{2}{3} y^{\frac{3}{2}} ln (y) - \frac{4}{9} y^{\frac{3}{2}}]_{1}^{9}

Compute the boundaries:

36 ln (3) - \frac{104}{9}

= 36 ln (3) - \frac{104}{9}

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What is the integral from 1 to 9 of y^{1/2}ln(y) ?
The integral from 1 to 9 of y^{1/2}ln(y) is 36ln(3)-104/9