Answer
$\displaystyle \frac{8}{3}\ln(2)-\frac{7}{9} ≈ 1.070614$
Work Step by Step
$\displaystyle \int_1^2w^2\ln(w)dw$
______________________________
Let $f = \ln(w)$ and $g' = w^2$
thus $\displaystyle f' = \frac{1}{w}$ and $\displaystyle g = \frac{w^3}{3}$
______________________________
$\displaystyle \int_1^2w^2\ln(w)dw = [\frac{w^3}{3}\ln(w)]_1^2 - \int_1^2\frac{w^3}{3}(\frac{1}{w})dw$
$\displaystyle [\frac{8}{3}\ln(2)-\frac{1}{3}\ln(1)] - \int_1^2\frac{w^2}{3}dw$
$\displaystyle [\frac{8}{3}\ln(2)-0]-\frac{w^3}{9}|_1^2$
$\displaystyle \frac{8}{3}\ln(2)-(\frac{8}{9}-\frac{1}{9})$
$\displaystyle \frac{8}{3}\ln(2)-\frac{7}{9} ≈ 1.070614$
