Answer:
18 ln(9) − ¹⁰⁴/₉
Explanation:
∫₁⁹ √y ln(y) dy
If u = ln(y), then du = dy/y.
If dv = √y dy, then v = ⅔ (y)^³/₂.
∫ u dv = uv − ∫ v du
= ln(y) (⅔ (y)^³/₂) − ∫ (⅔ (y)^³/₂) (dy/y)
= ln(y) (⅔ (y)^³/₂) − ∫ (⅔ (y)^½) dy
= ln(y) (⅔ (y)^³/₂) − ⁴/₉ (y)^³/₂ + C
= (⅔ (y)^³/₂) (ln(y) − ⅔) + C
Evaluate between y=1 and y=9.
[(⅔ (9)^³/₂) (ln(9) − ⅔)] − [(⅔ (1)^³/₂) (ln(1) − ⅔)]
[18 (ln(9) − ⅔)] − [(⅔) (-⅔)]
18 ln(9) − 12 + ⁴/₉
18 ln(9) − ¹⁰⁴/₉