How do you do this question?

18 ln(9) − ¹⁰⁴/₉

Step-by-step 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

  b.   the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation:

since the length of the interval is the same in each case, we only need to compare the change in y-value to see which is greater.

so the change in y-value for function f is (6 )) = 9.

the table tells us

so the change in y-value for function g is (31 -1) = 30.

the change in y-value of f(x) on the interval is less than the change in y-value of g(x) on the same interval, so the average rate of change of f(x) is less than the average rate of change of g(x).

maybe showing the graph would

step-by-step explanation:

-387129 iq