Hi, guys. Here's an integration problem.

f(t)= tln(1+0,53t) cm giving the quantity of snow on the ground t days after the first snow.

Whats the mean level of snow on the ground on 0<t<25? (I hope its translated correctly, Im not English)

= =

(the two = refers to <)

I need to integrate, then calculate with the integral of the function when t=25 and

then substract by the integral of the function when t=0

S f(25) - S f(0), and then divide by the number of days on the interval (25)

(S f(25) - S f(0) ) / 25 = answer to the problem

Right? Im sure it is, but, still, just in case.....

My problem is, how do I integrate ln? I checked many tutorials on the web,

but what my teacher wrote on our sheet is not the same. Here's what he wrote:

At question 13, you will need this: S uln(u)du= (u^2/4)(2ln(u)-1)+C

Do I just calculate with the latter and just put 25 instead of u, or do I

put (1+0,53t) instead of u? I tried both, and it gave me something that makes more sense when I replaced u by 25:

(25^2/4)(2ln(25)-1) (no constant, because we got a defined interval), then divide by 25

=33,9859 cm on average on the ground on the interval 0 to 25, which would make sense, seeing as on one of the other questions of the problem, I had a variation of 3,01 cm per day at the start of the third week.

When I replaced u by (1+0,53t), it gave me 8,759 cm on average.

Can you guys help me out? And point out at any errors I could have done (btw, we never saw how to integrate ln, thats why Im having difficulties here, because, basically, one other question of the homework was the same thing, but it was easy, because there was no ln)