[SOLVED] on a defined interval, integrating ln to find the answer

**kalossimitar** · December 10th 2008, 09:56 AM

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)

**Ziaris** · December 10th 2008, 11:03 AM

Integrate the logarithm by parts.

$\int\ln t dt=\int (1\cdot\ln t)dt = t\ln t - t + C$

To integrate $t\ln t$ with respect to $t$ , you can integrate by parts again without knowing the antiderivative of $\ln t$ .

$\int t\ln t dt=\frac{1}{2}t^2\ln t-\frac{1}{2}\int\frac{t^2}{t}dt=\frac{1}{2}t^2\ln t-\frac{t^2}{4}.$

**kalossimitar** · December 10th 2008, 11:21 AM

Thank you very much! Now, I know where what my teacher wrote comes from. One last question still pending though, would you use t=25 right away, or t=(1+0,53t) in the integrated form of the equation and then replace the t in it by 25?

example:

(1/2(25)^2)x ln (25) - ((25)^2/4) =answer/25= 33.something cm

OR

((1/2(1+0,53t)^2)x ln(1+0,53t) - ((1+0,53t)^2/4), then replace t in i by 25: ((1/2(1+0,53(25))^2........etc. =answer/25= 9.something cm?

I THINK that I just need to replace t by 25 right away (=33 cm), because its what I did in the similar problem i spoke about in first post, and I had the right answer, but I just wanna be sure, you know

(mostly because a friend of mine who get better marks int he course said that he thought he would use (1+0,53t).

**Grandad** · December 10th 2008, 11:21 AM

Hi -

Here's some info that should help:

$\int\text{ln }u.du=u\text{ ln }u-u+c$

(If you want to check this, just differentiate the RHS using the product rule.)

You were given that

$\int u\text{ ln } u.du=\frac{u^2}{4}(2\text{ ln }u-1)+c$

and you need to find:

$I=\int_0^{25}t \text{ ln }(1+0.53t).dt$

before dividing the answer by 25, to get the average depth of snow.

To find the value of the integral I, use the substitution:

$1+0.53t=u$

So: $t=\frac{u-1}{0.53}$

and: $dt = \frac{du}{0.53}$

When you do the substitution, replace the limits of the integral by the corresponding values of u. Thus:

$I=\int_1^{14.25}\frac{u-1}{0.53}\text{ ln }u.\frac{du}{0.53}$

You'll see now that you've got to integrate $(u-1)\text{ ln }u$ .

Do this as two separate terms: $u\text{ ln }u$ and $-\text{ ln }u$ , using the integral formulae above.

If I've done the arithmetic correctly, the average depth comes out to be about 27.7 cm (but it would be good if you could check that!).

I hope that helps.

Grandad

**kalossimitar** · December 10th 2008, 11:35 AM

I got an exam today, and another tomorrow. I'll be back tomorrow or after tomorrow to try this out. Please don't forget about my thread.

**kalossimitar** · December 10th 2008, 11:41 AM

Originally Posted by Grandad

Hi -

Here's some info that should help:

$\int\text{ln }u.du=u\text{ ln }u-u+c$

(If you want to check this, just differentiate the RHS using the product rule.)

You were given that

$\int u\text{ ln } u.du=\frac{u^2}{4}(2\text{ ln }u-1)+c$

and you need to find:

$I=\int_0^{25}t \text{ ln }(1+0.53t).dt$

before dividing the answer by 25, to get the average depth of snow.

To find the value of the integral I, use the substitution:

$1+0.53t=u$

So: $t=\frac{u-1}{0.53}$

and: $dt = \frac{du}{0.53}$

When you do the substitution, replace the limits of the integral by the corresponding values of u. Thus:

$I=\int_1^{14.25}\frac{u-1}{0.53}\text{ ln }u.\frac{du}{0.53}$

You'll see now that you've got to integrate $(u-1)\text{ ln }u$ .

Do this as two separate terms: $u\text{ ln }u$ and $-\text{ ln }u$ , using the integral formulae above.

If I've done the arithmetic correctly, the average depth comes out to be about 27.7 cm (but it would be good if you could check that!).

I hope that helps.

Grandad

If someone in the meantime can come up with the equation to do so (the bold part), I would appreciate it, because, I understand until there, and I dont want the answer as much as I want to understand, because, we didnt see that in class, and my other option would be to make a complaint about the teacher giving things at the exam that we didnt see in class (the final exam should be nearly the same as this whole work I have to do, because it was like that at mid-semester (the mid-term exam was nearly identical to the homework he gave us.)

**kalossimitar** · December 10th 2008, 07:42 PM

any taker?

**Grandad** · December 11th 2008, 05:13 AM

Hello kalossimitar -

$I=\int_1^{14.25}\frac{u-1}{0.53}\text{ ln }u.\frac{du}{0.53}$

$= \frac{1}{0.53^2}\int_1^{14.25}(u-1)\text{ ln }u.du$

$= \frac{1}{0.53^2}\int_1^{14.25}(u\text{ ln }u-\text{ ln }u).du$

$= \frac{1}{0.53^2}\Big[\frac{u^2}{4}(2\text{ ln }u-1)-(u\text{ ln }u-u)\Big]_1^{14.25}$ (using the integrals that I gave you in my original posting)

The rest is arithmetic. I've checked my original answer by putting the original function, $D=t \text{ ln }(1+0.53t)$ , into a spreadsheet, and working out 250 values of $D$ (depth) between $t=0$ and $t=25$ at intervals of 0.1, summing the values and dividing by 250, to get an average. The answer I got this way was 27.8. (My original answer using integration was 27.7, so I reckon this is about right!)

OK?
Grandad

**kalossimitar** · December 12th 2008, 01:37 PM

yes, thank you, it helped. Its still much more advanced maths than what we saw in the course. The teacher's assistant said this should not be in the final exam, but still, I needed to know how to do it for the homework which counts for 10% of the semester. Thank you.

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