Few Questions Needed Help With

[bgallz]

Hi there, I have a few problems that I am struggling with. The answers in the back of the textbook don't give explanations and my answer comes out different.

Homework Statement

a. (From 0 to 1) \intln(x) dx
I took this as [1/x]. So I did 1/1 - 1/0 = 1. Is that correct?

b. (From 0 to 3) \int1 / (x-1)² dx
I used u substitution.

u = x-1
du = dx

New integral: (From 0 to 3)\int1 / u² du.
Taking the integral: u^-2 = - 1 / u => -1 / x-1. Then I took it from 0 to 3.
(- 1 / 3 - 1) - (-1 / 0 - 1) = (-1/2) - (1/1) = -3/2. I got a 7/10 on this one. :\

c. (From 1 to e)\intx²*ln(x) dx
I turned this into integration by parts like this:

u = ln(x)
du = 1/x dx
dv = x²
v = (x³)/3

So uv - \intv du = ln(x)*(x³/3) - \int(x³/3)*1/x dx. I eventually got down to:

1/3[x³ln(x)] - 1/9[e³ - 1] + C

Is that correct? This problem isn't given an answer but my original try was definitely wrong. lol

Thank you so much for any help!

 

[tiny-tim]

welcome to pf!

hi bgallz! welcome to pf!

1. erm … the derivative of lnx is 1/x, isn't it?

2. would be ok, except that everything is infinite at x = 1

1/3[x³ln(x)] - 1/9[e³ - 1] + C

that isn't finished

(and how can a definite integral have a C ?)

 

[bgallz]

Hey thanks for the reply.

1. I thought that taking the integral of ln(x) is also 1/x. But it looks like this right here proves that wrong:
http://math2.org/math/integrals/more/ln.htm

2. Ah yes I totally overlooked that. So I would have to break it up into 2 integrals right? One from 0 to 1 and one from 1 to 3? I forget how that works.

3. Do you mean that I have to calculate e^3 - 1? And yea I don't know why I put the C there.

Thanks!

 

[tiny-tim]

hi bgallz!

2. Ah yes I totally overlooked that. So I would have to break it up into 2 integrals right? One from 0 to 1 and one from 1 to 3? I forget how that works.

yes (though if they're ±∞, the sum would be indeterminate)

3. Do you mean that I have to calculate e^3 - 1?

no, i mean you still have x in the first part

 

[bgallz]

Can you explain that first part about breaking it up?

And for the 3rd one, I have to run that uv part from 1 to e as well?

 

[tiny-tim]

Can you explain that first part about breaking it up?

if you draw the graph, you can see that it obviously has a discontinuity, where it goes off to ±∞

And for the 3rd one, I have to run that uv part from 1 to e as well?

yes, everything in integration by parts goes between the same limits