Calculus Homework Help

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jai6638

Golden Member
Apr 9, 2004
1,790
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Originally posted by: hypn0tik
Originally posted by: jai6638
for ln[x]

I got xln(x)-x+c which seems correct

For xln[x]

I got 1/4 x^2(2lnx-1) which seems correct

However, (ln[x])^2 ... im not sure


I set u=ln x , u'=1/x dx
v'=lnxdx, v=lnx

Since its uv-int(lnx.1/x dx ) , how do i get antiderivatiev of lnx .. I suppose my substitutions are wrong..
Click to expand...

Careful. Look at question 1 to find v from v'.

I have a meeting to run to. I'll be back in a few hours if you still need help.
Click to expand...


so would v be lnx(x)-x?
 
H

hypn0tik

Diamond Member
Jul 5, 2005
5,866
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0
Originally posted by: jai6638
Originally posted by: hypn0tik
Originally posted by: jai6638
for ln[x]

I got xln(x)-x+c which seems correct

For xln[x]

I got 1/4 x^2(2lnx-1) which seems correct

However, (ln[x])^2 ... im not sure


I set u=ln x , u'=1/x dx
v'=lnxdx, v=lnx

Since its uv-int(lnx.1/x dx ) , how do i get antiderivatiev of lnx .. I suppose my substitutions are wrong..
Click to expand...

Careful. Look at question 1 to find v from v'.

I have a meeting to run to. I'll be back in a few hours if you still need help.
Click to expand...


so would v be lnx(x)-x?
Click to expand...

Bingo.
 
J

jai6638

Golden Member
Apr 9, 2004
1,790
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0
damn it.. not gettin the right answer.

am stuck at (lnx)(lnx)(x)-x - int( lnx-1 dx ).
do I plug in anti derivative of ln x again? I tried that it just made it messy.. doh

I apologize for delaying you. Thanks much for your help.. really appreciate it :)
 
H

hypn0tik

Diamond Member
Jul 5, 2005
5,866
2
0
Originally posted by: jai6638
damn it.. not gettin the right answer.

am stuck at (lnx)(lnx)(x)-x - int( lnx-1 dx ).
do I plug in anti derivative of ln x again? I tried that it just made it messy.. doh

I apologize for delaying you. Thanks much for your help.. really appreciate it :)
Click to expand...

No worries.

You're offline but you may check this thread again.

u = lnx
dv = lnx dx

So, du = 1/x dx
v = xlnx - x

Now, int (u*dv) = u*v - int (v*du)
so, int (lnx * lnx) = lnx * (xlnx - x) - int [(xlnx - x)*1/x dx]
= lnx*(xlnx -x) - int[(lnx - 1)dx]

Now,

int[(lnx - 1)dx] = int (lnx dx) - int (dx)

Question 1 dealt with int (lnx dx). int (dx) is trivial.
 
J

jai6638

Golden Member
Apr 9, 2004
1,790
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0
Hey..

In int(lnx-1), what happened to the 1? doesnt that need to be taken into accuont?

thanks

EDIT: nevermind.. you distributed.
 
H

hypn0tik

Diamond Member
Jul 5, 2005
5,866
2
0
Originally posted by: jai6638
Hey..

In int(lnx-1), what happened to the 1? doesnt that need to be taken into accuont?

thanks
Click to expand...

Int [(lnx-1)dx] = int[lnx dx] + int [-1 dx] = int [lnx dx] - int[dx]
 
J

jai6638

Golden Member
Apr 9, 2004
1,790
0
0
Originally posted by: hypn0tik
Originally posted by: jai6638
Hey..

In int(lnx-1), what happened to the 1? doesnt that need to be taken into accuont?

thanks
Click to expand...

Int [(lnx-1)dx] = int[lnx dx] + int [-1 dx] = int [lnx dx] - int[dx]
Click to expand...

lol.... I see it now.. so the final answer i get without simplifying is:

((lnx)(x)-x)lnx + ((lnx)(x)-x)-x


is this right?
 
H

hypn0tik

Diamond Member
Jul 5, 2005
5,866
2
0
Originally posted by: jai6638
Originally posted by: hypn0tik
Originally posted by: jai6638
Hey..

In int(lnx-1), what happened to the 1? doesnt that need to be taken into accuont?

thanks
Click to expand...

Int [(lnx-1)dx] = int[lnx dx] + int [-1 dx] = int [lnx dx] - int[dx]
Click to expand...

lol.... I see it now.. so the final answer i get without simplifying is:

((lnx)(x)-x)lnx + ((lnx)(x)-x)-x


is this right?
Click to expand...

Make sure you have your signs correct.

int (u*dv) = u*v - int (v*du)
Note the sign ^^^^
 
H

hypn0tik

Diamond Member
Jul 5, 2005
5,866
2
0
Originally posted by: jai6638
Originally posted by: hypn0tik
Originally posted by: jai6638
x((lnx-1)lnx-(lnx-1)+1))
Click to expand...

Not quite what I had in mind.

Originally posted by: jai6638

((lnx)(x)-x)lnx - ((lnx)(x)-x)+x
Click to expand...

(xlnx -x) (lnx -1) + x
Click to expand...

what'd you do?
Click to expand...

You have:
((lnx)(x)-x)lnx - ((lnx)(x)-x)+x

Let ((lnx)(x) -x) = Z
So your expression simplifies to:
Zlnx - Z + x
= Z (lnx -1) + x
 
J

jai6638

Golden Member
Apr 9, 2004
1,790
0
0
Originally posted by: hypn0tik
Originally posted by: jai6638
Originally posted by: hypn0tik
Originally posted by: jai6638
x((lnx-1)lnx-(lnx-1)+1))
Click to expand...

Not quite what I had in mind.

Originally posted by: jai6638

((lnx)(x)-x)lnx - ((lnx)(x)-x)+x
Click to expand...

(xlnx -x) (lnx -1) + x
Click to expand...

what'd you do?
Click to expand...

You have:
((lnx)(x)-x)lnx - ((lnx)(x)-x)+x

Let ((lnx)(x) -x) = Z
So your expression simplifies to:
Zlnx - Z + x
= Z (lnx -1) + x
Click to expand...

wow. I dunno how I didnt get that.
 
H

hypn0tik

Diamond Member
Jul 5, 2005
5,866
2
0
Originally posted by: jai6638
Originally posted by: hypn0tik
Originally posted by: jai6638
Originally posted by: hypn0tik
Originally posted by: jai6638
x((lnx-1)lnx-(lnx-1)+1))
Click to expand...

Not quite what I had in mind.

Originally posted by: jai6638

((lnx)(x)-x)lnx - ((lnx)(x)-x)+x
Click to expand...

(xlnx -x) (lnx -1) + x
Click to expand...

what'd you do?
Click to expand...

You have:
((lnx)(x)-x)lnx - ((lnx)(x)-x)+x

Let ((lnx)(x) -x) = Z
So your expression simplifies to:
Zlnx - Z + x
= Z (lnx -1) + x
Click to expand...

wow. I dunno how I didnt get that.
Click to expand...

My math teacher always said, "Common factoring is the #1 skill in math." I must say he's right.
 
J

jai6638

Golden Member
Apr 9, 2004
1,790
0
0
hey guys.. need some help once again:

The following are the questoins and my solutions:

Find the integral using the tables:

.....


Thanks much
 
U

Udel

Senior member
Sep 2, 2005
892
0
0
OMG I just called your professor and he said that you will not be expelled. Heh. Actually, I'm not so hot at Calculus, just got by with a C, so I can't much help you buddy. Good luck.
 
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