Question

I need someone to look this over and see if I did it right. please :-)
integral lower limit is 0 upper limit is 1
u=3x-2
dx=du/3
1/3integral lower limit -2 upper limit 1
answer is 1/3 [-ln2]

Answer 1

what is your integrand (what are you integrating)?

Answer 2

Your u substitution is right :)
so you are left to evaluate\[\frac{ 1 }{ 3 } \ln(3x-2)\] from 0 to 1.

Answer 3

the initial problem is 1/(3x-2)
from 0 to 1.

Answer 4

So you did your u substition and you know have:\[\int\limits_{}^{}\frac{ 1 }{ u } (1/3) du\] right?

Answer 5

yes. and then I refigured my limits and got -2 to 1

Answer 6

Oh you don't have to do it that way, but I guess you could. That way seems harder.
Just take your indefinite integral in this form and then substitute back in your 3x-2, then you can use your original limits.

Answer 7

\[\frac{1 }{ 3 } \int\limits_{}^{}\frac{ 1 }{ u }du=\frac{1 }{ 3 }\left[ \ln u \right]=\frac{1 }{ 3 }\left[ 3x-2 \right]_{0}^{1}\]

Answer 8

Oops that last part should be
ln(3x-2) in the brackets.

Answer 9

so you are saying that my answer should be 1/3 ln(3x-2)

Answer 10

This integral doesn't converge, therefore cannot have a definite answer. ie if you sub in your values, you'll end up with log of a negative number which cannot be evaluated.

Answer 11

\( \frac 1 {3x-2} \) is not continuous on [0,1]

Answer 12

ln (-2) is not defined

Answer 13

But if I set it up with |u| doesn't that take care of the negative

Answer 14

ok I see now