1. ## integral

Hi everyone,

Could someone tell me if this is correct so far, please?

Integral 0 to 1 lnx/sq. x

lim t goes to 1 from the left lnx/sq. x

u=lnx
dv=sq. x
du=1/x
v=1/2xto the -1/2

lim t goes to 1 from the left lnx(1/2x^-1/2)-int.(1/2x^-1/2)(1/x)

Could I break this up into two integrals?
-1/2int. x^-1/2-1/2int. 1/x

Thank you very much

2. Originally Posted by chocolatelover
Hi everyone,

Could someone tell me if this is correct so far, please?

Integral 0 to 1 lnx/sq. x

lim t goes to 1 from the left lnx/sq. x

u=lnx
dv=sq. x
du=1/x
v=1/2xto the -1/2

lim t goes to 1 from the left lnx(1/2x^-1/2)-int.(1/2x^-1/2)(1/x)

Could I break this up into two integrals?
-1/2int. x^-1/2-1/2int. 1/x

Thank you very much
the function is undefined at 0, so you would find $\displaystyle \lim_{t \to 0} \int_{t}^{1} \frac {\ln x}{\sqrt {x}}~dx$

Let $\displaystyle u = \ln x$ and $\displaystyle dv = \frac 1{\sqrt {x}} = x^{-1/2}$

3. Does -1/2 look right?

At the end I got the lim as t goes to 1-[(-1^1/2)-[ln(0)(0^1/2)/2)-0^1/2]

So, that would just be -1/2, right?

Thank you very much

4. Originally Posted by chocolatelover
Does -1/2 look right?

At the end I got the lim as t goes to 1-[(-1^1/2)-[ln(0)(0^1/2)/2)-0^1/2]

So, that would just be -1/2, right?

Thank you very much
no. we spoke about this already. where is your t? and i told you that the limit is as t goes to zero. why is it you have trouble following clear instructions?

you should have: $\displaystyle \lim_{t \to 0} \left[ 2 \sqrt {x} \ln x - 4 \sqrt {x} \right]_{t}^{1}$

5. oops!

6. Is it 0 from the left or 0 from the right?

7. Originally Posted by chocolatelover
Is it 0 from the left or 0 from the right?
from the right, this is what the graph looks like

this is my 41th post!!!

8. I got -4. Is that correct?

Thank you

9. Originally Posted by chocolatelover
I got -4. Is that correct?

Thank you
yup, that's what i got. good job