- #1
fstam2
- 10
- 0
Confused, but tried it this way:
Use u-substitution to show that (for y a positive number and x>0)
[tex]\int_{x}^{xy} \frac{1}{t} dt = \int_{1}^{y} \frac{1}{t} dt [/tex]
so, u=t and du=dt
if x=1
t=xy u=y(1)=y
t=x u=1
or
u=1/t and du/ln [t] = dt
if x=1
t=xy u=1/y
t=x y=1
Thanks for your help
Todd
Use u-substitution to show that (for y a positive number and x>0)
[tex]\int_{x}^{xy} \frac{1}{t} dt = \int_{1}^{y} \frac{1}{t} dt [/tex]
so, u=t and du=dt
if x=1
t=xy u=y(1)=y
t=x u=1
or
u=1/t and du/ln [t] = dt
if x=1
t=xy u=1/y
t=x y=1
Thanks for your help
Todd