## anonymous 4 years ago I have question in substation rule

1. anonymous

2. anonymous

in substitution rule

3. myininaya

all those are z's right?

4. anonymous

yep

5. myininaya

did you let u=the bottom

6. myininaya

$u=e^z +z => du=(e^z+1 ) dz$

7. anonymous

yep i did in that way

8. myininaya

$\int\limits_{e^0+0}^{e^1+1}\frac{du}{u}$

9. myininaya

$\ln|u||_1^{e^1+1}$

10. myininaya

$\ln|e^1+1|-\ln|1|=\ln(e+1)-0=\ln(e+1)$

11. anonymous

yeah that is the right answer Thank you so much :)

12. myininaya

Do you understand all the steps I took?

13. anonymous

i didn't understand one part of it which is the one before last step of it

14. myininaya

you mean when I plugged in the upper limit and then minus plugged in the lower limit?

15. anonymous

yeah why did you put e to 1 + 1 and e to 0 +0?

16. myininaya

so before the substitution the limits were 0<z<1 we wanted everything in terms of u after we made substitution if u=e^z+z, then e^0+0<z<e^1+1 ---- if z=0, then u=e^0+0 if z=1, then u=e^1+1

17. anonymous

Oh I see Thank you so much that was very helpful :)

18. anonymous

I understood each step that you had done

19. myininaya

Awesome! :)