Question

I have question in substation rule

Answer 1

in substitution rule

Answer 2

all those are z's right?

Answer 3

yep

Answer 4

did you let u=the bottom

Answer 5

\[u=e^z +z => du=(e^z+1 ) dz\]

Answer 6

yep i did in that way

Answer 7

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

Answer 8

\[\ln|u||_1^{e^1+1}\]

Answer 9

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

Answer 10

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

Answer 11

Do you understand all the steps I took?

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

I understood each step that you had done

Answer 18

Awesome! :)