Question

Can you please help me solve this ?
https://www.dropbox.com/s/u2j3kptjtn6vvxz/Screenshot%202013-11-04%2008.00.03.jpg

Answer 1

did you try 1-e^-x as u ?

Answer 2

1-e^-x = t^2......sunstitution

Answer 3

subs**

Answer 4

why t^2 ? and not t ?

Answer 5

coz of root.....it will blow out sqrt

Answer 6

I get int from 0 to 0 of du/u

Answer 7

when x = infinity , e^-inf = 0

Answer 8

yes thats why I say 0 to 0

Answer 9

but t^2 will make dt complicated

Answer 10

1-e^-x = 1

Answer 11

1-e^-x as u

Answer 12

result will be same....i prefer t^2 @hartnn

Answer 13

ohh, if result same than ok :)

Answer 14

hartnn ?

Answer 15

the function is 1-e^-x = u
so the new limits are 0 to 1

Answer 16

wait if u = 1 - e^-x

Answer 17

u 1 - e^-0

Answer 18

u = 1 - 1

Answer 19

u = 0

Answer 20

for x= 0, u=0
for x =infinity, u=1

Answer 21

u = 1 -e-inf

Answer 22

u = 1 - 0 aah

Answer 23

and integrate du/sqrt u from 0 to 1

Answer 24

ln|1| - ln|0|

Answer 25

if you are having prob in values of e^x then remember its graph as sign of nike

Answer 26

sqrt u = u^(1/2)
you won't get du/u
you get du/sqrt u

Answer 27

ln|sqrt(1)| - ln|sqrt(0)|

Answer 28

how are you even getting ln ?

Answer 29

for 1/sqrt u
use
integral x^n formula

u^(-1/2) du

Answer 30

so its 2sqrt(1) - 2sqrt(0) ?

Answer 31

oh wait I'm confused now , lol

Answer 32

n= -1/2
n+1 = ... ?

Answer 33

my answer with 3 in the denominators

Answer 34

see mine one was good @hartnn

Answer 35

you can show your method too :)

Answer 36

n= -1/2
n+1 = ... ?
@Christos

Answer 37

n= -1/2
n+1 = 1/2

Answer 38

so, integral u^(-1/2) du = (1/2) sqrt u
got this ?

Answer 39

http://www.wolframalpha.com/input/?i=integral+of+1%2Fsqrt%28x%29

Answer 40

oops, its divided by (1/2)
so yeah

Answer 41

with a 2 as a coeff right ?

Answer 42

yea, so my answer was correct ??

Answer 43

so its 2sqrt(1) - 2sqrt(0) ? <<<<< correct

Answer 44

Nice :D

Answer 45

divu, you can show your method now if you want to :)

Answer 46

btw why are those labelled as "improper integrals" ??