quick integral question.

[lethalasian]

s(t)=s(0)+intergal [t,0] (60- e^-x) dx
s(0) = 40
i got s(t) = 40+60t+e^-t - 1
is this correct?

 

[Deleted member 4993]

s(t)=s(0)+intergal [t,0] (60- e^-x) dx
s(0) = 40
i got s(t) = 40+60t+e^-t - 1
is this correct?

If 't' was the upper-limit of your integration and '0' was the lower limit - then it is correct.

 

[lethalasian]

thank you! yay i got the -1 but the answer sheet skipped the -1 part

 

[topsquark]

s(t)=s(0)+intergal [t,0] (60- e^-x) dx

i got s(t) = 40+60t+e^-t - 1

Notation tip: It's not something that can really be misinterpreted but you should be writing your exponentials as e^(-t). It's just a good practice to get into.

-Dan

 

[Steven G]

Any reason you did not say 39 instead of 40 - 1??
I would have taken a point off for not finishing.

 

[lethalasian]

haha i left the work out so people can see i got 40 and then -1