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How do you do the hence using substitution bit in part iii thanks
hiya.PNG
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Do you mean
(1) you've not been taught/learnt substitution, so this is all new
(2) you need help with just this substitution
(3) other
Reply 2
Avatar for Kalabamboo
Kalabamboo
OP
14
Original post by Ann Kittenplan
Do you mean
(1) you've not been taught/learnt substitution, so this is all new
(2) you need help with just this substitution
(3) other


I mean I am stuck on part III - so just this and not the whole of part III just the hence bit
(edited 5 years ago)
Reply 3
Avatar for Kalabamboo
Kalabamboo
OP
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Can somebody help please?
Original post by Kalabamboo
I mean I am stuck on part III - so just this and not the whole of part III just the hence bit


Have you been taught how to do substitutions?
Reply 5
Avatar for vc94
vc94
12
Original post by Kalabamboo
Can somebody help please?


The substitution is u=x-pi, so du/dx=1, so dx=du. Replace dx with du in the integration.
Also f(x) = f(u+pi) for which you can use the first bit of (iii)
Reply 6
Avatar for Kalabamboo
Kalabamboo
OP
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Original post by vc94
The substitution is u=x-pi, so du/dx=1, so dx=du. Replace dx with du in the integration.
Also f(x) = f(u+pi) for which you can use the first bit of (iii)


Ah thanks a lot for replying -what is confusing me here is how the limits are changed in this case. Could you pls let me know?:smile:
so
how you get
Attachment not found
(edited 5 years ago)
Original post by Kalabamboo
Ah thanks a lot for replying -what is confusing me here is how the limits are changed in this case. Could you pls let me know?:smile:
so
how you get
Attachment not found


you have to change the limits in terms of x, to in terms of u as you are integrating with respect to u.
using u=x-pi
so 2pi become 2p-pi = pi
and pi becomes pi-pi = 0
Reply 8
Avatar for Kalabamboo
Kalabamboo
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Original post by luckybillion
you have to change the limits in terms of x, to in terms of u as you are integrating with respect to u.
using u=x-pi
so 2pi become 2p-pi = pi
and pi becomes pi-pi = 0

Thanks a lot!!!:smile: just a bit confused with this bit -could you please help?
bee.PNG
Original post by Kalabamboo
Thanks a lot!!!:smile: just a bit confused with this bit -could you please help?
bee.PNG


so i would use the information in the first part of the question
and use the same expansion for f(x+pi) = -e^(-1/5pi)f(x)
as you have the same function form
so f(u+pi)= -e^(-1/5pi)f(u)

and as -e^(-1/5pi) is a constant you can take it outside of the integral
Original post by luckybillion
so i would use the information in the first part of the question
and use the same expansion for f(x+pi) = -e^(-1/5pi)f(x)
as you have the same function form
so f(u+pi)= -e^(-1/5pi)f(u)

and as -e^(-1/5pi) is a constant you can take it outside of the integral

Ahh thanks a lot!!!!:smile: And just the very last bit just the interpret this result geographically bit -could you please help me with that? :smile:
can anyone help with the very last part of iii please ?the graphically bit
(edited 5 years ago)
Original post by Kalabamboo
can anyone help with the very last part of iii please ?the graphically bit


Well firstly note that the integral of f(u) du can be replaced with the integral of f(x) dx, since u or x or whatever is just a dummy variable. Thus the result tells us that the area under the graph between pi and 2pi is e^(-1/5pi) times the area between 0 and pi, with the negative sign reflecting the fact that the area is below the x-axis, such that the integral will work out as negative (but area is always positive).

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