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    How do you do the hence using substitution bit in part iii thanks
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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
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    (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
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    Can somebody help please?
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    (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?
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    (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)
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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?
    so
    how you get
    Attachment 749052
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    (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?
    so
    how you get
    Attachment 749052
    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
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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!!! just a bit confused with this bit -could you please help?
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    (Original post by Kalabamboo)
    Thanks a lot!!! just a bit confused with this bit -could you please help?
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    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
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    (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!!!! And just the very last bit just the interpret this result geographically bit -could you please help me with that?
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    can anyone help with the very last part of iii please ?the graphically bit
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    (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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