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    Hi guys ,
    I'm trying to integrate -4e^-t , isn't this in the form of y=e^f(x) hence dy/dx=f'(x)e^f(x) ? the answer comes out as +4e^-t but somehow i just can't get to that any help would be much appreciated
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    (Original post by Alen.m)
    Hi guys ,
    I'm trying to integrate -4e^-t , isn't this in the form of y=e^f(x) hence dy/dx=f'(x)e^f(x) ? the answer comes out as +4e^-t but somehow i just can't get to that any help would be much appreciated
    What answer did you get? And could you post your working/thoughts to explain how you got it?
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    (Original post by notnek)
    What answer did you get? And could you post your working/thoughts to explain how you got it?
    first i diffrentiated -t which gave me -t^2/2 and then put in the form of dy/dx=f'(x)e^f(x) which again gave me -4(-t^2/2 e^-t) and this is just far away from what the text book said
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    (Original post by Alen.m)
    first i diffrentiated -t which gave me -t^2/2 and then put in the form of dy/dx=f'(x)e^f(x) which again gave me -4(-t^2/2 e^-t) and this is just far away from what the text book said
    Do you know that \displaystyle \frac{\mathrm{d} }{\mathrm{d} t}\left ( e^{-t} \right ) = -e^{-t}? If you differentiate the result you obtained, it does not give your initial function.

    Recall that \displaystyle \int e^{x} \ \mathrm{d}x = e^{x} + \mathrm{C}.
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    (Original post by Alen.m)
    first i diffrentiated -t which gave me -t^2/2 and then put in the form of dy/dx=f'(x)e^f(x) which again gave me -4(-t^2/2 e^-t) and this is just far away from what the text book said
    You have integrated when you should have differentiated

    \displaystyle \int -t \ dt = -\frac{1}{2}t^2 + c
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    (Original post by Alen.m)
    first i diffrentiated -t which gave me -t^2/2 and then put in the form of dy/dx=f'(x)e^f(x) which again gave me -4(-t^2/2 e^-t) and this is just far away from what the text book said
    Differentiating -t with respect to t will give you -1.
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    (Original post by aymanzayedmannan)
    Do you know that \displaystyle \frac{\mathrm{d} }{\mathrm{d} t}\left ( e^{-t} \right ) = -e^{-t}? If you differentiate the result you obtained, it does not give your initial function.

    Recall that \displaystyle \int e^{x} \ \mathrm{d}x = e^{x} + \mathrm{C}.
    To be honest the text bok that im revising has got nothing about the formula you nust mentioned and that was why i kept trying wrong ways i guess but yours absoultely makes sense thanks
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    (Original post by Alen.m)
    To be honest the text bok that im revising has got nothing about the formula you nust mentioned and that was why i kept trying wrong ways i guess but yours absoultely makes sense thanks
    This formula dy/dx=f'(x)e^f(x) does work.

    You just made the mistake of integrating -t when you should have differentiated it to give f'(x) = -1.
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    (Original post by notnek)
    This formula dy/dx=f'(x)e^f(x) does work.

    You just made the mistake of integrating -t when you should have differentiated it to give f'(x) = -1.
    Well actually the question is on M2 where you have to integrate certain expression to get to another expeesion for displacement, acceleration and velocity . Im half sure that i had to integrate certain expression which had -4e^-t in it as well to get to another expression. I'll send you the full question once i got home
 
 
 

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