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    Can't do this qu. the shape of the curve the frustrating part. Diagram attached.

    Diagram shows the curve x = 2t, y=3/t

    a. Find areas A and B


    for working out A, i plugged in x =1 and x =3 into x = 2t to find two values of t, which i used as limits. I then used the standard formula

    A = (curly f) t2 on top, t1 on bottom: y dy/dx dt

    but when it came to integrating, it all came out as zero. Where have i gone wrong?

    thanks. also rep x
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    Could you write out some workings for us, so we can see where you've gone wrong.
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    Just lol now.
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    the problem is probably that you got the formula wrong. I don't know how to use LaTeX, but it should be:

    Int: y (dx/dt) dt with the t values as the limits.

    not (dy/dx)
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    Can you not just change from parametric by eliminating t, to get equation in usual format? i.e. get:

     \int^3_1 \frac{6}{x} dx

    Must admit bit rusty on parametrics so this could be a no no - please let me know if so!

    EDIT

    Checked with this formula
    Int: y (dx/dt) dt with the t values as the limits.
    Result is the same
 
 
 
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Updated: January 25, 2010
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