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    I got stuck on some of these questions
    For question 13

    I found t = 0, -2 and 2 on part a
    for part c, I integrate part c and get (2t^5)/5 - (8t^3)/3 = f(x)

    Then I subs t = 2 and t = -2 into f(x)
    then take f(2) - f(-2) and get the ans -17/1/15? Is this the right ans?

    For question 6, I dont know how to find value of x after getting all the 4 terms?

    Thanks
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    (Original post by prettyboy_bn)
    For question 6, I dont know how to find value of x after getting all the 4 terms?
    1-3x=97

    solve that and substitute it in
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    1-3x=97

    solve that and substitute it in
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    Do u mean x = (1 - 97)/3 ?
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    Anyone?
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    (Original post by prettyboy_bn)
    Do u mean x = (1 - 97)/3 ?
    no - you've just expanded (1-3x)^{\frac{3}{2}} so if you find the solution to 1-3x=97 and substitute it into your expansion, you'll find an approximation to 97^{\frac{3}{2}}
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    no - you've just expanded (1-3x)^{\frac{3}{2}} so if you find the solution to 1-3x=97 and substitute it into your expansion, you'll find an approximation to 97^{\frac{3}{2}}
    ok so the solution to 1-3x=97 is  -32 and I subs it into the expansion (4 terms) and I got - 51,695 which is the approximation to 97^{\frac{3}{2}} ?
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    (Original post by prettyboy_bn)
    ok so the solution to 1-3x=97 is  -32 and I subs it into the expansion (4 terms) and I got - 51,695
    ergo, your expansion is incorrect
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    ergo, your expansion is incorrect
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    I just checked with the answer booklet, I got the expansion right. Their answer for x = 0.01?

    Anything for question 13?

    Thanks
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    (Original post by prettyboy_bn)
    I just checked with the answer booklet, I got the expansion right. Their answer for x = 0.01?

    Anything for question 13?

    Thanks
    letting x=0.01 gives an approximation to 0.97^{\frac{3}{2}}, not 97^{\frac{3}{2}}
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    For question 13 dont you intergrate within t = 0 and t=2?
    O.o I could be wrong though
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    (Original post by Snapshot13)
    For question 13 dont you intergrate within t = 0 and t=2?
    O.o I could be wrong though
    At t=0, the point is (1,0), at t=2 the point is (-3,0). The issue is, if you integrate from t=0 to t=2, you don't get the area of the full graph.

    So, the OP is correct integrating from t=-2 to t=2
 
 
 
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