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    I got 1/2 but the answer is 1/12...can anyone help? Thanks!!
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    (Original post by FlyingPooMan)
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    I got 1/2 but the answer is 1/12...can anyone help? Thanks!!
    The integrand is a translated version of y = x^2. If you think about the position that this translated parabola would have to be in so that the area enclosed by it, x = 0 ans x = 1, the answer should be fairly obvious. If you can see this position, then you have the value of a. From there, I couldn't think of any shortcut other than to actually do the integration (I expanded the bracket, but it would have probably been quicker to have done it without doing this), and the answer does indeed pop out to be 1/12.
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    (Original post by Pangol)
    The integrand is a translated version of y = x^2. If you think about the position that this translated parabola would have to be in so that the area enclosed by it, x = 0 ans x = 1, the answer should be fairly obvious. If you can see this position, then you have the value of a. From there, I couldn't think of any shortcut other than to actually do the integration (I expanded the bracket, but it would have probably been quicker to have done it without doing this), and the answer does indeed pop out to be 1/12.
    Thank you! I got it now, a was 1/2 and gives 1/12
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    (Original post by FlyingPooMan)
    Thank you! I got it now, a was 1/2 and gives 1/12
    I agree with that!
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    (Original post by Pangol)
    The integrand is a translated version of y = x^2. If you think about the position that this translated parabola would have to be in so that the area enclosed by it, x = 0 ans x = 1, the answer should be fairly obvious. If you can see this position, then you have the value of a. From there, I couldn't think of any shortcut other than to actually do the integration (I expanded the bracket, but it would have probably been quicker to have done it without doing this), and the answer does indeed pop out to be 1/12.
    Alternatively, you can compute the integral to find I(a), then minimise this w.r.t. a. The geometric approach is slightly quicker though, I'd say.

    [edit: it's worth bearing in mind too that since the function is symmetric in the required region of integration, we can translate it back by -1/2 and evaluate \int_{-1/2}^{1/2} x^2 \ dx = 2 \int_0^{1/2} x^2 \ dx instead, which is a bit easier.]
 
 
 
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