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    By the way, I think you're all over complicating the solution to 1997 4c).

    The integral of the upper curve from 0 to b could easily shown to be (pi/2)sin(b). This region could be divided into three: the shaded region, the triangle like region, and a square. Call them A, B, C respectively.

    We wish to find A, which is given by (pi/2)sin(b)-B-C. We could easily see that C=b^2, and B is simply the integral from b to pi/2 of the lower curve with respect to y. This is simply (pi/2)sin(y) with limits b and pi/2, so B=(pi/2)-(pi/2)sin(b).

    Combining all the above we have A=(pi/2)sin(b)-(pi/2-(pi/2)sin(b))-b^2=pi(sinb)-pi/2-b^2 as required.

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    Do any of you know how to solve Q5 of the 2000 paper (cube made of rods)? a) and b) are easy but is there a way to solve c) without counting all the possible paths?
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