Oxford PAT 2015, Question 4

2015_paper__page4

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9 thoughts on “Oxford PAT 2015, Question 4

  1. Hi,

    Just trying to make sense of this question. I’m trying to work through C3 (and then C4) before November 2nd and seeing stuff like this is both scary but I suppose expected.

    What chapter is this? I mean I’ve never done anything vaguely like this and I can sort of follow it tbf but there is no way I would have done anything like this.

  2. Okay, this is kind of….a useless answer. Simpy put, you have skipped an entire section. You have gone from DU/DX straight to DU. That’s the equivalent of writing the equation F=MA then simply writing down the answer.
    You wouldn’t get the full marks, nor would it actually help anybody.

    1. Let’s go through the answer line by line:

      Line 1 — introduces the substitution we are going to use: u = (x-4)(x-2)
      Line 2 — derives the value of du/dx
      Line 3 and line 4 — work out the upper and lower limits of the resulting definite integral
      Line 5 — states that given the previous four lines we can transform the integral with respect to x into its simpler equivalent with respect to u
      Line 6 — does the integral and calculates its value between the limits

      Can you explain which steps you think are missing?

      1. You have shown what du/dx is, but not what the value of du (2x-6dx) or the value of dx (du/2x-6) is.

        Also, wouldn’t it simply be more efficient to substitute du/2x-6 into dx, hence showing that you can ‘cancel’ out the 2x-6, leaving you simply with root(U)^1/2dx? Right now, you are kind of leaving the du (2x-6dx) in and not really showing why you don’t use that further.
        For somebody who understands U-substitution this would suffice but for somebody trying to figure out how this question works its rather vague.

        1. Why would I want to show what the value of du (2x-6dx) or dx (du/2x-6) is??

          Let’s just go through the key point again. I have introduced a substitution u = (x-4)(x-2), and shown that du/dx = 2x – 6. And therefore Integral (2x-6) √((x-4)(x-2)) dx equals Integral du/dx √u dx which equals Integral √u du/dx dx which equals Integral √u du. What is the problem?

    1. In the first and second lines of the answer, we show that (2x-6) = du/dx. Then, in the fifth line of the answer, the (2x-6) disappears because (2x-6) dx is equal to (du/dx) dx, which is equal to du. So when the integral with respect to x is turned into an integral with respect to u, the (2x-6) term vanishes. If you are not familiar with this ‘integration by substitution’ approach, then there is some tutorial material at http://www.mathtutor.ac.uk/integration/integrationbysubstitution/text that might be helpful.

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