# Oxford PAT 2015, Question 4

## 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.

1. This is a C3 thing I think. Not sure which chapter because it might depend on the exam board. But look for ‘integration by substitution’.

2. Confused says:

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. Confused says:

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?