OCR Soloman question help (integration) C4

Hey guys, I'm working through soloman papers and on this paper I struggled even on the first question!

https://5c59854d0ccd29d489c9e5e689a8bbadf49aa0f0.googledrive.com/host/0B1ZiqBksUHNYREhxMHhfam1IQm8/for-OCR/Solomon%20J%20QP%20-%20C4%20OCR.pdf

question 1, how do I solve this? I tried doing integration by parts but I just got myself confused.

Reply 1

physicskid123

Try reverse chain ruling it, with the knowledge that the derivative of x^2 is 2x

Reply 2

HapaxOromenon3

Original post by ollieui

Do a substitution, u = x^2 - 4.

Reply 3

HapaxOromenon3

Original post by physicskid123

Try reverse chain ruling it, with the knowledge that the derivative of x^2 is 2x

No such thing as "reverse chain rule". The chain rule for derivatives does not work in reverse for integrals except where the inner function is linear.

Reply 4

physicskid123

Original post by HapaxOromenon3

No such thing as "reverse chain rule". The chain rule for derivatives does not work in reverse for integrals except where the inner function is linear.

reverse chain ruling it would give us (x/2x) * (2/3) (x^2-4)^3/2

aka 2x/6x so 1/3 (x^2-4)^3/2

and put the limits into that and see what you get

Reply 5

HapaxOromenon3

Original post by physicskid123

reverse chain ruling it would give us (x/2x) * (2/3) (x^2-4)^3/2

aka 2x/6x so 1/3 (x^2-4)^3/2

and put the limits into that and see what you get

Consider the integral of e^(x^2) dx. Using the "reverse chain rule" would give an answer of e^(x^2)/2x, but in fact differentiate that and you find it doesn't work, and in fact there is no function whose derivative is e^(x^2). As I said, the only reason your "reverse chain rule" works here is because y=x is a linear function. What you are really doing is performing the substitution in your head and trying to be clever, but in fact there is no need to be unnecessarily fancy in this case.

Reply 6

physicskid123

Original post by HapaxOromenon3

what do you mean? I get the correct answer of 8 sqrt(3) using my method, so I am assuming it is correct.

Reply 7

HapaxOromenon3

Original post by physicskid123

what do you mean? I get the correct answer of 8 sqrt(3) using my method, so I am assuming it is correct.

The point is the only reason your answer is correct is the fact that x has degree 1 and (x^2-4) has degree 2, so the difference in degrees is 1.
(Similarly your method would work for something like 7x^2*(x^3+9)^5, as the difference in degrees is 3-2=1).
However, for something like 7x^2*(x^4+9)^5, it wouldn't work. Thus since your method is really a trick that is only applicable in a small class of problems, it's better to just do the substitution properly.

Reply 8

physicskid123

Original post by HapaxOromenon3

So it does work for this given example? I don't see the problem here officer. how would you do it?

Reply 9

HapaxOromenon3

Original post by physicskid123

So it does work for this given example? I don't see the problem here officer. how would you do it?

The point is that your method is a trick only applicable to a small class of problems. Therefore it's not good mathematics, and it would be better simply to substitute u = x^2-4, dx = du/2x, to give int of x*sqrt(u)*du/2x = int of 1/2*sqrt(u) du, which can be trivially integrated and thus the answer found.