Originally Posted by martin
just in case anyone still doesn't understand have a look at this (rather stupid) example.
int(x-6)dx
if you forgot that you could just do each bit separately you could let u=x-6 then dx=du
so int(x-6)dx
= int(u)du
= 1/2 u^2 + C
= 1/2 (x-6)^2 +C
= 1/2 x^2 - 6x + 18 + C
and here you would just write it as 1/2*x^2 -6x + K (noting that K = 18 + C).
This seems pretty stupid but it can come up in more subtle ways, I'll see if I can find an example.
subtle examples usually involve integrals in (ln) form, esp. ones that involve trig in them as well, since you can transform them into many forms which can produce constants in the process.
anyways, the example you gave isn't "stupid" at all. in fact it's exact how a lot of ppl would actually end up with constants when integrating indefinite integrals:
whether or not a constant comes up is dependent on the way most ppl approach the integration - and as such most ppl doing the HSC (esp. in 2 unit) would follow the way most textbooks would specify or define the antiderivative for common cases (eg. yellow fitzpatrick):
one of the first things ppl learn from HSC textbooks (even before proper integral calculus is taught) is that ---> the
primitive or antiderivative of expressions of the form
(ax+b)^n is
((ax+b)^(n+1))/(a(n+1)) .where a,b are real constants and n a positive integer.
so when n=1, then by following that rule most 2u ppl would automatically end up with (1/2)(x-6)^2 from integrating (x-6), which then produces a constant upon expansion.
so constants appear in both definite and indefinite integrals depending on the way one would approach the integration i suppose.
P.S. of course the question remains as to why one would go on to expand (1/2)(x-6)^2 to check for constants in the first place. the integral is more elegantly left in factorised form and there won't be a question in the HSC where one would need to intentionally eliminate constants in their integral. there's no need to expand which is in fact a waste of time in an exam.
