Intergration Constants (1 Viewer)

[Robbo_m8]

Right in an intergration question using substitution it would be possible to end up with a constant term in the intergral along with the constant of intergration C.
eg x + 6 + C
This doesnt really happen except if substitution is used, but is this a correct answer? Obviously the 6 can be implied in the C, but are marks lost for chucking that fella in there, if it has legitimately come out of the working?

 

[m_isk]

it wouldn't hurt to say X+6+C = X+ M (where M is a constant equal to 6+C).

 

[SteaLaZ]

*agrees with m_isk

 

[Sirius Black]

How would u get a "6" in an indefinite integration? It must be a definite integral then it doesn't require the "+c" thingie

 

[Emma-Jayde]

How? You can't integrate and get a known constant can you? Unless it IS a definate integral

 

[Robbo_m8]

say if u are asked to integrete sin^2(x-pi/4). I attacked this by saying let x-pi/4 = u and obviously du=dx, so its rather simple. When the function of x is subbed in later pi/8 occurs as a constant.

 

[Sirius Black]

I didn't get a constant in the end but I didn't use the ur substitution method

 

[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.

 

[who_loves_maths]

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.