Simple question(s) (1 Viewer)

[hscishard]

Integrate (x+1)^2 via:
a) Expanding the brackets
b) Reverse chain rule/sub method/ w/e

Then tell me if the C in a) is different or not different to the C in b).
I think it isn't..

 

[random-1005]

Integrate (x+1)^2 via:
a) Expanding the brackets
b) Reverse chain rule/sub method/ w/e

Then tell me if the C in a) is different or not different to the C in b).
I think it isn't..

lol what, the c isnt different

integral (ax+b)^n = (ax+b)^(n+1) / (a*(n+1)) +C

just like a question i asked my maths teacher in yr 11, if integral 1/x = lnx +C, shouldnt integral 5/x be 5lnx +5C, well yes and no

because in the second case each "C" is only worth 1/5 of the C in the first part so to speak, all that matters is there is a constant

 

[MrSparky]

im assuming ur talking about the +c or the constant that comes from integrating
if u integrate (x+1)^2 this does not necessarily mean its equal to the integral of (x+1)^2, why? because they may differ by constant so to answer your questions, using those 2 methods, the c's you get will be different, however if your boundaries for both integrals are the same, then there will be no +c, meaning that the 2 integrals will be equal

 

[hscishard]

If you expand the primitive in b), you will get a constant.
I know the question doesn't really make sense.

But if they are the exact same primitives, the Cs are different, right?

 

[random-1005]

If you expand the primitive in b), you will get a constant.
I know the question doesn't really make sense.

But if they are the exact primitives, the Cs are different, right?

the Cs can be anything, they can be equal, they dont have to be though

dy/dx of (x+1)^2 +5 is the same as dy/dx of (x+1)^2 +1

C's can be different or equal, but they would never ask something like that in exam

 

[hscishard]

No like..

Ok. Let's say they both pass through (1,1)
The + C in a) will have a different value to +C in b). Is that right?

 

[random-1005]

No like..

Ok. Let's say they both pass through (1,1)
The + C in a) will have a different value to +C in b). Is that right?

no, why should they be different?

if u evaluate the C (trivial calculation), you will see they are the same

 

[MrSparky]

if u have a set of values that are common to both functions then the +c will be the same when your trying to find that 'c'

 

[hscishard]

Ummm. I was thinking:

x^2+2x+1
(x^3)/3 +x^2 + x + C

(x+1)^3/3 + C
(x^3)/3+ x^2 + x + 1/3 + C

the "c's won't be equal right? If the two primitives are equal

 

[random-1005]

Ummm. I was thinking:

x^2+2x+1
(x^3)/3 +x^2 + x + C

(x+1)^3/3 + C
(x^3)/3+ x^2 + x + 1/3 + C

the "c's won't be equal right?

1/3 ( 1(x^3) + 3x^2 +3x +1)


expansion is correct, you will still get same c, dont worry, no ok, the C in the second one will be 1/3 less than the c in the first one,

but the total constant in expression 1 will still equal total constant in expression 2, so you are sort of right

 

[hscishard]

Yea cool. That's what I was trying to verfiy.

 

[tommykins]

explicit value of C's can only be found given initial conditions.

Doesn't really matter what it is, as long as a C is there to represent some arbitary constant.