# Working out +C from integration

• Jun 3rd 2009, 02:15 PM
Envy
Working out +C from integration
Hey I was doing an integration question and I got the answer:

${y}=6{x}-2{x}^2-{x}^3+c$ the final marking point on the question was to work out c. The point went through the origin so the end answer (which I had to look up) was ${y}=6{x}-2{x}^2-{x}^3$. How do I work out c if say the curve ${y}=6{x}-2{x}^2-{x}^3+c$ went through $(3,2)$?
• Jun 3rd 2009, 02:29 PM
e^(i*pi)
Quote:

Originally Posted by Envy
Hey I was doing an integration question and I got the answer:

${y}=6{x}-2{x}^2-{x}^3+c$ the final marking point on the question was to work out c. The point went through the origin so the end answer (which I had to look up) was ${y}=6{x}-2{x}^2-{x}^3$. How do I work out c if say the curve ${y}=6{x}-2{x}^2-{x}^3+c$ went through $(3,2)$?

Sub in x=3 and y=2 and solve for c.

So ${y}=6{x}-2{x}^2-{x}^3+c$ went through $(3,2)$

becomes ${2}=6(3)-2(3)^2-(3)^3+c \:, \: c = 2+27+18-18= 29$
• Jun 3rd 2009, 02:32 PM
Envy
rearrange?
so literally rearrange to make c the subject and substitute in the x and y values?
• Jun 3rd 2009, 02:33 PM
e^(i*pi)
Quote:

Originally Posted by Envy
so literally rearrange to make ${c}$ the subject and substitute the ${x}$ and ${y}$?

Yeah, that's all there is to it.
• Jun 3rd 2009, 02:34 PM
Envy
Thanks
Cheers, it seems so obvious now :)
• Jun 3rd 2009, 02:36 PM
e^(i*pi)
Quote:

Originally Posted by Envy
Cheers, it seems so obvious now :)

Yeah, it's even simpler when it's the origin because it usually means that c=0. It only really gets harder to decide when you get an initial condition