When we’re integrating sometimes it seems we need to do the +c thing and sometimes we don’t. What’s the story? Do we need it or not?
So the short answer is: Yes, you need it.
The slightly longer answer is: Yes, you need it for the indefinite integral but you can safely ignore it if you are working on a definite integral.
Ok, but why?
Good question. So let’s look at why we need it in the first place by going back and differentiating a few functions:
So all of these different functions have the same derivative (
So there are, in fact, an infinite number of functions that have a derivative of
If I’m now going to go back in the opposite direction and integrate
And so I include a Constant of Integration, which is commonly written as 
I know there is some constant there, some number, but I don’t know what it is. It could be 0 or -37 or 892000000 or any of an infinite number of other possibilities. Without more information, I simply don’t know.
What we just did was an indefinite integral and the constant of integration is essential in representing it as what it is: not just a single function, but an infinite set of possible functions.
When it comes to calculating a definite integral what we are doing is evaluating an indefinite integral between a given set of limits. For example:
![\large{\boldsymbol{\int_1^3\! 6x \,\mathrm{d}x=\left[3x^2+c\right]_1^3}}](http://s0.wp.com/latex.php?latex=%5Clarge%7B%5Cboldsymbol%7B%5Cint_1%5E3%5C%21+6x+%5C%2C%5Cmathrm%7Bd%7Dx%3D%5Cleft%5B3x%5E2%2Bc%5Cright%5D_1%5E3%7D%7D&bg=ffffff&fg=000&s=0 "\large{\boldsymbol{\int_1^3\! 6x \,\mathrm{d}x=\left[3x^2+c\right]_1^3}}")
You’ve probably been told that you can just ignore the constant now, you don’t need it. That’s true, but let’s leave it in there for now and step through the calculation to see why it is that we can ignore it.
![\large{\boldsymbol{\int_1^3\! 6x \,\mathrm{d}x=\left[3x^2+c\right]_1^3}}\\ \large{\boldsymbol{\quad\quad\quad\quad =\left(3(3)^2+c\right)-\left(3(1)^2+c\right)}}\\ \large{\boldsymbol{\quad\quad\quad\quad =\left(27+c\right)-\left(3+c\right)}}\\ \large{\boldsymbol{\quad\quad\quad\quad =27+c-3-c}}\\ \large{\boldsymbol{\quad\quad\quad\quad =24+c-c}}\\ \large{\boldsymbol{\quad\quad\quad\quad =24}}](http://s0.wp.com/latex.php?latex=%5Clarge%7B%5Cboldsymbol%7B%5Cint_1%5E3%5C%21+6x+%5C%2C%5Cmathrm%7Bd%7Dx%3D%5Cleft%5B3x%5E2%2Bc%5Cright%5D_1%5E3%7D%7D%5C%5C+%5Clarge%7B%5Cboldsymbol%7B%5Cquad%5Cquad%5Cquad%5Cquad+%3D%5Cleft%283%283%29%5E2%2Bc%5Cright%29-%5Cleft%283%281%29%5E2%2Bc%5Cright%29%7D%7D%5C%5C+%5Clarge%7B%5Cboldsymbol%7B%5Cquad%5Cquad%5Cquad%5Cquad+%3D%5Cleft%2827%2Bc%5Cright%29-%5Cleft%283%2Bc%5Cright%29%7D%7D%5C%5C+%5Clarge%7B%5Cboldsymbol%7B%5Cquad%5Cquad%5Cquad%5Cquad+%3D27%2Bc-3-c%7D%7D%5C%5C+%5Clarge%7B%5Cboldsymbol%7B%5Cquad%5Cquad%5Cquad%5Cquad+%3D24%2Bc-c%7D%7D%5C%5C+%5Clarge%7B%5Cboldsymbol%7B%5Cquad%5Cquad%5Cquad%5Cquad+%3D24%7D%7D&bg=ffffff&fg=000&s=0 "\large{\boldsymbol{\int_1^3\! 6x \,\mathrm{d}x=\left[3x^2+c\right]_1^3}}\\ \large{\boldsymbol{\quad\quad\quad\quad =\left(3(3)^2+c\right)-\left(3(1)^2+c\right)}}\\ \large{\boldsymbol{\quad\quad\quad\quad =\left(27+c\right)-\left(3+c\right)}}\\ \large{\boldsymbol{\quad\quad\quad\quad =27+c-3-c}}\\ \large{\boldsymbol{\quad\quad\quad\quad =24+c-c}}\\ \large{\boldsymbol{\quad\quad\quad\quad =24}}")
The two
No matter which of our infinite possibilities that
So we take a little shortcut, and just leave it out altogether.