I am faced with the following integral:

$\int(x-a) dx$

I can think of two approaches to solve it:

1) Separating into two terms as follows:

$\int x dx -a\int 1 dx$

From which the result would be:

$\int (x-a) dx = x^2/2 - ax$

2) Substituting $u =x-a$; $du=dx$

$\int u du = u^2 / 2 = (x - a)^2 / 2$

If we expand this solution:

$\int (x-a) dx = x^2/2 -ax + a^2/2$

Now, clearly, $x^2/2 -ax+a^2/2$ is not equal to $x^2/2 - ax$. So is either method invalid for some reason? Or am I making a mistake elsewhere?

I am aware this is probably a very dumb question, so I thank you very much for your attention and help!

  • 3
    $\begingroup$ Well that $a^2/2$ term is a constant. Usually just a big $+C$. If you differentiate both, you get the same answer. $\endgroup$ – combinatorialist46Carey2 Jun 20 at 19:31

Don't keep forgetting the constant of integration please :P

$$\int(x-a)dx = x^2/2 - ax + C_1$$ $$\int u dx = u^2/2 = (x-a)^2/2 = x^2/2-ax+a^2/2 + C_2$$ and these two are as equal as they can be, as $a^2/2$ is a constant you can pick $C_1 = a^2/2 + C_2$.

  • 1
    $\begingroup$ Should you not label the constant of integration in both differently like C1, C2? $\endgroup$ – Rishi Jun 21 at 5:43
  • $\begingroup$ @Rishi you can if you wish. The point is that it should be there. I will make an edit. $\endgroup$ – Steve Jun 21 at 13:55

Remember that for an indefinite integral you must add a constant of integration. The two integrals are then the same in terms of their dependence on $x$, up to a constant. If you would replace the indefinite integral with a definite integral, the difference would work out in how the bounds of the integral change when you make the change of variables.


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