Brainfreeze, can't integrate $x-1$

Question

If $f(x)=x-1$, I want to find $$ \int (x-1) \ dx $$. I make a variable substitution z = x-1, dx = dz so it becomes $$ \int z \ dz \\ = \frac{z^2}{2} + C $$. If I now substitute back $z=x-1,$ I get: $$ \int x-1 \ dx = \frac{(x-1)^2}{2} + C $$

Now if I do it another way without variable substitution I get: $$ \int x \ dx - \int 1 \ dx = \frac{x^2}{2} - x + C$$

What is going on here since these answers seem different to me? Are they both right but the constans $C$ are different or what am I missing here?

Two antiderivatives differs in a constant. Just expand the 1st solution — FormerMath, Aug 19 '21 at 01:13
One way to avoid such ambiguity is to work with definite integrals instead, e.g., $$F_0(x)=\int_0^x (x'-1),dx'=\frac12 x^2-x,\quad F_1(x) =\int_1^x (x'-1),dx'=\frac12 (x-1)^2.$$ Note that $F_1(x)-F_0(x)=\int_0^1 (x-1),dx = -1/2$, so the two indeed differ by a constant. Whether this is an improvement will, in part, depend on your stomach for dummy variables. — Semiclassical, Aug 19 '21 at 01:42
@Deane I absolutely agree. Every time I see a $\int$ without a domain of integration I cringe - what does it mean geometrically? If one wants an arbitrary integral they should write $\int_a^x (t-1),\textrm{d}t$ for an arbitrary constant $a$. — , Aug 19 '21 at 01:53
Check the derivatives of both the answers, i.e, another way to check if they differ by a constant. — Aman Kushwaha, Aug 19 '21 at 03:20
Does this answer your question? Getting different answers when integrating using different techniques — Arjun Vyavaharkar, Aug 19 '21 at 18:10

score 5 · Answer 1 · answered Aug 19 '21 at 01:24

5

Both answers are correct. Indefinite integrals are defined up to a constant C.

answered Aug 19 '21 at 01:24

Sidharth Ghoshal

18,359

score 3 · Accepted Answer · answered Aug 19 '21 at 01:25

3

Both answers are correct.

$$\begin{align}\frac{(x-1)^2}{2}&=\frac{x^2-2x+1}{2}\\ &=\frac{x^2}2-x+\frac{1}{2} \end{align}$$

So your two solutions differ by a constant.

answered Aug 19 '21 at 01:25

Thomas Andrews

186,215

Which equivalent answer is considered the most simplified or generally preferred form? – user946772 Aug 19 '21 at 01:30
2

There isn’t really a preferred answer, as far as I know. I think $x^2/2-x+C$ solution is simpler to derive, and generalizes to all polynomials, but there is an elegance to the substitution method, too, and might even be preferred when integrating something like $(x-1)^3.$ – Thomas Andrews Aug 19 '21 at 01:34

bjcolby15 · Answer 3 · 2021-08-19T01:36:51.250

1

First solution: $${\dfrac {(x-1)^2}{2} + C = \dfrac {(x^2-2x+1)}{2}+C = \dfrac {x^2}{2}-x + \dfrac {1}{2}+C}.$$ The $\dfrac {1}{2} +C$ is also constant; thus, both solutions are correct.

edited Aug 19 '21 at 01:36

answered Aug 19 '21 at 01:29

bjcolby15

4,210

Which equivalent answer (entailing both solutions differing by a constant of integration) is considered the most simplified or generally preferred form? The most expanded vs factorized one? – user946772 Aug 19 '21 at 01:31
1

Why write $\rightarrow$ when $=$ will do? – Thomas Andrews Aug 19 '21 at 01:35
It depends on how much you want it simplified. If you prefer your solution neater, use the first method of substitution and do not expand your solution; otherwise, you can integrate term separately. – bjcolby15 Aug 19 '21 at 01:36
1

@ThomasAndrews You know, I had that first and then thought arrows would be better, but you're right - equals signs work much better. I've corrected my post. – bjcolby15 Aug 19 '21 at 01:37

Brainfreeze, can't integrate $x-1$

3 Answers3