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I am confiused when it comes to these two formulas:

$\int 1 dx=x +C$

and

$\int a dx=ax +C$

which one would apply to $\int x\ dx$ ? and why

Can someone explain the difference?

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    $\begingroup$ Dear person that voted to close this question: how on earth is this question off topic? The OP clearly stated the question, and it doesn't look like it's homework. If you aren't going to explain to any of us why you voted to close, why bother voting to close? $\endgroup$
    – layman
    Oct 10, 2015 at 3:02

6 Answers 6

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Well, ask yourself:

  • Is "$x$" the same as "1"?

  • Is "$x$" the same as "$a$"?

If so, then that one applies. If neither is true, then neither apply, and you need some other rule . . .

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  • $\begingroup$ "$x$" could be the same as "$a$". You have to highlight for the OP that $a$ represents a number, not a variable. Otherwise, the OP can't answer your second question, because to them $x$ and $a$ are both letters... $\endgroup$
    – layman
    Oct 10, 2015 at 2:59
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None of them is applicable. The second one is exactly the first one when you take $a=1$.

None of them contain $x$. You need $\int xdx=x^2/2+C$.

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This is a great question.

$\int 1 \,dx$ is just notation for the antiderivative of $1$, i.e., the function that, when you differentiate it, gives you $1$. Well, we know the derivative of $x$ is $1$. So is the derivative of $x + 1$, $x + 2$, $x + \pi$, $x - 500$, $x + \dfrac{\sqrt{7}}{e}$. So, since all of these have derivative $1$, we say that the antiderivative can be any of them, i.e., it is $x + C$ for any constant real number $C$.

Now, if $a$ is any number (not a variable), then $\int a \,dx = \int a \cdot 1 \,dx = a \int 1 dx = a(x + C) = ax + C'$. Remember that we are allowed to pull constant numbers out of the integral; that's why I pulled out the $a$.

Now, to figure out $\int x \,dx$ you can't use any of the information above, because when we were working above, we were finding the antiderivative of a number. Here, we are finding the antiderivative of a variable $x$. We are asking for the function such that its derivative is $x$. This turns out to be $\dfrac{x^{2}}{2}$, but also $\dfrac{x^{2}}{2} + 1$, $\dfrac{x^{2}}{2} + 5$, $\dfrac{x^{2}}{2} -19$, $\dfrac{x^{2}}{2} + \pi$, etc. So, the antiderivative's general form is $\dfrac{x^{2}}{2} + C$ for a constant $C$.

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Neither applies directly, but you could apply integration by parts and use the first formula in the process.

$$\int xdx=x\int dx - \int xdx$$

directly leads to

$$\int xdx=\frac{x^2}{2}+C.$$

Can you fill in the intermediate steps?

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Initial commentary:

In my opinion, the notation for "antiderivation" and integration is among the worst of mathematics, alongside the notation for partial derivatives. All doubts of students regarding this notation are justified. For an answer regarding the notation for integration, see this answer I gave on another question.


Back to the matter at hand: What does this notation mean? It holds on itself a lot of dense information. Let's step back and try to define things again. Forget what you know. We take functions with domain $\mathbb{R}$ instead of an interval just for the sake of cleanness. Also, we always assume functions to be continuous.

Definition: An anti-derivative of a function $f: \mathbb{R} \rightarrow \mathbb{R}$ is a function $F$ such that $F'=f$. We denote such an $F$ by $\displaystyle \int f$.

It may not be obvious at first, but it follows from the mean value theorem that all such $F$'s are related by the fact that they differ by a constant. Hence, if we find one anti-derivative, every other one can be found by adding a constant. Therefore, it does us no harm to denote $\displaystyle \int f=F+c$, for instance. Non-uniqueness is not a big issue. Just keep in mind that we are considering $c$ to be a constant, and the equation is more to be interpreted as "every anti-derivative is of this form" than an actual equation. Note that no "dummy variables" or anything of the sort was introduced anywhere.

For example: Take the function

$$f: \mathbb{R} \rightarrow \mathbb{R}$$ $$x \mapsto x^2.$$

A straight-forward calculation leads us to the fact that the function

$$F: \mathbb{R} \rightarrow \mathbb{R}$$ $$x \mapsto \frac{x^3}{3}$$

is an anti-derivative for $f$. Hence, we have that $\displaystyle \int f =F+c$, or $\displaystyle (\int f) (x)=\frac{x^3}{3}+c$.

Now, what means $\int x dx$? First of all, you have to interpret this as:

$$\int x dx$$

$$\int \quad (x) \quad (dx)$$

The first thing to understand is that the second parenthesis gives you what is being taken to be the "input". The first parenthesis is the "expression $P$ in terms of the input". Therefore, what you are really evaluating, in our previous notation, is

$$\int (x \mapsto P(x)).$$

This justifies why:

$$\int 1 dx=x+c$$

After all, $x \mapsto 1$ is the constant function $\equiv 1$. Hence, $x$ is an antiderivative.

Also,

$$\int x dx= \frac{x^2}{2}+c$$

since $x \mapsto x$ has the function $x \mapsto \frac{x^2}{2}$ as anti-derivative. Note now that:

$$\int a dx=ax+c$$

Since $x \mapsto a$ is the constant function $\equiv a$, having $ax$ as anti-derivative.

The point is: whether you are taking an antiderivative or integrating, you are always working with functions. You are operating functions. This mainstream notation for antiderivative and integration just helps on immortalizing the confusing, "insightless" and naive interpretation of functions as expressions, a notion which, when finally recognized to be inherently limited, lead to considerable progress in mathematics.

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Neither applies to $\int x dx$. Both $\int 1 dx$ and $\int a dx$ are integrals of constant functions: $f(x) = 1$ and $g(x) = a$ respectively. With $\int x dx$ you're integrating a linear function, $h(x) = x$. What's a function whose derivative is $h$? Well, $ {x^2} / 2$. So: $$ \int x dx = \frac {x^2} 2 + C$$

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