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Let me begin with an example. If we were to integrate the indefinite integral of $(2x)^2$ with respect to $x$ with u-substitution, we would first say that $u=2x$ and therefore $du=2dx$. In order to substitute $du$ in for $dx$, we have to multiply the inside of the integral by $2$ and the outside by $0.5$. Then we have the indefinite integral of $u^2 du$. We say this equals $(u^3)/6 + C$ which gives us a final answer of $(4x^3)/3 + C$.

Now, I understand why in a definite integral $dx$ and $f(x)$ are multiplied. I believe it is because the definition of the definite integral from $a$ to $b$ is the limit as "$n$" goes to infinity of the sum of "$n$" rectangles under the curve, each with area $f(x)$ times $dx$. Notice in the definition $f(x)$ and $dx$ are multiplied so in a definite integral $f(x)$ and $dx$ should also be multiplied.

In the indefinite integral example, the only way we can substitute in $du$ is if the $dx$ and $(2x)^2$ in the original integral were being multiplied. This is what confuses me: why in the indefinite integral are $f(x)$ and $dx$ multiplied?

Note: I had origianlly asked this question at What does multiplying the integrand by $dx$ mean in an indefinite integral? but I feel that my original question there was not accurately portraying what I was asking.

ryang
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Peter Blood
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  • The argument you gave for definite integrals makes proper sense in context of non-standard analysis. In the usual study of Riemann integrals the $dx$ is just a notation used for historical reasons and some amount of practical convenience. The same goes for indefinite integrals. No one is multiplying $f(x) $ and $dx$. – Paramanand Singh Nov 13 '18 at 15:27

2 Answers2

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Let me begin with an example. If we were to integrate the indefinite integral of (2x)^2 with respect to x with u-substitution, we would first say that u=2x and therefore du=2dx. In order to substitute du in for dx, we have to multiply the inside of the integral by 2 and the outside by 0.5.

However, why in the indefinite integral are f(x) and dx multiplied?

The multiply and divide is helpful to make the substitutions clear.   $u\gets 2x$ and $\mathsf d u\gets 2\mathsf d x$. $$\begin{align}\int (2x)^2\cdot\tfrac 22\mathsf d x &= \int (u)^2\cdot\tfrac 22~\mathsf d x \\&= \int (u)^2\cdot\tfrac 12~\mathsf d u \\&= \dfrac {u^3}{2\cdot 3}+c\\&=\dfrac{(2x)^3}{2\cdot 3}+c\\&=\dfrac {4x^3}3+c\end{align}$$

Community
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Graham Kemp
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Indefinite integral $$\int f(x) dx$$ is by definition an anti-derivative of $f(x)$ and the $dx$ simply indicates that it is the anti-derivative with respect to $x$.

One does not split the above notation into $\int $, $f(x)$, and $dx$

You may drop the $dx$ if there is no confusion. $$\int f(x) $$ is as good as $$\int f(x) dx$$

For example $$\int x^2+1 = x^3/3 +x +c $$ is quite acceptable and it is the same as $$\int (x^2+1) dx = x^3/3 +x +c $$

Mohammad Riazi-Kermani
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    High school teachers might have a different opinion about omitting the $dx$. – R zu Nov 13 '18 at 05:06
  • @ Rzu I am not suggesting to drop $dx$ but I am saying that with or without $dx$ the indefinite integral means an anti-derivative and $dx$ is there to tell that the anti-derivative is with respect to $x$ – Mohammad Riazi-Kermani Nov 13 '18 at 05:14
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    @Rzu: the use of $dx$ is not only to appease high school teachers but it offers some practical advantages as far as tricks of the trade are concerned. – Paramanand Singh Nov 13 '18 at 15:33