# Questions about Indefinite Integrals and U - Substitution

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 $$(4x^3)/3 + C$$.

I'm able to do u-substitution, but I am confused as to one aspect of it. In the above 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 are $$f(x)$$ and $$dx$$ being multiplied together in the indefinite integral.

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.

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

Note: I had origianlly asked this question on "Questions about U-Substitution and Integration" but I felt my original question there was not accurately portraying what I was asking.

Thank you for helping.

• 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

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 $$\int (x^2+1) dx = x^3/3 +x +c$$

• High school teachers might have a different opinion about omitting the $dx$. – R zu Nov 13 '18 at 5:06
• @ 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 5:14
• @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

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}