Question

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If integral of dx is x

Then what is the integral of 1/dx

And what is the integral of (dx)²


Maths aryabhatta's a great question for uh pupil ‼️​

Answer 1

Answer:

$\large\boxed{\sf{\int dx = x + c}}$

Step-by-step explanation:

To find the integral of,

$\displaystyle \int dx$

We can writ this as,

$= \displaystyle \int1.dx$

Or, we can further write as,

$= \displaystyle \int {x}^{0} dx$

But, we know that,

• $\displaystyle \int {x}^{n} = \dfrac{ {x}^{n} }{n + 1}$

Therefore, we will get,

$= \dfrac{ {x}^{0 + 1} }{0 + 1} + c \\ \\ = x + c$

Where, c is an arbitrary constant.

Now, to find the integral of 1/dx.

Actually, 1/dx doesn't have any meaning in mathematics.

It's an invalid expression or function.

So, we can't find it's integral.

Similarly, to find the integral of (dx)²

It's also, an invalid expression or function in mathematics.

So, we can't find it's integral.

Keep in mind, both of these expressions doesn't hold any meaning in mathematics.

Answer 2

Well the first statement seems totally wrong!

$\begin{minipage}{10cm}``\textsf{ x , added with a constant or not, is the integral of \mathbf{1} with respect to x , not of dx .}"\end{minipage}$

Because we see that,

$\displaystyle\longrightarrow\sf{\int dx=\int1\ dx}$

$\sf{dx}$ exists there alone due to that $\sf{1}$ !

$\displaystyle\longrightarrow\sf{\int dx=x+c}$

And if the integral constant is zero, then simply,

$\displaystyle\longrightarrow\sf{\int dx=x}$

And how the statement is possible?

We know the formula,

$\displaystyle\longrightarrow\sf{\int x^n\ dx=\dfrac{x^{n+1}}{n+1}+c,\quad n\neq-1}$

Hence we see that,

$\displaystyle\longrightarrow\sf{\int dx=\int1\ dx}$

$\displaystyle\longrightarrow\sf{\int dx=\int x^0\ dx}$

$\displaystyle\longrightarrow\sf{\int dx=\dfrac{x^{0+1}}{0+1}+c}$

$\displaystyle\longrightarrow\sf{\int dx=\dfrac{x^{1}}{1}+c}$

$\displaystyle\longrightarrow\sf{\int dx=x+c}$

If $\sf{c=0,}$

$\displaystyle\longrightarrow\sf{\int dx=x}$

Remember, we write some variables alone if their coefficients are 1 right?!

$``\textsf{ x+y is the same as 1x+1y .}"$

Likewise, $\displaystyle\mathsf{\int 1\ dx}$ is simply written as $\displaystyle\mathsf{\int dx,}$ which is wrongly taken as "the integral of $\sf{dx}$" here. That's the matter!

$\begin{minipage}{10cm}\displaystyle``\textsf{Integral of x wrt x is written symbolically as \mathsf{\int x\ dx} .}"\\\\``\textsf{Similarly, integral of dx wrt dx should be written as \mathsf{\int dx\ dx} .}"\end{minipage}$

I just took the integral here "with respect to $\sf{x,}$" though the first statement $``\textsf{if the integral of dx is x}"$ does not imply that $dx$ is integrated about which variable, $x$ or $y$ or $2x\ !$

By the way what does the term $\sf{dx}$ actually mean?

The infinitesimally small sections like $\sf{dx,\ dy,\ dt,\ d\theta,\ etc.}$ according to which the given function has to be integrated are called the integral operators.

$``\textsf{Only the \mathbf{functions} about the mentioned variables are able to be integrated}"$

But integral operators are no longer functions. They represent the very small variation about them in the particular function as we go along it. Then how $\sf{dx}$ can be integrated?!

Similarly, how $\sf{\dfrac{1}{dx}}}$ and $\sf{(dx)^2}$ can be integrated?! They could be if and only if the following functions were defined.

• $\sf{f(x)=dx}$

• $\sf{f(x)=\dfrac{1}{dx},\ \ dx=0}$

• $\sf{f(x)=(dx)^2}$