Where does the $dy$ go in the process of integration?

$\begingroup$

I'm currently looking at integration (calculus 1).

For example $y = \int 3x dx$

I don't understand how we ended up with just "$y$" on the left hand side.

For example $dy/dx = 3x$, then $dy = 3x dx$

so $\int dy = \int 3x dx$ which is not the same. Is $\int dy$ somehow the same as $y$?

I get confused as some use $dx$ as just notation while others actually use it as a term that can be moved around.

$\endgroup$ 1

4 Answers

$\begingroup$

Yes, ∫dy is somehow the same as y!

imagine dy as a bit of y and ∫ as a sum over the bits.

$\endgroup$ 2 $\begingroup$

I agreed with Dr. S. Graubner. Just see the fact that, $\displaystyle \frac{dy}{dx}=3x$ so $dy=3xdx$ and integrating $\displaystyle \int dy=\int 3xdx\implies y=\int 3xdx+C\implies y=\frac{3}{2}x^2+C+C'=\frac{3}{2}x^2+C''$ where $C''=C+C'$, being the constant of integration.

$\endgroup$ $\begingroup$

Assuming you're comfortable with the idea of functions, I think the best way for you to understand what's happening with integration and differentiation is as an operator on a function. What this means is that we're transforming one function into another function, just like how addition is an operator to turn one number into another one.

Take the graph of $y=x^2$. We can modify the graph in different ways. For instance, $x^2\mapsto x^2+1$ would raise up the graph $1$ unit or $x^2 \mapsto (x-1)^2$ would shift the graph $1$ unit to the right. But in the same way, when we differentiate: $x^2\mapsto \frac{\mathrm{d}}{\mathrm{d}x}x^2$ we are transforming the graph into another one ($2x$). I.e. using a specific operation to one function into another. The same is true for integration (but beware the constant of integration).

I don't understand how we ended up with just "y" on the left hand side.

When we say $y=\int3x\ \mathrm{d}x$, the LHS is defined as a function of $x$ using the expression on the RHS. So the reason we have "just y" on the LHS is the same reason as in $y=mx+c$. In either case, we choose particular values of $x$ (e.g. $x=3$) to put into the function by substituting into the RHS and we get back a value of $y$ (e.g. $y=5$).

So, we start with a function ($3x$), then transform our function with the integration operator $3x\mapsto\int3x\ \mathrm{d}x$. Then $y$ is defined as the transformed function.

For example dy/dx = 3x, then dy = 3x dx

The problem here is that you're doing something which in some ways isn't allowed and is slightly controversial. $\mathrm{d}y$ and $\mathrm{d}x$ aren't really separate terms that can have specific values so $\frac{\mathrm{d}y}{\mathrm{d}x}$ isn't really a fraction, it's an operation ($\frac{\mathrm{d}}{\mathrm{d}x}$) on a function ($y$). So $y\mapsto \frac{\mathrm{d}}{\mathrm{d}x}y$. You can get away with treating $\mathrm{d}y$ and $\mathrm{d}x$ as individual terms if you use differentials but that's beyond the scope of this answer but discussed elsewhere on the site.

Is ∫dy somehow the same as y?

Yes, with a caveat. Again, think of integration as an operator on a function: $\int\cdot\ \mathrm{d}y$ but this time, the variable in the function is $y$, not $x$. What is the function we're integrating? Well it's just the constant $1$ but for simplicity, it's not been written. This is slightly nuanced because $1$ may not seem like a function. But in fat it's the constant function that returns $1$ for any input. I.e. a flat horizontal graph.

So starting with our constant function, $1$, we transform it with the integration operator $1\mapsto\int 1\ \mathrm{d}y$. We can simplify the RHS using the rule of indices: $1=y^0$ and then the power rule for integration. So, we have $\int 1\ \mathrm{d}y=\int y^0\ \mathrm{d}y=\frac{y^1}{1}+C=y+C$.

$C$ is the caveat I mentioned earlier and is called the constant of integration. When we apply the integration operator, it turns out we don't just get transformed into one function, we could have $y+1$ or $y+2$ or $y-3$, etc. So, this is concisely expressed by the notation $y+C$, where $C$ is any constant. Again, this is discussed elsewhere on the site.

$\endgroup$ $\begingroup$

Because integration is the inverse process of differentiation and you are integrating derivative of $y$ which is same as derivative of $y+c$, so they cancelled out and you will remain with $y+c$, where $c$ is constant of integration. Also we use $dx$ just as notation when we are working with definite integral, because after putting limits that variable will vanish but not in the case of indefinite integral.

$\endgroup$

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy