Another Way to Express Calculus

[Jason76]

\(\displaystyle u = 5x\)

\(\displaystyle du = 5dx \rightarrow\)

\(\displaystyle \dfrac{1}{5}du = dx\)

Let's say I had:

\(\displaystyle \int 5x \dx \rightarrow\)

\(\displaystyle \int u \dfrac{1}{5}du \rightarrow\)

\(\displaystyle \dfrac{1}{5}\int u \rightarrow\)

\(\displaystyle \dfrac{1}{5} u + C \rightarrow\)

\(\displaystyle \dfrac{1}{5} 5x + C \rightarrow\)

\(\displaystyle x + C\)

This is a different way from what I was doing, but perhaps it makes things more clear. This is called back-substitution by some forum members on here.

 

[Deleted member 4993]

\(\displaystyle u = 5x\)

\(\displaystyle du = 5dx \rightarrow\)

\(\displaystyle \dfrac{1}{5}du = dx\)

Let's say I had:

\(\displaystyle \int 5x \dx \rightarrow\)

\(\displaystyle \int u \dfrac{1}{5}du \rightarrow\)

\(\displaystyle \dfrac{1}{5}\int u \rightarrow\) .................. where did du go?

\(\displaystyle \dfrac{1}{5} u + C \rightarrow\) .................... how is that? Does not follow from above even after replacing missing du

\(\displaystyle \dfrac{1}{5} 5x + C \rightarrow\)

\(\displaystyle x + C\)

This is a different way from what I was doing, but perhaps it makes things more clear. This is called back-substitution by some forum members on here.

What is the problem that you are trying solve in this example? Is it:

\(\displaystyle \displaystyle \int 5\ \ x \ \ dx\) ?

Please review your post for errors prior to posting.

 

[mmm4444bot]

Jason, x+c is not an antiderivative for 5x.

This ought to be obvious!

The derivative of x+c is 1.

How do you get 5x from that?!

Again, I'm wondering why you do not spend your resources studying algebra. You could be preparing for your precalculus class. Instead, you push symbols around threads without caring about the meaning or standards, in future subjects that you have yet to receive instruction no less...baffling, to me...like goofing off, to me... :cool:

 

[Jason76]

Jason, x+c is not an antiderivative for 5x.

This ought to be obvious!

The derivative of x+c is 1.

How do you get 5x from that?!

Again, I'm wondering why you do not spend your resources studying algebra. You could be preparing for your precalculus class. Instead, you push symbols around threads without caring about the meaning or standards, in future subjects that you have yet to receive instruction no less...baffling, to me...like goofing off, to me... :cool:

I'm just trying to analyze what a poster (sorabon) sent to me via private message. That was his way of doing it. If he wants to come on and clarify his method, then that would be helpful. Actually, I was doing things a different way, until I got the message. But I do appreciate his help, but I need to clarify this stuff.

 

[mmm4444bot]

I'm just trying to analyze what a poster (sorabon) sent to me via private message.

That is not what you wrote, Jason. You posted this thread as "another way to express calculus".

How are readers of your threads supposed to determine when you're trying to figure out portions of private messages, unless you say so?

You presented the antiderivative in your original post as an exercise, followed by incorrect steps. Soroban did not send you those steps, and he never told you that x+c is an antiderivative for 5x.

The rest of us cannot determine what soroban sent you because it's private. If you want to discuss a private message, you may do that privately with the author. If you have a specific question about something you see off the boards, then start a thread about what you're actually looking at and ask your specific questions.