The reason is simple : Unlike indefinite integrals, definite integrals will give us results that does not involved in terms of the variable that will be integrated with respect to (in other words, functions that are not involved in terms of the variable that will be integrated with respect to). That's why, regardless of the variables that will be integrated with respect to, the result of a definite integral is still the same.
Could you elaborate the "in other words" part ? What do you mean by "functions that aren't involved in terms if the variable that will be integrated in respect to)
People get confused because they think we are re-substituting the "u substitution". We are not doing that, we just take the u and turn into x without any relation.
It was strange to me that people didn't think you could just use whatever variables in definite integrals, series, products, etc.. Each one you evaluate is just a number, regardless of what letter you use.
Awesome video! Derivatives and integrals have been in my mathematical toolkit for quite some time, but one thing that I never understood was why they are miraculously linear functions. It's not intuitive to me why the integral of 1/(1+x^2) + e^-x is actually the sum of the two separate integrals. Perhaps that is a video for another time!
The way i understand it is that for definite integrals, the answer is just a number. Regardless of what letter you use, you will get the same number back.
I still don't understand because it seems like saying the equation x^m (x+1)^n = x^n(x+1)^m which is false (except for m=n). I tried integrating over the interval (0;3) x/(x+1)² and upon applying what you said I obtained (x-1)/x² but the later function isn't defined at x=0
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The hardest part of Calculus is accepting things that seem too good to be true
Exactly!
Nicest comment! 🙂
The reason is simple : Unlike indefinite integrals, definite integrals will give us results that does not involved in terms of the variable that will be integrated with respect to (in other words, functions that are not involved in terms of the variable that will be integrated with respect to). That's why, regardless of the variables that will be integrated with respect to, the result of a definite integral is still the same.
This is exactly how i reasoned it, glad you made this comment :)
Could you elaborate the "in other words" part ?
What do you mean by "functions that aren't involved in terms if the variable that will be integrated in respect to)
People get confused because they think we are re-substituting the "u substitution". We are not doing that, we just take the u and turn into x without any relation.
But if we already have draw the relation like x=1/t , then after some steps how can we put x in place of t in the equation .?
at this point hes just flexing the amount of pens he can hold and draw with
It was strange to me that people didn't think you could just use whatever variables in definite integrals, series, products, etc.. Each one you evaluate is just a number, regardless of what letter you use.
Awesome video! Derivatives and integrals have been in my mathematical toolkit for quite some time, but one thing that I never understood was why they are miraculously linear functions. It's not intuitive to me why the integral of 1/(1+x^2) + e^-x is actually the sum of the two separate integrals. Perhaps that is a video for another time!
The way i understand it is that for definite integrals, the answer is just a number. Regardless of what letter you use, you will get the same number back.
Great video!! getting ready for grad school and had to understand why this was the way it was. You hit the nail right on the head, excellent job!
Very convincing explanation. Thanks
Can you make a video explain wtf is a double integral or triple integral or something like partial derivative?
pretty simple to proof! more such kind of videos, more proofs!
I don't get why it would matter...
I still don't understand because it seems like saying the equation x^m (x+1)^n = x^n(x+1)^m which is false (except for m=n). I tried integrating over the interval (0;3) x/(x+1)² and upon applying what you said I obtained (x-1)/x² but the later function isn't defined at x=0
Was there anyone at all that didn't understand this???
891537 people raised their hands.....
Nice bro!!!!
Excelent!
thank you so fucking much