For 94Q: I think it's integral from 0 to *pi* of sin(x)dx. -cos(x)] from 0 to pi -> -cos(pi) - -cos(0) = 1 + 1 = 2 I think the trick is to divide the inside constant by the outside one(inside and outside the actual functions in the limits), right?
At 95 for example we don't have to start with 0. It is actually easier to start with 1 and go to 4. So x_i = 1 + deltax * i and our integrating function becomes sqrt(x) instead of sqrt(1+x). I found that pretty interesting. I should have watched on.... At Q100 we could have done it to and would just have lnx as a function. Wow I really like Riemann Summs now.
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"It's 2022, don't write Δx just write dx"
One of my favorites quotes from you
For 94Q: I think it's integral from 0 to *pi* of sin(x)dx.
-cos(x)] from 0 to pi -> -cos(pi) - -cos(0) = 1 + 1 = 2
I think the trick is to divide the inside constant by the outside one(inside and outside the actual functions in the limits), right?
At 95 for example we don't have to start with 0. It is actually easier to start with 1 and go to 4. So x_i = 1 + deltax * i and our integrating function becomes sqrt(x) instead of sqrt(1+x).
I found that pretty interesting.
I should have watched on....
At Q100 we could have done it to and would just have lnx as a function.
Wow I really like Riemann Summs now.
Wow! This is fabulous - thank you.
It will be nice to apply this to *Riemann zeta function* somehow😎
Thank you!
19:40 . I saw you think iπ =ln(-1) 🤣🤣
very informative :)
I am so sorry, if I didi something wrong
=(