Simple question(s) that needs an answer (1 Viewer)

turtle_2468

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Riemann sum = 1 way of doing integrals.
You know how you approximate area by using really small rectangles = integration? That essentially means that you add up those rectangles then see the limit as the width of the triangles goes to zero (ie no. rectangles goes to infinity).. when you do that, you take the limit of a sum, (you can see this in any 4U textbook). That sum is called the Riemann sum.

Miscellaneous info:
There is more than one way to do integrals, namely Lebesgue and Riemann integration. Seeing Lebesgue integration only gets taught in 3rd year uni, and I still haven't taken the course, I can only tell you that it's supposedly more flexible. (ie you can integrate around curves and crazy stuff!)
 

turtle_2468

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In principle btw, the safer the better. But then, if you spend say 1 minute writing those 2 lines of tricky notation, that's 1 less minute spent doing another q/at least scabbing part marks for things you can't do...
 

underthesun

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It's also what my teacher says. But i thought i've seen somewhere that when you get a correct answer to "find the volume or stuff", you get full marks?

The reason i'll write it down is because I can never be sure of the full mark stuff..
 

Richard Lee

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The scale measures the mass which doesn't change no matter where it is, not weight which is determined by the gravity where it is. Even around the Earth, u will find the weight of object will change due to the position since the Earch is not a real round.

If not clear, contact me.
If any question, go to the thread "Any harder question!"
 

Fosweb

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For answer, go to quote on next line:

Originally posted by underthesun
Great, now the answer to number 3 came from the person It should have come from :D
 

turtle_2468

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the point is, though, the scale measures WEIGHT, and assuming that gravity is constant uses that weight to convert to mass...

On earth, the weight changes, but as it very closely approximates to a sphere, g only varies from 9.8...(whatever the value is) by a v small amount, so the approximation is valid. Whereas g on the moon is 1.6 or so, causing a big inaccuracy in this method (see my post above..)
 

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