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Fp2 : series and integrals

I really don't get this topic :frown:
Can someone somehow explain it to me .
1i first of all. I dont get the over and underestimate part.
http://m.imgur.com/RfhVAQC
Thanks
Reply 1
Avatar for Zacken
Zacken
22
Original post by Super199
I really don't get this topic :frown:
Can someone somehow explain it to me .
1i first of all. I dont get the over and underestimate part.
http://m.imgur.com/RfhVAQC
Thanks


Why don't you sketch the curve and then think about drawing rectangles where the right part of the rectangle meets the curve and where the left part of the rectangle meets the curve. Sketch these on two different diagrams, can you see which is an overestimate and which is an underestimate?
Reply 2
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Super199
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Original post by Zacken
Why don't you sketch the curve and then think about drawing rectangles where the right part of the rectangle meets the curve and where the left part of the rectangle meets the curve. Sketch these on two different diagrams, can you see which is an overestimate and which is an underestimate?

The left is under and the right is over. I still dont get why you dont go upto x=2 for the under?
Reply 3
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Zacken
22
Original post by Super199
The left is under and the right is over. I still dont get why you dont go upto x=2 for the under?


I'm not sure what you mean? Could you post your working? You should have 4 rectangles that go up to x=2 I think.
Reply 4
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Super199
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Original post by Zacken
I'm not sure what you mean? Could you post your working? You should have 4 rectangles that go up to x=2 I think.


Idk tbh. But when we were doing it in class i remember doing
0.5×(1+0^3)^1/2 + 0.5(1+0.5^3)^1/2+0.5(1+1^3)^1/2 + 0.5(1+1.5^3)^1/2 but not including 0.5(1+2^3)^1/2 not sure why?
Reply 5
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Zacken
22
Original post by Super199
Idk tbh. But when we were doing it in class i remember doing
0.5×(1+0^3)^1/2 + 0.5(1+0.5^3)^1/2+0.5(1+1^3)^1/2 + 0.5(1+1.5^3)^1/2 but not including 0.5(1+2^3)^1/2 not sure why?


Sketch the triangle and you'll see why. It's a 'left'-triangle and the height is determined by the left-part of the rectangle touching the curve. So the rectangle 'left' part touches the curve at x=1.5 and then has a width of 0.5 which covers the region to x=2. Wait, let me sketch it real quick.
Reply 6
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Zacken
22
Original post by Super199
Idk tbh. But when we were doing it in class i remember doing
0.5×(1+0^3)^1/2 + 0.5(1+0.5^3)^1/2+0.5(1+1^3)^1/2 + 0.5(1+1.5^3)^1/2 but not including 0.5(1+2^3)^1/2 not sure why?


You're not sketching your graphs, are you?



Can you see how the height of the fourth rectangle is determined by sqrt(1+1.5^3)?
Reply 7
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Super199
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Original post by Zacken
You're not sketching your graphs, are you?



Can you see how the height of the fourth rectangle is determined by sqrt(1+1.5^3)?


oo is it where the left bit of the rectangle hits the curve. which is at x=1.5?
Reply 8
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Zacken
22
Original post by Super199
oo is it where the left bit of the rectangle hits the curve. which is at x=1.5?


Yes, and that's the left-rectangle sum thingy. This is why you need to skeeeetch.
Reply 9
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Super199
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Original post by Zacken
Yes, and that's the left-rectangle sum thingy. This is why you need to skeeeetch.


For the right-rectangle you don't include zero because the right bit doesn't touch it?
Reply 10
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Zacken
22
Original post by Super199
For the right-rectangle you don't include zero because the right bit doesn't touch it?


Indeed.

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