You're not going to die! Don't worry!
Here's the big secret:
You're adding up areas of rectangles. The big scary hard part is finding a value for the height of each rectangle.
First, decide how many rectangles you want. The more rectangles, the more accurate the answer, but the more work!
Let's do four.
Your interval is 2 units wide, 2/4 = .5, so your rectangles will be .5 units wide.
Your rectangles go from:
1 to 1.5, 1.5 to 2, 2 to 2.5, 2.5 to 3
Now choose a value in each of those four intervals to be a "representative" value of that interval. You can choose the smallest number (Left Endpoint), largest number (Right endpoint), middle number (Midpoint), or anything you want.
Let's use the Left Endpoints.
The Left Endpoints of each of our rectangles are:
1, 1.5, 2 and 2.5
How high is the graph above each of these points?
The way to figure this out is to plug each of these numbers into your function, 2x + 1:
1, 1.5, 2 and 2.5 become:
3, 4, 5 and 6
So our four rectangles each have a width of .5, and heights of 3,4,5 and 6, giving us:
3*.5 + 4*.5 + 5*.5 + 6*.5 = 9
Notice that this answer is close to the exact answer of 10, but not equal. The more rectangles you use, the better your answer.
See if you can do this problem again with 8 rectangles and Left Endpoints. Your answer will be closer to 10
(Hint: the widths of your rectangles will be .25; your rectangles will go from what number to what number?)
2007-01-21 13:19:26
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answer #1
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answered by Anonymous
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A Riemann Sum of f over [a, b] is the sum
Sum(i=1,n) f(xi)*(x2-x1)
f(x) = 2x+1
The question is, into how many rectangles are you going to subdivide this area? The more rectangles you subdivide this by, the closer you will get to the real answer.
Think of it this way: graph the line. Realize that you're attempting to compute the area of a trapezoid. Riemann sums makes you APPROXIMATE the total area by breaking up the real area into RECTANGLES. So, by just drawing your graph, you'll see that for each rectangle you draw, you'll have a tiny triangle left over (the summation of all the triangles' areas equals the total error in your approximation).
The more rectangles you have, the smaller the triangles, therefore, the smaller the area.
Why calculus is brilliant is because it makes the triangles' areas infinitesimal.
Now, back to the story: if you set n = 4, that means your rectangles will be .5 wide.
Your sum will look like:
A1 + A2 + A3 + A4
where A1 is the area of your first Riemann rectangle, and so forth.
f(x)=2x+1
x0 = 1
x1 = 1.5
x2 = 2
x3 = 2.5
x4 = 3
so, f(x1) = 2*1.5+1 = 4
and x1 - x0 = .5
so A1 = 4*.5 = 2
By doing this successively for all 4 areas, you'll see that each rectangle has an area of 2 and the total area will be 8. Not bad for an approximation.
Now do the same calculation assuming that you break the areas up into n=10 rectangles.
A1 would be .2 wide and, judging the height off of the right side of the rectangle: f(1.2)=2*1.2+1 = 3.4 ==> A1 = .68
Do this for the remainder...
2007-01-21 21:10:08
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answer #2
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answered by mjatthebeeb 3
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it's been very long time since i left this riemann sum many years ago. really hate it. so i want to forget about it
all i can say is the answer can't be 10:
i give you a hint:
when you integrate (2x+1)dx, you'll get x^2 + x + C; where C is a constant. you missed this C my friend.
Unless C=0.
need to find C value
2007-01-21 21:30:16
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answer #3
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answered by ¥op 6
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Even though I finished both calculus in single variable and multivariables, I still can't do Riemann's sums. Believe it or not, integration makes your life alot easier. Or probably just ask the Prof. that will make your life alot easier.
2007-01-21 20:47:47
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answer #4
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answered by Lu 1
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I can't even understand the question but what I can I would quess
(41-7+65=56802)
2007-01-21 20:46:28
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answer #5
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answered by ? 2
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First of all, You are not going to die if you don't know the answer!
Second of all, I have NO idea!!
2007-01-21 20:46:42
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answer #6
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answered by ♥Sasukeluver♥ 3
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I hate math, I actually thought it was a serious question about life. Thanks for the two points!
2007-01-21 20:53:54
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answer #7
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answered by Jerrysberries 4
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wow sounds hard im only in pre cal so i cant help ya there
2007-01-21 20:45:49
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answer #8
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answered by Anonymous
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why you go to deth for it. keer control mind. and then you can solve the problem
2007-01-21 20:47:20
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answer #9
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answered by keral 6
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sorry looks hard i am kinda smart but i don't no that
sorry
2007-01-21 20:49:07
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answer #10
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answered by mini 2
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