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April 16th, 2017, 09:55 AM   #1
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Find the Riemann Sum

Find the Riemann sum for f(x) = x2 + 3x over the interval [0, 8], where
x0 = 0, x1 = 1, x2 = 2, x3 = 7, and x4 = 8,
and where
c1 = 1, c2 = 2, c3 = 5, and c4 = 8.

I'm confused. What do the c's and x's mean again? I know, I'm dumb and have no clue on where to begin.
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April 16th, 2017, 04:06 PM   #2
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Without further context, they could mean anything.
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April 16th, 2017, 04:41 PM   #3
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Originally Posted by mathman View Post
Without further context, they could mean anything.
Well I learned about it, I just forgot cause it's been a week. Could you lead me in the right direction with the problem though?
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April 16th, 2017, 07:22 PM   #4
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Originally Posted by nbg273 View Post
Find the Riemann sum for $f(x) = x^2 + 3x$ over the interval [0, 8], where
$x_0 = 0, x_1 = 1, x_2 = 2, x_3 = 7, and x_4 = 8$,
and where
$c_1 = 1, c_2 = 2, c_3 = 5, and c_4 = 8.$

I'm confused. What do the c's and x's mean again? I know, I'm dumb and have no clue on where to begin.
There are also different variations of the Riemann sum. Is this the full question?

It's likely the x's are your known nodes. i.e., $f(x_0) = f(0) = 0$. etc.
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April 16th, 2017, 08:33 PM   #5
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There are also different variations of the Riemann sum. Is this the full question?

It's likely the x's are your known nodes. i.e., $f(x_0) = f(0) = 0$. etc.
Well there's a graph, but I can't show you that cause it's on my e-homework. But it's literally just a graph, no other information. It looks like the (x, y) coordinates from the interval [0, 8] are (0, 0), (2, 10), (4, 25), (6, 50), (8, 85)... I don't think that's 100% accurate, but that's what I see and all the information I can get from the graph.
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April 18th, 2017, 04:46 AM   #6
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Surely you have a textbook that deals with the 'Riemann sum"? The Riemann sum is a way to approximate the area under the graph of a function and so the integral of the function. Given f(x) and x from a to b, divide the interval, [a, b], into n sub-intervals. Those are the "x" values you are given: x0= 0, x1= 1, x2= 2, x3= 7, and x4= 8. Now, construct rectangles over each interval, [0, 1], [1, 2], [2, 7], and [7, 8] to approximate the area under the curve and above each interval. Take the height of each rectangle to be the value, f(x), at some point in that interval. It looks to me like those points are the "c"s you are given. That is. with c1= 1, we imagine a rectangle with base from 0 to 1 and height f(1)= 1^2+ 3(1)= 4 (notice the "^" to indicate the power- "x^2" is clearer than just "x2"). That rectangle, with base 1- 0= 1 and height 4 has area 4.

With c2= 2, the rectangle over the interval [1, 2] has height f(2)= 2^2+ 3(2)= 10 so area (2- 1)(10)= 10.

With c3= 5, the rectangle over the interval [2, 7] has height f(5)= 5^2+ 3(5)= 40 so area (7- 2)(40)= 200.

With c4= 8, the rectangle over the interval [7, 8] has height f(8 )= 8^2+ 3(8 )= 88 so area (8- 7)(88 )= 88.

The Riemann sum, approximating the area under this curve, above the x-axis, between 0 and 8 is 4+ 10+ 200+ 88= 302.

Last edited by Country Boy; April 18th, 2017 at 04:48 AM.
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