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RHS HELP!!! (Reimann Sums) (Note that I have NOT learned about integrals yet) Here's the Q: I'm given a curve where 0<=x<=10, and there are 5 rectangles, so n=5. I calculated delta x to be 2. I'm really confused how to use these two formulas to find both the LHS and the RHS. I've typed the two formulas below, in my reply. Please help!! I AM REALLY CONFUSED! Thank you!
\[\ \huge \text {LHS:} \sum_{i=0}^{n-1} f(x_i) Δ x \]
\[\ \ \huge \text {RHS:} \sum_{i=1}^{n} f(x_i) Δ x \]
well delta x = (b-a)/n
and what is your f(x)?
I'm not sure what you're asking. I need your function value.
\(\ \Huge \text{Oh, and: } x_i = a+iΔx \)
if you just started integrals you shouldve went over sigma notation im pretty sure, and you just expand it
This isn't helpful without the function. Otherwise the answer will be just theoretical stuff.
I don't have a function value... I'm only given a graph in my textbook and am told to find the lower estimate and an upper estimate.
And, we haven't even discussed what an integral is in class...
Could you give a description of what your graph is?
It looks something like:|dw:1360464858577:dw|
My confusion is I'm not sure how to use the two summation equations....
And then use the values I know from the graph
Okay, that's better. Well you first need to partition the interval into 5 equal intervals. For the left Riemann: for your values x = 0,2,4,6,8, draw a vertical line that TOUCHES the graph of the function. This vertical line is the height of the rectangle that you are estimating the curve with. Each rectangle's width is 2. |dw:1360465074815:dw| For the right hand do the same except for now, use let the x values be x = 2,4,6,8,10 |dw:1360465192357:dw|
Note how the left hand Riemann is an underestimation of area while the right hand Riemann is an over estimation of area.
@khoala4pham How do I use the sigma equations? Where do they come into this?
The sigma equations represent area. They mean that you sum up the areas of each individual rectangle. That is all that they mean. For the left hand Riemann, you take 2*(f(0) + f(2) + f(4) + f(6) + f(8)) For the right, 2*(f(2) + f(4) + f(6) + f(8) + f(10)) Why that though? Because for the RHR, the first rectangle has height f(0) and its width is 2. The second rectangle has height f(2) and width 2. So forth and so on.
How do you know to choose 0,2,4,6,8 and 2,4,6,8,10?
Well your a is zero in this case. a is the place where your interval begins. From your equation of the left hand Riemann, you see that x_i = a + i*(deltax) i runs from 0 to n-1. Well, n here is 5--the domain is partitioned into 5 intervals. That means that i = 0,1,2,3,4. Consequently x will be x = 0 + 2(0), 0 + 2(1), 0+2(2),0+2(3),0+2(4) x = 0,2,4,6,8. The same applies to the right Riemann.
THAT MAKES SO MUCH MORE SENSE! THANK YOU, @khoala4pham! I was getting confused with the \(\ \Huge \sum \text{ notation!} \)
The notation is highly ambiguous if you are not used to it. The x_i is a number. Deltax is also a number. You are summing numbers across indexes. How you found your index is obtained above.
Can you explain the RHS for "Well your a is zero in this case. a is the place where your interval begins. From your equation of the left hand Riemann, you see that x_i = a + i*(deltax) i runs from 0 to n-1. Well, n here is 5--the domain is partitioned into 5 intervals. That means that i = 0,1,2,3,4. Consequently x will be x = 0 + 2(0), 0 + 2(1), 0+2(2),0+2(3),0+2(4) x = 0,2,4,6,8. The same applies to the right Riemann." I dont really understand the RHS...
How you got the numbers 2,4,6,8,10
If i runs from 1 to n, so that means i runs 1 to 5...
I'm getting like 2, 5 ...? x_1= (a=0) + (1)(delta x = 2) = 2 x_2 = (a=1) + (2)( delta x = 2) =5
@khoala4pham Why do you keep 0 constant? Consequently x will be x = 0 + 2(0), 0 + 2(1), 0+2(2),0+2(3),0+2(4) x = 0,2,4,6,8.
your equation of x_i = a + i*(delta x). a = 0 by default. a is the value that you start your interval with. if your interval were from (20,50) then a = 20. If your interval were (-3,24), then a = -3. In your initial problem you told me that 0 <= x < = 10. Thus a = 0.