How do I find the area under the curve using right end rectangles for f(x) = x?

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The discussion focuses on calculating the area under the curve for the function f(x) = x over the interval [0, 1] using right end rectangles with 10 partitions. The method involves partitioning the interval into segments I_{j}=[(j-1)/10, j/10] for j=1 to 10. The area A_{sub,10} is computed using the Riemann Sum formula, resulting in A_{sub,10} = (1/10^2) * Σ(j-1) from j=1 to 10, which simplifies to a specific numerical value.

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QuantumTheory
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After reading my Calculus II book, I am having problems actually applying what I've learned to a problem. It's very difficult for me to learn off just reading books, (and then applying what I've learned).

So here's the problem, find the area under the curve using right end rectangles for: f(x) = x; [0,1]. With 10 rectangles, A(sub10) = ?

Thank you. I just understand it but I cannot apply it to what I've learned, that is what is hard. I don't have a math teacher; I'm doing this on my own accord.
 
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See this page for an explanation of partitioning over a closed interval and then determining the "Riemann Sum" of the partitions a similar problem to what you gave - I think they use f(x) = x^2 and the interval is [0,2].

Perion
 
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Quantumtheory:
1) Partition your interval as follows:
[tex]I_{j}=[\frac{j-1}{10},\frac{j}{10}],j=1,2...10[/tex]
2) The minimum value of f(x)=x on [tex]I_{j}[/tex] is obviously [tex]\frac{j-1}{10}[/tex]
3) Hence, since interval length is 1/10, you get:
[tex]A_{sub,10}=\sum_{j=1}^{10}\frac{j-1}{10}\frac{1}{10}=\frac{1}{10^{2}}\sum_{j=1}^{10}(j-1)[/tex]
 
Last edited:

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