# Finding the area under a graph using Riemann Sum question

• Nov 7th 2009, 08:06 PM
br1tt204
Finding the area under a graph using Riemann Sum question
I would appreciate any help I can get with this problem! My prof usually gives us a few practice problems we should know. One of them this week was finding the area under the graph y=x^3 using Riemann Sum. Even in high school, I have had a hard time grasping how to find the area using Riemann Sum. I do not think it was explained to me right the first time.

Problem: find the area underneath the graph y=x^3 using eight subdivisions and rectangles underneath y.

The range is from 0 to 1

in my prof's example (of y=x^2) he used the following information:
the area of each rectangle is i^2/n^3 where i=(n-1)

so: [0^2]/n^3 + [1^2]/n^3 + [2^2]/n^3.... [(n-1)^2]/n^3

and because: [m(m+1)]/2=1+2+3...m
and 1^2+2^2...m^2=m(m+1)(2m+1)/6

then: [0^2]/n^3 + [1^2]/n^3 + [2^2]/n^3.... [(n-1)^2]/n^3 = (n-1)(n)2n-1)/

I know that the above is confusing (at least in IMO) but I really would appreciate any guidance at all. Let me know if left any information out or was not clear enough
• Nov 8th 2009, 02:25 AM
HallsofIvy
Quote:

Originally Posted by br1tt204
I would appreciate any help I can get with this problem! My prof usually gives us a few practice problems we should know. One of them this week was finding the area under the graph y=x^3 using Riemann Sum. Even in high school, I have had a hard time grasping how to find the area using Riemann Sum. I do not think it was explained to me right the first time.

Problem: find the area underneath the graph y=x^3 using eight subdivisions and rectangles underneath y.

The range is from 0 to 1

Quote:

In my prof's example (of y=x^2) he used the following information:
the area of each rectangle is i^2/n^3 where i=(n-1)
Okay, he is one end of the subinterval as "x" for the entire interval: if the length of the entire interval is 1, as it is in your problem, and we divide it into n subintervals, then each subinterval has length 1/n and the endpoint of the \$\displaystyle i^{th}\$ interval is given by x= i/n. The height of the rectangle is f(x)= f(i/n)= \$\displaystyle i^2/n^2\$. The base of the rectangle is the length of the subinterval, 1/n, so the area is "base times height" \$\displaystyle (1/n)(i^2/n^2)= i^2/n^3\$.

("where i= (n-1)" is incorrect. i goes from 0 to n-1. n-1 is the final value of i.)

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so: [0^2]/n^3 + [1^2]/n^3 + [2^2]/n^3.... [(n-1)^2]/n^3
and this is equal to [0^2+ 1^2+ 2^2+ ...+ (n-1)^2]/n^3

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and because: [m(m+1)]/2=1+2+3...m
Not relevant here.

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and 1^2+2^2...m^2=m(m+1)(2m+1)/6
This is the formula you use.

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then: [0^2]/n^3 + [1^2]/n^3 + [2^2]/n^3.... [(n-1)^2]/n^3 = (n-1)(n)2n-1)/
[0^2+ 1^2+ 2^2+ ...+ (n-1)^2]/n^3= [(n-1)((n-1)+1)(2(n-1)+1)]/6n^3= (n-1)n(2n-1)/6n^3. you dropped the denominator.

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I know that the above is confusing (at least in IMO) but I really would appreciate any guidance at all. Let me know if left any information out or was not clear enough
Okay, since you are dividing the interval from 0 to 1 into 8 subintervals, each subinterval has length 1/8 and the "\$\displaystyle i^{th}\$" interval has left endpoint i/8. At x= i/8, f(x)= x^3= (i/8)^3= i^3/8^3.

Each rectangle has base 1/8 and height i^3/8^3, it has area (1/8)(i^3/8^3)= i^3/8^4. Add those up: i= 0 to 7. The sum can be written as
0/8^4+ 1/8^4+ 2^3/8^4+ 3^3/8^4+ 4^3/8^4+ 5^3/8^4+ 6^3/8^4+ 7^3/8^4= [0+ 1+ 8+ 27+ 64+ 125+ 216+ 343]/4096. Now it's just arithmetic.
• Nov 22nd 2009, 10:41 PM
br1tt204
Thank you so so much! That was it! (Clapping)