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Homework Help: Finding the area of a region between curve and horizontal axis

  1. Jan 24, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the area under y = 6x3 - 2 for 5 <_ x <_ 10

    2. Relevant equations

    3. The attempt at a solution

    I'm not sure how it wants us to do it; do we use sigma notation or the general right/left-hand sums?

    Using sigma notation, delta T = 10-5/n

    T0= 5/n (?)
    T1= 10/n
    Ti = 5i/n

    5/n[tex]\sum6(5i/n)^3 -2[/tex]

    750/n^4 -2 (Sigma) i^3

    I'm not sure if I can bring the 2 out like that, and I'm also not sure what i^3 equals (is it (n(n+1)/2)^3?)
    Last edited: Jan 24, 2010
  2. jcsd
  3. Jan 24, 2010 #2
    Without a bit more information it's difficult to know how they expect you to find the answer, because there are numerous ways to do so. You could simply take the definite integral, but it looks like (from your work) you're expected to use the limit definition and Riemann Sums.

    Your delta t is correct, but t0 for these problems is usually equal to the lower boundary. t1 and your other subintervals would only be correct if the lower boundary was zero. Think about that.

    And no, you're not allowed to pull out the -2 like that. Try to take a look at your Sigma rules to see what you can do to simplify the problem once you find the value of f(ti).

    The sum of i3, with i = 1 to n is [n(n+1)/2]2.
  4. Jan 24, 2010 #3
    Looking back, a T0 of 5 does make a lot more sense. T1, then, would be 5 +5/n, T2 would be 5+10/n, so Ti would be 5+5i/n?


    I've started simplifying this and it doesn't look right.
  5. Jan 24, 2010 #4
    Well, what were you trying and what doesn't look right? :)
  6. Jan 31, 2010 #5
    Sorry for the late response.

    It just seemed too "long" for a generic textbook problem; asked the instructor, turns out we were supposed to use a calculator on those ones (other included cos and natural logs inside the summation which I had no idea how to pull out).

    Used a calculator and the antiderivative, came to the same answer for both. Thank you for helping even though it turned out to be a waste of time.
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