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Problem with Very Easy Integral

  1. Nov 25, 2009 #1
    1. The problem statement, all variables and given/known data

    Evaluate the integral

    [tex]\int^{2}_{1}3u^{2} du[/tex]

    2. Relevant equations

    None

    3. The attempt at a solution

    Ok...I can solve this problem fine but I am worried my work up to the answer isn't even close to what the answer book shows. So I am wondering if I should be doing it like the answer book shows, or if my way is fine. I've gotten all of the answers correct so far using my method.

    Here is how I solved it.

    [tex]\int^{2}_{1}3u^{2} du = \frac{3u^{3}}{3} = u^{3} = 2^{3}-1^{3} = 7[/tex]

    Here is where I get confused and how the answer book solves it.

    [tex]\int^{2}_{1}3u^{2} du = 3\int^{2}_{1}u^{2} du = 3\left[\int^{2}_{0}u^{2} du - \int^{1}_{0}u^{2} du\right][/tex]

    And then just plug and chug...

    [tex]= 3\left(\left[\frac{2^{3}}{3} - \frac{0^{3}}{3}\right] - \left[\frac{1^{3}}{3} - \frac{0^{3}}{3}\right]\right)[/tex] this eventually simplifies down to 7.

    This probably all doesn't make sense as I am sure the context of where we are in class would determine how one solves this. We just learned Riemann Sums and this section we are doing definate integrals.

    From what you can gather does my way look okay? And where does the [tex]\int^{1}_{0}u^{2} du\right][/tex] come from? I think that is where I am most confused since [0, 1] isn't in the interval.

    Let me know if you need anymore information.
     
  2. jcsd
  3. Nov 25, 2009 #2

    Mark44

    Staff: Mentor

    If you look at the integral in terms of area beneath the curve y = 3u2, what your book does makes sense. The area beneath this curve from u = 1 to u = 2 equals the area beneath the curve from u = 0 to u = 2 minus the area beneath the curve from u = 0 to u = 1.
    Or in terms of definite integrals,
    [tex]\int_{u = 1}^2 3u^2~du~=~\int_{u = 0}^2 3u^2~du~-~\int_{u = 0}^1 3u^2~du [/tex]

    If you add [tex]\int_{u = 0}^1 3u^2~du[/tex] to both sides of the equation above, you get a new equation with the total area (the integral on the right) being equal to the two constituent areas (the integrals on the left).
     
  4. Nov 25, 2009 #3
    So is the method I used wrong?

    The way I did it is a lot easier (at least to me), but if I should be doing it the book way I'll start doing that. It seems like the professor taught the wrong section or something because we didn't go over any of the assignment stuff.
     
  5. Nov 25, 2009 #4

    Mark44

    Staff: Mentor

    Your way is fine. I don't know why the book did it their way - I was just explaining what they were doing, not justifying it or saying that it was better.
     
  6. Nov 25, 2009 #5
    That makes sense. Thanks for the help, it's appreciated.
     
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