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Homework Help: Definite Integral question

  1. Mar 18, 2010 #1
    1. The problem statement, all variables and given/known data

    The letter W is defined as 33.gif . The value of W, correct to the nearest hundredth, is?

    2. Relevant equations

    3. The attempt at a solution

    Now, I had this question on a test and got it wrong. I get 0.93, which I think is correct. Apparently the answer is 0.24. I call BS on my teacher.
  2. jcsd
  3. Mar 18, 2010 #2


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    Well, you're incorrect, as any calculus capable calculator will confirm. So, instead of directing profanities towards your teacher, perhaps you should post your calculations so we can point out where your error(s) are.
  4. Mar 18, 2010 #3
    How am I wrong? The function is negative from pi/2 to 2 and the area below the x-axis has to be calculated separately as an absolute value and added to the area above the x-axis. Of course an integral calculator will get the question wrong, they can't handle area below the x-axis.
  5. Mar 18, 2010 #4
    Who said anything about the area under a curve? It says compute a definite integral, not "Find the area under the curve."
  6. Mar 18, 2010 #5
    The definite integral is the area under the curve...
  7. Mar 18, 2010 #6
    That's a common misconception. The definite integral can get you the area under the curve, but that doesn't mean it is.
  8. Mar 18, 2010 #7
    So what you're saying is that if it doesn't ask for me to compute the area for a definite integral, that I should just calculate it straight away and disregard if it is negative?
  9. Mar 18, 2010 #8

    Remember the fundamental theorem of Calculus:

    For a sufficiently nice function,

    \int_a^b f(x) dx = F(b) - F(a)

    Where a and b are the bounds and F is the anti-derivative.

    It doesn't care about negative or positive. All it cares about is the value of F at b and a.
  10. Mar 18, 2010 #9
    Alright. That makes sense. Thank you.
  11. Mar 18, 2010 #10
    The definite integral is the signed, net area under the curve. That's quite easy to forget.

    That's why

    [tex] \int_0^{2\pi} \cos x \ dx[/tex]

    evaluates to zero. There is the same amount of area above the x-axis as there is below it.
  12. Mar 18, 2010 #11


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    That's not the definition of a definite integral. When in doubt, you should always turn to the definitions.In this case, the definition involves a limit of a summation. Thinking of a definite integral as being an area under a curve is only strictly true when the curve is above the x-axis over the entire interval.

    If you were asked to find the area between the curve [itex]f(x)=4\sin x\cos x[/itex] and the x-axis on the interval [itex]x\in [1,2][/itex], then your method would be correct. However, that wasn't what you were asked.
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