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Why is this integral positive or negative

  1. Jan 24, 2006 #1
    I have been asked to explain without evaluating the integrals why

    1) the integral of x cos x from 0 to pi/2 is positive and the integral of x cos x from pi/2 to pi is negative. And also would I expect x cos x from 0 to pi to be positive, zero or negative??? And why ???

    2) to find the indefinite integral of x cos x dx and hence the exact values of values of x cos x from 0 to pi/2, cos x from pi/2 to pi and x cos x from 0 to pi?

    Any help would be must appreciated. I know I haven't done anything so far so no need to have a go at me for it but I just wanted some help with it and even better the solutions obviously as this blimming thing is for in a few days time. Many thanks in advance

  2. jcsd
  3. Jan 24, 2006 #2


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    1) What are your thoughts on this? Have you drawn the graph?

    2) Have you learnt about integration by parts?
  4. Jan 24, 2006 #3
    [tex]\int x\cos(x)dx[/tex]

    By parts, u=x dv=cos(x)
    du = 1 v = sin(x)

    [tex]\int udv= uv - \int vdu[/tex]

    You can take it from there.
  5. Jan 25, 2006 #4


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    #1, What can you say about the integral of [tex]\int \limits_{\alpha} ^ {\beta} f(x) \ dx[/tex] (i.e greater or less than or equal to 0), if:
    (i) [itex]f(x) > 0 , \ \forall x \in ( \alpha, \ \beta )[/itex]
    (ii) [itex]f(x) < 0 , \ \forall x \in ( \alpha, \ \beta )[/itex]?
    You can draw a graph to see this. Remember that definite integral of some function from a to b will give you the area under the graph of that function from x = a to x = b.
    #2, as others have pointed out, this should be done by Integration by parts.
    You can either read the article there, or look at your text book. There should be something about Integration by parts.
    Can you go from here? If you still have problems, just shout out. :smile:
    Last edited: Jan 25, 2006
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