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Use the properties of integrals to verify the inequality

  1. Aug 23, 2011 #1
    1. The problem statement, all variables and given/known data
    ∫(from pi/4 to pi/2)sin x/x ≤ 1/√2.


    2. Relevant equations



    3. The attempt at a solution

    I know the pi/4≤x≤pi/2 and so 1/√2 ≤ sin x ≤ 1 and i have tried to manipulate this to no end and it has annoyed the living daylights out of me
     
  2. jcsd
  3. Aug 23, 2011 #2

    Mark44

    Staff: Mentor

    sin(x) is increasing on your interval, and so is x. What about sin(x)/x? Is this function increasing, decreasing, or neither? Can you find upper and lower bounds on the values of sin(x)/x on the given interval?
     
  4. Aug 23, 2011 #3
    Yes I tried that,
    the derivative of the function sin(x)/x would be: [x*cos(x)-sin(x)]/[x^2]

    the denominator will be bigger than 0. sin(x) is between 1/sqrt(2) and 1 cos(x) is between 0 and 1/sqrt(2). If I was sure that x is smaller that on or equal to one then I could conclude that the derivative is negative because sinx >= cos x on the interval. But x is between pi/4(smaller than one) and pi/2(bigger than one)
     
  5. Aug 23, 2011 #4
    YES NOW I GOT IT!!!!!

    [x*cos(x)-sin(x)]/[x^2] will be bigger than 0 because:

    if 0<=x<=pi/2 then x<=tan(x) so x<=sin(x)/cos(x) xcos(x)<=sin(x) xcos(x)-sin(x)<=0

    thus the derivative is negative or 0 on the interval
    so sin(x)/x <= sin(pi/4)/(pi/4) because it is decreasing

    so ∫(from pi/4 to pi/2)sin x/x <= ∫(from pi/4 to pi/2)1/sqrt(2)*4/pi = 1/sqrt(2)*4/pi*(pi/2-pi/4)= 1/sqrt(2)



    YEEEEEEEEEEEEE. I have been working on this stupid problem for 3 days.

    Thank you for helping out.
     
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