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To compare an integral with an identity

  1. Feb 26, 2008 #1
    Integral inequality and comparison

    1. The problem statement, all variables and given/known data

    Prove the inequality

    [tex]\frac{2}{(n+1) \cdot \pi} \leq a_n \leq \frac{2}{n \pi}}[/tex]

    where [tex]a_n = \int_{0}^{\pi} \frac{sin(x)}{n \cdot \pi +x} dx[/tex]

    and [tex]n \geq 1[/tex]

    3. The attempt at a solution


    If n increased the left side of inequality because smaller every time it increases while the integral becomes larger, then by camparison the right side of the inequality always will become larger, than the integral.

    Is this proof enough?

    Last edited: Feb 26, 2008
  2. jcsd
  3. Feb 26, 2008 #2
    well try to use the fact that

    [tex]n\pi\leq \pi n+x \leq n\pi+\pi[/tex], then

    [tex]\frac{1}{n\pi+\pi}\leq\frac{1}{n\pi+x}\leq\frac{1}{n\pi}[/tex], multiply both sides by sinx, we get, notice that x goes only from 0 to pi, so sinx is always positive, so

    [tex]\frac{sin x}{n\pi+\pi}\leq\frac{sin x}{n\pi+x}\leq\frac{sin x}{n\pi}[/tex], now

    [tex]\int_0^{\pi}\frac{sin x }{n\pi+\pi}dx\leq\int_0^{\pi}\frac{sin x}{n\pi+x}dx\leq\int_0^{\pi}\frac{sin x}{n\pi}dx[/tex],
    [tex]\frac{1}{n\pi+\pi}\int_0^{\pi}sin xdx\leq\int_0^{\pi}\frac{sin x}{n\pi+x}dx\leq\frac{1}{n\pi}\int_0^{\pi}\sin xdx[/tex],
    I think this will work!!
  4. Feb 26, 2008 #3
    I shouldn't have done the whole thing for you, since you are on the homework forum, and you are supposed to show your work, however, i hope u got it now!
  5. Feb 26, 2008 #4
    I get it now. I will present something that I have been working later I hope you will review.
  6. Feb 26, 2008 #5
    well, someone will!!
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