Using comparison theorem for integrals to prove an inequality











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The problem states to show that $int_{0}^{frac{pi}{2}} frac{sin(x)}{x(x+5)} dx < frac{pi}{10}$ using the fact that $int_{a}^{b}f(x)dx<int_{a}^{b}g(x)dx$ and $sin(x)<x$.



I attempt to set up the inequality as such, but I am having a hard time finding a continuous g(x) on the bounds that is easily integrable so that the inequality holds. Maybe I can find some $c in [a,b]$ s.t. $g(c) > f(c)$ which would give me a constant that is easily integrable, but I am not sure if this will prove that $int_{0}^{frac{pi}{2}} frac{sin(x)}{x(x+5)} dx < frac{pi}{10}$. If someone who is more creative than me could help me out that would be awesome.






real-analysis integration







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    up vote
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    down vote

    favorite












    The problem states to show that $int_{0}^{frac{pi}{2}} frac{sin(x)}{x(x+5)} dx < frac{pi}{10}$ using the fact that $int_{a}^{b}f(x)dx<int_{a}^{b}g(x)dx$ and $sin(x)<x$.



    I attempt to set up the inequality as such, but I am having a hard time finding a continuous g(x) on the bounds that is easily integrable so that the inequality holds. Maybe I can find some $c in [a,b]$ s.t. $g(c) > f(c)$ which would give me a constant that is easily integrable, but I am not sure if this will prove that $int_{0}^{frac{pi}{2}} frac{sin(x)}{x(x+5)} dx < frac{pi}{10}$. If someone who is more creative than me could help me out that would be awesome.






    real-analysis integration







    share|cite|improve this question
























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      The problem states to show that $int_{0}^{frac{pi}{2}} frac{sin(x)}{x(x+5)} dx < frac{pi}{10}$ using the fact that $int_{a}^{b}f(x)dx<int_{a}^{b}g(x)dx$ and $sin(x)<x$.



      I attempt to set up the inequality as such, but I am having a hard time finding a continuous g(x) on the bounds that is easily integrable so that the inequality holds. Maybe I can find some $c in [a,b]$ s.t. $g(c) > f(c)$ which would give me a constant that is easily integrable, but I am not sure if this will prove that $int_{0}^{frac{pi}{2}} frac{sin(x)}{x(x+5)} dx < frac{pi}{10}$. If someone who is more creative than me could help me out that would be awesome.






      real-analysis integration







      share|cite|improve this question













      The problem states to show that $int_{0}^{frac{pi}{2}} frac{sin(x)}{x(x+5)} dx < frac{pi}{10}$ using the fact that $int_{a}^{b}f(x)dx<int_{a}^{b}g(x)dx$ and $sin(x)<x$.



      I attempt to set up the inequality as such, but I am having a hard time finding a continuous g(x) on the bounds that is easily integrable so that the inequality holds. Maybe I can find some $c in [a,b]$ s.t. $g(c) > f(c)$ which would give me a constant that is easily integrable, but I am not sure if this will prove that $int_{0}^{frac{pi}{2}} frac{sin(x)}{x(x+5)} dx < frac{pi}{10}$. If someone who is more creative than me could help me out that would be awesome.






      real-analysis integration




      real-analysis integration






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      asked Nov 29 at 3:31









      hkj447

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      244






















          1 Answer
          1






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          up vote
          1
          down vote



          accepted










          HINTS:



          Note that for $xin [0,pi/2]$, we have



          $$0le frac{sin(x)}{x}le 1$$



          and



          $$0le frac{1}{x+5}le frac15$$






          share|cite|improve this answer





















          • ah thank you, I was trying to put the x on top and getting nasty integrals.
            – hkj447
            Nov 29 at 3:39










          • You're welcome. My pleasure.
            – Mark Viola
            Nov 29 at 3:43











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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          1
          down vote



          accepted










          HINTS:



          Note that for $xin [0,pi/2]$, we have



          $$0le frac{sin(x)}{x}le 1$$



          and



          $$0le frac{1}{x+5}le frac15$$






          share|cite|improve this answer





















          • ah thank you, I was trying to put the x on top and getting nasty integrals.
            – hkj447
            Nov 29 at 3:39










          • You're welcome. My pleasure.
            – Mark Viola
            Nov 29 at 3:43















          up vote
          1
          down vote



          accepted










          HINTS:



          Note that for $xin [0,pi/2]$, we have



          $$0le frac{sin(x)}{x}le 1$$



          and



          $$0le frac{1}{x+5}le frac15$$






          share|cite|improve this answer





















          • ah thank you, I was trying to put the x on top and getting nasty integrals.
            – hkj447
            Nov 29 at 3:39










          • You're welcome. My pleasure.
            – Mark Viola
            Nov 29 at 3:43













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          HINTS:



          Note that for $xin [0,pi/2]$, we have



          $$0le frac{sin(x)}{x}le 1$$



          and



          $$0le frac{1}{x+5}le frac15$$






          share|cite|improve this answer












          HINTS:



          Note that for $xin [0,pi/2]$, we have



          $$0le frac{sin(x)}{x}le 1$$



          and



          $$0le frac{1}{x+5}le frac15$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 29 at 3:35









          Mark Viola

          130k1273170




          130k1273170












          • ah thank you, I was trying to put the x on top and getting nasty integrals.
            – hkj447
            Nov 29 at 3:39










          • You're welcome. My pleasure.
            – Mark Viola
            Nov 29 at 3:43


















          • ah thank you, I was trying to put the x on top and getting nasty integrals.
            – hkj447
            Nov 29 at 3:39










          • You're welcome. My pleasure.
            – Mark Viola
            Nov 29 at 3:43
















          ah thank you, I was trying to put the x on top and getting nasty integrals.
          – hkj447
          Nov 29 at 3:39




          ah thank you, I was trying to put the x on top and getting nasty integrals.
          – hkj447
          Nov 29 at 3:39












          You're welcome. My pleasure.
          – Mark Viola
          Nov 29 at 3:43




          You're welcome. My pleasure.
          – Mark Viola
          Nov 29 at 3:43


















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