On the bound of an integral of a differentiable function with bounded derivative

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Let $f:[0,1]to mathbb R$ be a differentiable function such that $sup_xin[0,1]|f'(x)|$ exists.



If $int_0^1xf(x)dx=0$, then how to show that $36|int_0^1x^2f(x) dx|le sup_xin[0,1]|f'(x)|$ ?






real-analysis integration derivatives continuity riemann-integration







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    up vote
    1
    down vote

    favorite












    Let $f:[0,1]to mathbb R$ be a differentiable function such that $sup_xin[0,1]|f'(x)|$ exists.



    If $int_0^1xf(x)dx=0$, then how to show that $36|int_0^1x^2f(x) dx|le sup_xin[0,1]|f'(x)|$ ?






    real-analysis integration derivatives continuity riemann-integration







    share|cite|improve this question























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Let $f:[0,1]to mathbb R$ be a differentiable function such that $sup_xin[0,1]|f'(x)|$ exists.



      If $int_0^1xf(x)dx=0$, then how to show that $36|int_0^1x^2f(x) dx|le sup_xin[0,1]|f'(x)|$ ?






      real-analysis integration derivatives continuity riemann-integration







      share|cite|improve this question













      Let $f:[0,1]to mathbb R$ be a differentiable function such that $sup_xin[0,1]|f'(x)|$ exists.



      If $int_0^1xf(x)dx=0$, then how to show that $36|int_0^1x^2f(x) dx|le sup_xin[0,1]|f'(x)|$ ?






      real-analysis integration derivatives continuity riemann-integration




      real-analysis integration derivatives continuity riemann-integration






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 4 hours ago









      user521337

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      351113




















          1 Answer
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          Hint. For any real $a$, by integration by parts,
          $$beginalignint_0^1x^2f(x) dx&=int_0^1(x^2-ax)f(x) dx\
          &=left[left(fracx^33-fracax^22right)f(x)right]_0^1-int_0^1left(fracx^33-fracax^22right)f'(x)dx\
          &=left(frac13-fraca2right)f(1)+int_0^1left(fracax^22-fracx^33right)f'(x)dxendalign$$
          Now choose an appropriate value for $a$. Can you take it from here?






          share|cite|improve this answer






















          • I'm not sure whether you did that integration by parts correctly or not ...
            – user521337
            3 hours ago










          • Sorry, I forgot a term...
            – Robert Z
            3 hours ago










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          1 Answer
          1






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          oldest

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          active

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          up vote
          4
          down vote



          accepted










          Hint. For any real $a$, by integration by parts,
          $$beginalignint_0^1x^2f(x) dx&=int_0^1(x^2-ax)f(x) dx\
          &=left[left(fracx^33-fracax^22right)f(x)right]_0^1-int_0^1left(fracx^33-fracax^22right)f'(x)dx\
          &=left(frac13-fraca2right)f(1)+int_0^1left(fracax^22-fracx^33right)f'(x)dxendalign$$
          Now choose an appropriate value for $a$. Can you take it from here?






          share|cite|improve this answer






















          • I'm not sure whether you did that integration by parts correctly or not ...
            – user521337
            3 hours ago










          • Sorry, I forgot a term...
            – Robert Z
            3 hours ago














          up vote
          4
          down vote



          accepted










          Hint. For any real $a$, by integration by parts,
          $$beginalignint_0^1x^2f(x) dx&=int_0^1(x^2-ax)f(x) dx\
          &=left[left(fracx^33-fracax^22right)f(x)right]_0^1-int_0^1left(fracx^33-fracax^22right)f'(x)dx\
          &=left(frac13-fraca2right)f(1)+int_0^1left(fracax^22-fracx^33right)f'(x)dxendalign$$
          Now choose an appropriate value for $a$. Can you take it from here?






          share|cite|improve this answer






















          • I'm not sure whether you did that integration by parts correctly or not ...
            – user521337
            3 hours ago










          • Sorry, I forgot a term...
            – Robert Z
            3 hours ago












          up vote
          4
          down vote



          accepted







          up vote
          4
          down vote



          accepted






          Hint. For any real $a$, by integration by parts,
          $$beginalignint_0^1x^2f(x) dx&=int_0^1(x^2-ax)f(x) dx\
          &=left[left(fracx^33-fracax^22right)f(x)right]_0^1-int_0^1left(fracx^33-fracax^22right)f'(x)dx\
          &=left(frac13-fraca2right)f(1)+int_0^1left(fracax^22-fracx^33right)f'(x)dxendalign$$
          Now choose an appropriate value for $a$. Can you take it from here?






          share|cite|improve this answer














          Hint. For any real $a$, by integration by parts,
          $$beginalignint_0^1x^2f(x) dx&=int_0^1(x^2-ax)f(x) dx\
          &=left[left(fracx^33-fracax^22right)f(x)right]_0^1-int_0^1left(fracx^33-fracax^22right)f'(x)dx\
          &=left(frac13-fraca2right)f(1)+int_0^1left(fracax^22-fracx^33right)f'(x)dxendalign$$
          Now choose an appropriate value for $a$. Can you take it from here?







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 3 hours ago

























          answered 3 hours ago









          Robert Z

          88.3k1056127




          88.3k1056127











          • I'm not sure whether you did that integration by parts correctly or not ...
            – user521337
            3 hours ago










          • Sorry, I forgot a term...
            – Robert Z
            3 hours ago
















          • I'm not sure whether you did that integration by parts correctly or not ...
            – user521337
            3 hours ago










          • Sorry, I forgot a term...
            – Robert Z
            3 hours ago















          I'm not sure whether you did that integration by parts correctly or not ...
          – user521337
          3 hours ago




          I'm not sure whether you did that integration by parts correctly or not ...
          – user521337
          3 hours ago












          Sorry, I forgot a term...
          – Robert Z
          3 hours ago




          Sorry, I forgot a term...
          – Robert Z
          3 hours ago

















           

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