How to treat a constant when integrating

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I am wondering what to do when coming across an integral like this where $a$ is a constant:



$$int^1000_a (x-a)1over1000dx $$



As far as I can see, it should be ok to do this:



$$ left.1over 1000 left(x^2over2 -axright)right|^1000_a$$



But the book I am using does:



$$ left.1over 1000 left((x-a)^2over2right)right|^1000_a$$



Which also seems correct, but is different. How does one decide which way to integrate? (The actual question in the book is regarding expected value of probability, and I can copy it in full if it makes a difference.)




calculus integration




share|cite|improve this question

























    up vote
    3
    down vote

    favorite












    I am wondering what to do when coming across an integral like this where $a$ is a constant:



    $$int^1000_a (x-a)1over1000dx $$



    As far as I can see, it should be ok to do this:



    $$ left.1over 1000 left(x^2over2 -axright)right|^1000_a$$



    But the book I am using does:



    $$ left.1over 1000 left((x-a)^2over2right)right|^1000_a$$



    Which also seems correct, but is different. How does one decide which way to integrate? (The actual question in the book is regarding expected value of probability, and I can copy it in full if it makes a difference.)




    calculus integration




    share|cite|improve this question























      up vote
      3
      down vote

      favorite









      up vote
      3
      down vote

      favorite











      I am wondering what to do when coming across an integral like this where $a$ is a constant:



      $$int^1000_a (x-a)1over1000dx $$



      As far as I can see, it should be ok to do this:



      $$ left.1over 1000 left(x^2over2 -axright)right|^1000_a$$



      But the book I am using does:



      $$ left.1over 1000 left((x-a)^2over2right)right|^1000_a$$



      Which also seems correct, but is different. How does one decide which way to integrate? (The actual question in the book is regarding expected value of probability, and I can copy it in full if it makes a difference.)




      calculus integration




      share|cite|improve this question













      I am wondering what to do when coming across an integral like this where $a$ is a constant:



      $$int^1000_a (x-a)1over1000dx $$



      As far as I can see, it should be ok to do this:



      $$ left.1over 1000 left(x^2over2 -axright)right|^1000_a$$



      But the book I am using does:



      $$ left.1over 1000 left((x-a)^2over2right)right|^1000_a$$



      Which also seems correct, but is different. How does one decide which way to integrate? (The actual question in the book is regarding expected value of probability, and I can copy it in full if it makes a difference.)





      calculus integration





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      share|cite|improve this question




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      edited Aug 3 at 7:47









      Mutantoe

      52039




      52039









      asked Aug 3 at 3:42









      agblt

      547




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          2 Answers
          2






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          accepted










          The two expressions are different only by a constant. When you take bounds, the constant vanishes, giving the same result.






          share|cite|improve this answer





















          • Thanks, my confusion was from the fact that I had made a different mistake, and mistakenly attributed the mistake to the method of integration, but now you cleared it up for me!
            – agblt
            Aug 3 at 3:59










          • @agblt Might be worth mentioning that the fact that there is an unknown constant $a$ makes no difference than if it were a known constant - e.g. $x+3$ could be integrated as $x^2/2+3x$ or $(x+3)^2/2$.
            – Carmeister
            Aug 3 at 7:13


















          up vote
          3
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          $$dfrac(x-a)^22-left(dfracx^22-axright)=?$$ is independent of $x$ hence arbitrary constant






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          • @agblt, If $f(x)=dfraca^22,$ $$f(1000)-f(a)=?$$
            – lab bhattacharjee
            Aug 3 at 3:54










          • I see it now, thank you.
            – agblt
            Aug 3 at 4:01










          Your Answer




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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          3
          down vote



          accepted










          The two expressions are different only by a constant. When you take bounds, the constant vanishes, giving the same result.






          share|cite|improve this answer





















          • Thanks, my confusion was from the fact that I had made a different mistake, and mistakenly attributed the mistake to the method of integration, but now you cleared it up for me!
            – agblt
            Aug 3 at 3:59










          • @agblt Might be worth mentioning that the fact that there is an unknown constant $a$ makes no difference than if it were a known constant - e.g. $x+3$ could be integrated as $x^2/2+3x$ or $(x+3)^2/2$.
            – Carmeister
            Aug 3 at 7:13















          up vote
          3
          down vote



          accepted










          The two expressions are different only by a constant. When you take bounds, the constant vanishes, giving the same result.






          share|cite|improve this answer





















          • Thanks, my confusion was from the fact that I had made a different mistake, and mistakenly attributed the mistake to the method of integration, but now you cleared it up for me!
            – agblt
            Aug 3 at 3:59










          • @agblt Might be worth mentioning that the fact that there is an unknown constant $a$ makes no difference than if it were a known constant - e.g. $x+3$ could be integrated as $x^2/2+3x$ or $(x+3)^2/2$.
            – Carmeister
            Aug 3 at 7:13













          up vote
          3
          down vote



          accepted







          up vote
          3
          down vote



          accepted






          The two expressions are different only by a constant. When you take bounds, the constant vanishes, giving the same result.






          share|cite|improve this answer













          The two expressions are different only by a constant. When you take bounds, the constant vanishes, giving the same result.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Aug 3 at 3:53









          Deepak

          15.9k11437




          15.9k11437











          • Thanks, my confusion was from the fact that I had made a different mistake, and mistakenly attributed the mistake to the method of integration, but now you cleared it up for me!
            – agblt
            Aug 3 at 3:59










          • @agblt Might be worth mentioning that the fact that there is an unknown constant $a$ makes no difference than if it were a known constant - e.g. $x+3$ could be integrated as $x^2/2+3x$ or $(x+3)^2/2$.
            – Carmeister
            Aug 3 at 7:13

















          • Thanks, my confusion was from the fact that I had made a different mistake, and mistakenly attributed the mistake to the method of integration, but now you cleared it up for me!
            – agblt
            Aug 3 at 3:59










          • @agblt Might be worth mentioning that the fact that there is an unknown constant $a$ makes no difference than if it were a known constant - e.g. $x+3$ could be integrated as $x^2/2+3x$ or $(x+3)^2/2$.
            – Carmeister
            Aug 3 at 7:13
















          Thanks, my confusion was from the fact that I had made a different mistake, and mistakenly attributed the mistake to the method of integration, but now you cleared it up for me!
          – agblt
          Aug 3 at 3:59




          Thanks, my confusion was from the fact that I had made a different mistake, and mistakenly attributed the mistake to the method of integration, but now you cleared it up for me!
          – agblt
          Aug 3 at 3:59












          @agblt Might be worth mentioning that the fact that there is an unknown constant $a$ makes no difference than if it were a known constant - e.g. $x+3$ could be integrated as $x^2/2+3x$ or $(x+3)^2/2$.
          – Carmeister
          Aug 3 at 7:13





          @agblt Might be worth mentioning that the fact that there is an unknown constant $a$ makes no difference than if it were a known constant - e.g. $x+3$ could be integrated as $x^2/2+3x$ or $(x+3)^2/2$.
          – Carmeister
          Aug 3 at 7:13











          up vote
          3
          down vote













          $$dfrac(x-a)^22-left(dfracx^22-axright)=?$$ is independent of $x$ hence arbitrary constant






          share|cite|improve this answer





















          • @agblt, If $f(x)=dfraca^22,$ $$f(1000)-f(a)=?$$
            – lab bhattacharjee
            Aug 3 at 3:54










          • I see it now, thank you.
            – agblt
            Aug 3 at 4:01














          up vote
          3
          down vote













          $$dfrac(x-a)^22-left(dfracx^22-axright)=?$$ is independent of $x$ hence arbitrary constant






          share|cite|improve this answer





















          • @agblt, If $f(x)=dfraca^22,$ $$f(1000)-f(a)=?$$
            – lab bhattacharjee
            Aug 3 at 3:54










          • I see it now, thank you.
            – agblt
            Aug 3 at 4:01












          up vote
          3
          down vote










          up vote
          3
          down vote









          $$dfrac(x-a)^22-left(dfracx^22-axright)=?$$ is independent of $x$ hence arbitrary constant






          share|cite|improve this answer













          $$dfrac(x-a)^22-left(dfracx^22-axright)=?$$ is independent of $x$ hence arbitrary constant







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Aug 3 at 3:50









          lab bhattacharjee

          214k14152263




          214k14152263











          • @agblt, If $f(x)=dfraca^22,$ $$f(1000)-f(a)=?$$
            – lab bhattacharjee
            Aug 3 at 3:54










          • I see it now, thank you.
            – agblt
            Aug 3 at 4:01
















          • @agblt, If $f(x)=dfraca^22,$ $$f(1000)-f(a)=?$$
            – lab bhattacharjee
            Aug 3 at 3:54










          • I see it now, thank you.
            – agblt
            Aug 3 at 4:01















          @agblt, If $f(x)=dfraca^22,$ $$f(1000)-f(a)=?$$
          – lab bhattacharjee
          Aug 3 at 3:54




          @agblt, If $f(x)=dfraca^22,$ $$f(1000)-f(a)=?$$
          – lab bhattacharjee
          Aug 3 at 3:54












          I see it now, thank you.
          – agblt
          Aug 3 at 4:01




          I see it now, thank you.
          – agblt
          Aug 3 at 4:01












           

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