how does an integral becoming negative effect limits of integration?

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In the following question I am struggling to understand how the limits of integration have be changed. Mainly the reasoning behind this part:



"This gives a sign change
(“dy = −dx”) which we incorporate by changing the order of the limits of integration."



enter image description here



In essence why when f(-y) and g(-y) become negative why can we move 'a' to the top of the integral and why does it become posotive.




calculus integration




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    up vote
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    In the following question I am struggling to understand how the limits of integration have be changed. Mainly the reasoning behind this part:



    "This gives a sign change
    (“dy = −dx”) which we incorporate by changing the order of the limits of integration."



    enter image description here



    In essence why when f(-y) and g(-y) become negative why can we move 'a' to the top of the integral and why does it become posotive.




    calculus integration




    share|cite|improve this question





















      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      In the following question I am struggling to understand how the limits of integration have be changed. Mainly the reasoning behind this part:



      "This gives a sign change
      (“dy = −dx”) which we incorporate by changing the order of the limits of integration."



      enter image description here



      In essence why when f(-y) and g(-y) become negative why can we move 'a' to the top of the integral and why does it become posotive.




      calculus integration




      share|cite|improve this question











      In the following question I am struggling to understand how the limits of integration have be changed. Mainly the reasoning behind this part:



      "This gives a sign change
      (“dy = −dx”) which we incorporate by changing the order of the limits of integration."



      enter image description here



      In essence why when f(-y) and g(-y) become negative why can we move 'a' to the top of the integral and why does it become posotive.





      calculus integration





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      asked Jul 23 at 16:26









      ojd

      424




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          We have $x=-a implies y=a$ and then



          $$int_-a^0 f(x) ,dx=int_a^0 f(-y) cdot(-1)cdot dy=-int_a^0 f(-y) dy=int_0^a f(-y) dy$$






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            This is because with the change of variables $x = -y$ the region of integration $[-a, 0]$ is no longer valid. We must change this according to the change of variable. Thus, we now integrate on $[0, a]$.






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              2 Answers
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              active

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              2 Answers
              2






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              active

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              up vote
              4
              down vote



              accepted










              We have $x=-a implies y=a$ and then



              $$int_-a^0 f(x) ,dx=int_a^0 f(-y) cdot(-1)cdot dy=-int_a^0 f(-y) dy=int_0^a f(-y) dy$$






              share|cite|improve this answer

























                up vote
                4
                down vote



                accepted










                We have $x=-a implies y=a$ and then



                $$int_-a^0 f(x) ,dx=int_a^0 f(-y) cdot(-1)cdot dy=-int_a^0 f(-y) dy=int_0^a f(-y) dy$$






                share|cite|improve this answer























                  up vote
                  4
                  down vote



                  accepted







                  up vote
                  4
                  down vote



                  accepted






                  We have $x=-a implies y=a$ and then



                  $$int_-a^0 f(x) ,dx=int_a^0 f(-y) cdot(-1)cdot dy=-int_a^0 f(-y) dy=int_0^a f(-y) dy$$






                  share|cite|improve this answer













                  We have $x=-a implies y=a$ and then



                  $$int_-a^0 f(x) ,dx=int_a^0 f(-y) cdot(-1)cdot dy=-int_a^0 f(-y) dy=int_0^a f(-y) dy$$







                  share|cite|improve this answer













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                  share|cite|improve this answer











                  answered Jul 23 at 16:28









                  gimusi

                  65.2k73583




                  65.2k73583




















                      up vote
                      1
                      down vote













                      This is because with the change of variables $x = -y$ the region of integration $[-a, 0]$ is no longer valid. We must change this according to the change of variable. Thus, we now integrate on $[0, a]$.






                      share|cite|improve this answer

























                        up vote
                        1
                        down vote













                        This is because with the change of variables $x = -y$ the region of integration $[-a, 0]$ is no longer valid. We must change this according to the change of variable. Thus, we now integrate on $[0, a]$.






                        share|cite|improve this answer























                          up vote
                          1
                          down vote










                          up vote
                          1
                          down vote









                          This is because with the change of variables $x = -y$ the region of integration $[-a, 0]$ is no longer valid. We must change this according to the change of variable. Thus, we now integrate on $[0, a]$.






                          share|cite|improve this answer













                          This is because with the change of variables $x = -y$ the region of integration $[-a, 0]$ is no longer valid. We must change this according to the change of variable. Thus, we now integrate on $[0, a]$.







                          share|cite|improve this answer













                          share|cite|improve this answer



                          share|cite|improve this answer











                          answered Jul 23 at 16:29









                          Sean Roberson

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