When change of variable makes an empty interval












14












$begingroup$


Please consider the following case:



$$I = int^1_{-1}x^2dx$$
$$u(x) = x^2 rightarrow du = 2x,dx$$
$$u(-1) = 1, u(1) = 1$$



So



$$I = int^1_1frac{u}{2sqrt u} du = 0$$



Obviously the problem here is to only consider the positive root of u. I don't know how to consider both roots. This example is trivial but I have another example where such substitution would be really helpful:



$$I = int^1_{-1}frac{x^2(1 - x^2)^frac{3}{2}}{3} - frac{x^2(1 - x^2)^frac{5}{2}}{5} - frac{x^4(1 - x^2)^frac{3}{2}}{3} dx$$



I don't want you to solve it for me using another method, I know how to use an integral solver online. My question is how to properly do the change of variable.






integration definite-integrals change-of-variable







share|cite|improve this question











$endgroup$












  • $begingroup$
    The derivative of $x^2$ isn't 2
    $endgroup$
    – orange
    Dec 9 '18 at 19:06










  • $begingroup$
    @orange Thanks, corrected
    $endgroup$
    – Winter
    Dec 9 '18 at 19:42










  • $begingroup$
    related: $u$-substitution always evaluates to $0$
    $endgroup$
    – Dando18
    Dec 10 '18 at 2:45
















14












$begingroup$


Please consider the following case:



$$I = int^1_{-1}x^2dx$$
$$u(x) = x^2 rightarrow du = 2x,dx$$
$$u(-1) = 1, u(1) = 1$$



So



$$I = int^1_1frac{u}{2sqrt u} du = 0$$



Obviously the problem here is to only consider the positive root of u. I don't know how to consider both roots. This example is trivial but I have another example where such substitution would be really helpful:



$$I = int^1_{-1}frac{x^2(1 - x^2)^frac{3}{2}}{3} - frac{x^2(1 - x^2)^frac{5}{2}}{5} - frac{x^4(1 - x^2)^frac{3}{2}}{3} dx$$



I don't want you to solve it for me using another method, I know how to use an integral solver online. My question is how to properly do the change of variable.






integration definite-integrals change-of-variable







share|cite|improve this question











$endgroup$












  • $begingroup$
    The derivative of $x^2$ isn't 2
    $endgroup$
    – orange
    Dec 9 '18 at 19:06










  • $begingroup$
    @orange Thanks, corrected
    $endgroup$
    – Winter
    Dec 9 '18 at 19:42










  • $begingroup$
    related: $u$-substitution always evaluates to $0$
    $endgroup$
    – Dando18
    Dec 10 '18 at 2:45














14












14








14


2



$begingroup$


Please consider the following case:



$$I = int^1_{-1}x^2dx$$
$$u(x) = x^2 rightarrow du = 2x,dx$$
$$u(-1) = 1, u(1) = 1$$



So



$$I = int^1_1frac{u}{2sqrt u} du = 0$$



Obviously the problem here is to only consider the positive root of u. I don't know how to consider both roots. This example is trivial but I have another example where such substitution would be really helpful:



$$I = int^1_{-1}frac{x^2(1 - x^2)^frac{3}{2}}{3} - frac{x^2(1 - x^2)^frac{5}{2}}{5} - frac{x^4(1 - x^2)^frac{3}{2}}{3} dx$$



I don't want you to solve it for me using another method, I know how to use an integral solver online. My question is how to properly do the change of variable.






integration definite-integrals change-of-variable







share|cite|improve this question











$endgroup$




Please consider the following case:



$$I = int^1_{-1}x^2dx$$
$$u(x) = x^2 rightarrow du = 2x,dx$$
$$u(-1) = 1, u(1) = 1$$



So



$$I = int^1_1frac{u}{2sqrt u} du = 0$$



Obviously the problem here is to only consider the positive root of u. I don't know how to consider both roots. This example is trivial but I have another example where such substitution would be really helpful:



$$I = int^1_{-1}frac{x^2(1 - x^2)^frac{3}{2}}{3} - frac{x^2(1 - x^2)^frac{5}{2}}{5} - frac{x^4(1 - x^2)^frac{3}{2}}{3} dx$$



I don't want you to solve it for me using another method, I know how to use an integral solver online. My question is how to properly do the change of variable.






integration definite-integrals change-of-variable




integration definite-integrals change-of-variable






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 19:41







Winter

















asked Dec 9 '18 at 17:12









WinterWinter

492421




492421












  • $begingroup$
    The derivative of $x^2$ isn't 2
    $endgroup$
    – orange
    Dec 9 '18 at 19:06










  • $begingroup$
    @orange Thanks, corrected
    $endgroup$
    – Winter
    Dec 9 '18 at 19:42










  • $begingroup$
    related: $u$-substitution always evaluates to $0$
    $endgroup$
    – Dando18
    Dec 10 '18 at 2:45


















  • $begingroup$
    The derivative of $x^2$ isn't 2
    $endgroup$
    – orange
    Dec 9 '18 at 19:06










  • $begingroup$
    @orange Thanks, corrected
    $endgroup$
    – Winter
    Dec 9 '18 at 19:42










  • $begingroup$
    related: $u$-substitution always evaluates to $0$
    $endgroup$
    – Dando18
    Dec 10 '18 at 2:45
















$begingroup$
The derivative of $x^2$ isn't 2
$endgroup$
– orange
Dec 9 '18 at 19:06




$begingroup$
The derivative of $x^2$ isn't 2
$endgroup$
– orange
Dec 9 '18 at 19:06












$begingroup$
@orange Thanks, corrected
$endgroup$
– Winter
Dec 9 '18 at 19:42




$begingroup$
@orange Thanks, corrected
$endgroup$
– Winter
Dec 9 '18 at 19:42












$begingroup$
related: $u$-substitution always evaluates to $0$
$endgroup$
– Dando18
Dec 10 '18 at 2:45




$begingroup$
related: $u$-substitution always evaluates to $0$
$endgroup$
– Dando18
Dec 10 '18 at 2:45










2 Answers
2






active

oldest

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12












$begingroup$

In general you need to use a one-to-one change of variable; thus for instance if you were to compute $int_{-1}^1 x^2 dx$ you would split into $int_{-1}^0 x^2 dx + int_0^1 x^2 dx$. Then each of those integrals, after substitution and simplification, becomes $int_0^1 frac{1}{2} u^{1/2} du$.



(That said, there are some situations where by basically breaking the rules twice you cancel out your mistakes and get the right answer; see Why can the trig sub $x=cosh(theta)$ be used to solve the integral: $int frac{1}{(x^2-1)^{3/2}}dx$? for instance.)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    It is not necessary for changes of variable to be one-to-one, see my comment here: math.stackexchange.com/a/1101421/5316
    $endgroup$
    – Matthew Towers
    Dec 9 '18 at 19:39






  • 1




    $begingroup$
    @MatthewTowers Yes, it is true that $int_a^b f'(x) dx$ is always $f(b) - f(a)$, regardless of injectivity, which means you can do various similar things also regardless of injectivity. What injectivity buys you is a lack of a singularity in $frac{dx}{du}$, which is a serious problem for, say, $u=x^2$ on $[a,b]$ with $a<0,b>0$.
    $endgroup$
    – Ian
    Dec 9 '18 at 19:52



















15












$begingroup$

While it's certainly true that $du = 2x,dx$ over the whole region of integration, it's not true that $x = sqrt{u}$ over the region. This is only true for $x > 0$, so you can't apply it to the whole integral. Instead, you need to split it apart and use the correct transformation $x = -sqrt{u}$ for the region $x < 0$.



In general, if your change of variables $u(x)$ has a multivalued inverse, you'll need to split the integral into the regions where each possible inverse function $x(u)$ applies.






share|cite|improve this answer











$endgroup$













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    2 Answers
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    12












    $begingroup$

    In general you need to use a one-to-one change of variable; thus for instance if you were to compute $int_{-1}^1 x^2 dx$ you would split into $int_{-1}^0 x^2 dx + int_0^1 x^2 dx$. Then each of those integrals, after substitution and simplification, becomes $int_0^1 frac{1}{2} u^{1/2} du$.



    (That said, there are some situations where by basically breaking the rules twice you cancel out your mistakes and get the right answer; see Why can the trig sub $x=cosh(theta)$ be used to solve the integral: $int frac{1}{(x^2-1)^{3/2}}dx$? for instance.)






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      It is not necessary for changes of variable to be one-to-one, see my comment here: math.stackexchange.com/a/1101421/5316
      $endgroup$
      – Matthew Towers
      Dec 9 '18 at 19:39






    • 1




      $begingroup$
      @MatthewTowers Yes, it is true that $int_a^b f'(x) dx$ is always $f(b) - f(a)$, regardless of injectivity, which means you can do various similar things also regardless of injectivity. What injectivity buys you is a lack of a singularity in $frac{dx}{du}$, which is a serious problem for, say, $u=x^2$ on $[a,b]$ with $a<0,b>0$.
      $endgroup$
      – Ian
      Dec 9 '18 at 19:52
















    12












    $begingroup$

    In general you need to use a one-to-one change of variable; thus for instance if you were to compute $int_{-1}^1 x^2 dx$ you would split into $int_{-1}^0 x^2 dx + int_0^1 x^2 dx$. Then each of those integrals, after substitution and simplification, becomes $int_0^1 frac{1}{2} u^{1/2} du$.



    (That said, there are some situations where by basically breaking the rules twice you cancel out your mistakes and get the right answer; see Why can the trig sub $x=cosh(theta)$ be used to solve the integral: $int frac{1}{(x^2-1)^{3/2}}dx$? for instance.)






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      It is not necessary for changes of variable to be one-to-one, see my comment here: math.stackexchange.com/a/1101421/5316
      $endgroup$
      – Matthew Towers
      Dec 9 '18 at 19:39






    • 1




      $begingroup$
      @MatthewTowers Yes, it is true that $int_a^b f'(x) dx$ is always $f(b) - f(a)$, regardless of injectivity, which means you can do various similar things also regardless of injectivity. What injectivity buys you is a lack of a singularity in $frac{dx}{du}$, which is a serious problem for, say, $u=x^2$ on $[a,b]$ with $a<0,b>0$.
      $endgroup$
      – Ian
      Dec 9 '18 at 19:52














    12












    12








    12





    $begingroup$

    In general you need to use a one-to-one change of variable; thus for instance if you were to compute $int_{-1}^1 x^2 dx$ you would split into $int_{-1}^0 x^2 dx + int_0^1 x^2 dx$. Then each of those integrals, after substitution and simplification, becomes $int_0^1 frac{1}{2} u^{1/2} du$.



    (That said, there are some situations where by basically breaking the rules twice you cancel out your mistakes and get the right answer; see Why can the trig sub $x=cosh(theta)$ be used to solve the integral: $int frac{1}{(x^2-1)^{3/2}}dx$? for instance.)






    share|cite|improve this answer









    $endgroup$



    In general you need to use a one-to-one change of variable; thus for instance if you were to compute $int_{-1}^1 x^2 dx$ you would split into $int_{-1}^0 x^2 dx + int_0^1 x^2 dx$. Then each of those integrals, after substitution and simplification, becomes $int_0^1 frac{1}{2} u^{1/2} du$.



    (That said, there are some situations where by basically breaking the rules twice you cancel out your mistakes and get the right answer; see Why can the trig sub $x=cosh(theta)$ be used to solve the integral: $int frac{1}{(x^2-1)^{3/2}}dx$? for instance.)







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 9 '18 at 17:20









    IanIan

    67.7k25387




    67.7k25387












    • $begingroup$
      It is not necessary for changes of variable to be one-to-one, see my comment here: math.stackexchange.com/a/1101421/5316
      $endgroup$
      – Matthew Towers
      Dec 9 '18 at 19:39






    • 1




      $begingroup$
      @MatthewTowers Yes, it is true that $int_a^b f'(x) dx$ is always $f(b) - f(a)$, regardless of injectivity, which means you can do various similar things also regardless of injectivity. What injectivity buys you is a lack of a singularity in $frac{dx}{du}$, which is a serious problem for, say, $u=x^2$ on $[a,b]$ with $a<0,b>0$.
      $endgroup$
      – Ian
      Dec 9 '18 at 19:52


















    • $begingroup$
      It is not necessary for changes of variable to be one-to-one, see my comment here: math.stackexchange.com/a/1101421/5316
      $endgroup$
      – Matthew Towers
      Dec 9 '18 at 19:39






    • 1




      $begingroup$
      @MatthewTowers Yes, it is true that $int_a^b f'(x) dx$ is always $f(b) - f(a)$, regardless of injectivity, which means you can do various similar things also regardless of injectivity. What injectivity buys you is a lack of a singularity in $frac{dx}{du}$, which is a serious problem for, say, $u=x^2$ on $[a,b]$ with $a<0,b>0$.
      $endgroup$
      – Ian
      Dec 9 '18 at 19:52
















    $begingroup$
    It is not necessary for changes of variable to be one-to-one, see my comment here: math.stackexchange.com/a/1101421/5316
    $endgroup$
    – Matthew Towers
    Dec 9 '18 at 19:39




    $begingroup$
    It is not necessary for changes of variable to be one-to-one, see my comment here: math.stackexchange.com/a/1101421/5316
    $endgroup$
    – Matthew Towers
    Dec 9 '18 at 19:39




    1




    1




    $begingroup$
    @MatthewTowers Yes, it is true that $int_a^b f'(x) dx$ is always $f(b) - f(a)$, regardless of injectivity, which means you can do various similar things also regardless of injectivity. What injectivity buys you is a lack of a singularity in $frac{dx}{du}$, which is a serious problem for, say, $u=x^2$ on $[a,b]$ with $a<0,b>0$.
    $endgroup$
    – Ian
    Dec 9 '18 at 19:52




    $begingroup$
    @MatthewTowers Yes, it is true that $int_a^b f'(x) dx$ is always $f(b) - f(a)$, regardless of injectivity, which means you can do various similar things also regardless of injectivity. What injectivity buys you is a lack of a singularity in $frac{dx}{du}$, which is a serious problem for, say, $u=x^2$ on $[a,b]$ with $a<0,b>0$.
    $endgroup$
    – Ian
    Dec 9 '18 at 19:52











    15












    $begingroup$

    While it's certainly true that $du = 2x,dx$ over the whole region of integration, it's not true that $x = sqrt{u}$ over the region. This is only true for $x > 0$, so you can't apply it to the whole integral. Instead, you need to split it apart and use the correct transformation $x = -sqrt{u}$ for the region $x < 0$.



    In general, if your change of variables $u(x)$ has a multivalued inverse, you'll need to split the integral into the regions where each possible inverse function $x(u)$ applies.






    share|cite|improve this answer











    $endgroup$


















      15












      $begingroup$

      While it's certainly true that $du = 2x,dx$ over the whole region of integration, it's not true that $x = sqrt{u}$ over the region. This is only true for $x > 0$, so you can't apply it to the whole integral. Instead, you need to split it apart and use the correct transformation $x = -sqrt{u}$ for the region $x < 0$.



      In general, if your change of variables $u(x)$ has a multivalued inverse, you'll need to split the integral into the regions where each possible inverse function $x(u)$ applies.






      share|cite|improve this answer











      $endgroup$
















        15












        15








        15





        $begingroup$

        While it's certainly true that $du = 2x,dx$ over the whole region of integration, it's not true that $x = sqrt{u}$ over the region. This is only true for $x > 0$, so you can't apply it to the whole integral. Instead, you need to split it apart and use the correct transformation $x = -sqrt{u}$ for the region $x < 0$.



        In general, if your change of variables $u(x)$ has a multivalued inverse, you'll need to split the integral into the regions where each possible inverse function $x(u)$ applies.






        share|cite|improve this answer











        $endgroup$



        While it's certainly true that $du = 2x,dx$ over the whole region of integration, it's not true that $x = sqrt{u}$ over the region. This is only true for $x > 0$, so you can't apply it to the whole integral. Instead, you need to split it apart and use the correct transformation $x = -sqrt{u}$ for the region $x < 0$.



        In general, if your change of variables $u(x)$ has a multivalued inverse, you'll need to split the integral into the regions where each possible inverse function $x(u)$ applies.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 10 '18 at 4:54

























        answered Dec 9 '18 at 17:33









        eyeballfrogeyeballfrog

        6,093629




        6,093629






























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