Solve the integral $int_0^1int^1_xy^4e^{xy^2}dydx$.












1












$begingroup$



Solve the integral $int_0^1int^1_xy^4e^{xy^2}dydx$.




I think that variables substituation is neede here. I've substitute
$$
\ left{begin{matrix}
u=xy^2\
v=y
end{matrix}right.
$$
and calculated
$$\J=begin{vmatrix}
y^2 & 2xy\
0 & 1
end{vmatrix}=y^2
$$
Then, the new integrand is $v^2e^u$. But what is the new domain? Thanks.






integration multivariable-calculus substitution jacobian







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Maybe try reversing order of integration?
    $endgroup$
    – Zachary Selk
    Jan 6 at 17:44






  • 2




    $begingroup$
    Obviously, $$x=u/v^2qquad v=y$$ hence the domain $$0<x<y<1$$ translates as $$0<u/v^2<v<1$$ that is, $$0<u<v^3<1$$
    $endgroup$
    – Did
    Jan 6 at 17:46












  • $begingroup$
    @ZacharySelk Ok, but then, what meaning does the $x$ in the lower bound of the external integral have: $int_x^1int_0^1 A dxdy$ (if $A$ is the integrand)?
    $endgroup$
    – J. Doe
    Jan 6 at 17:52






  • 1




    $begingroup$
    Draw a picture of the region.
    $endgroup$
    – Zachary Selk
    Jan 6 at 17:53






  • 1




    $begingroup$
    Yeah now it's totally doable.
    $endgroup$
    – Zachary Selk
    Jan 6 at 18:12
















1












$begingroup$



Solve the integral $int_0^1int^1_xy^4e^{xy^2}dydx$.




I think that variables substituation is neede here. I've substitute
$$
\ left{begin{matrix}
u=xy^2\
v=y
end{matrix}right.
$$
and calculated
$$\J=begin{vmatrix}
y^2 & 2xy\
0 & 1
end{vmatrix}=y^2
$$
Then, the new integrand is $v^2e^u$. But what is the new domain? Thanks.






integration multivariable-calculus substitution jacobian







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Maybe try reversing order of integration?
    $endgroup$
    – Zachary Selk
    Jan 6 at 17:44






  • 2




    $begingroup$
    Obviously, $$x=u/v^2qquad v=y$$ hence the domain $$0<x<y<1$$ translates as $$0<u/v^2<v<1$$ that is, $$0<u<v^3<1$$
    $endgroup$
    – Did
    Jan 6 at 17:46












  • $begingroup$
    @ZacharySelk Ok, but then, what meaning does the $x$ in the lower bound of the external integral have: $int_x^1int_0^1 A dxdy$ (if $A$ is the integrand)?
    $endgroup$
    – J. Doe
    Jan 6 at 17:52






  • 1




    $begingroup$
    Draw a picture of the region.
    $endgroup$
    – Zachary Selk
    Jan 6 at 17:53






  • 1




    $begingroup$
    Yeah now it's totally doable.
    $endgroup$
    – Zachary Selk
    Jan 6 at 18:12














1












1








1





$begingroup$



Solve the integral $int_0^1int^1_xy^4e^{xy^2}dydx$.




I think that variables substituation is neede here. I've substitute
$$
\ left{begin{matrix}
u=xy^2\
v=y
end{matrix}right.
$$
and calculated
$$\J=begin{vmatrix}
y^2 & 2xy\
0 & 1
end{vmatrix}=y^2
$$
Then, the new integrand is $v^2e^u$. But what is the new domain? Thanks.






integration multivariable-calculus substitution jacobian







share|cite|improve this question









$endgroup$





Solve the integral $int_0^1int^1_xy^4e^{xy^2}dydx$.




I think that variables substituation is neede here. I've substitute
$$
\ left{begin{matrix}
u=xy^2\
v=y
end{matrix}right.
$$
and calculated
$$\J=begin{vmatrix}
y^2 & 2xy\
0 & 1
end{vmatrix}=y^2
$$
Then, the new integrand is $v^2e^u$. But what is the new domain? Thanks.






integration multivariable-calculus substitution jacobian




integration multivariable-calculus substitution jacobian






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 6 at 17:43









J. DoeJ. Doe

14510




14510








  • 1




    $begingroup$
    Maybe try reversing order of integration?
    $endgroup$
    – Zachary Selk
    Jan 6 at 17:44






  • 2




    $begingroup$
    Obviously, $$x=u/v^2qquad v=y$$ hence the domain $$0<x<y<1$$ translates as $$0<u/v^2<v<1$$ that is, $$0<u<v^3<1$$
    $endgroup$
    – Did
    Jan 6 at 17:46












  • $begingroup$
    @ZacharySelk Ok, but then, what meaning does the $x$ in the lower bound of the external integral have: $int_x^1int_0^1 A dxdy$ (if $A$ is the integrand)?
    $endgroup$
    – J. Doe
    Jan 6 at 17:52






  • 1




    $begingroup$
    Draw a picture of the region.
    $endgroup$
    – Zachary Selk
    Jan 6 at 17:53






  • 1




    $begingroup$
    Yeah now it's totally doable.
    $endgroup$
    – Zachary Selk
    Jan 6 at 18:12














  • 1




    $begingroup$
    Maybe try reversing order of integration?
    $endgroup$
    – Zachary Selk
    Jan 6 at 17:44






  • 2




    $begingroup$
    Obviously, $$x=u/v^2qquad v=y$$ hence the domain $$0<x<y<1$$ translates as $$0<u/v^2<v<1$$ that is, $$0<u<v^3<1$$
    $endgroup$
    – Did
    Jan 6 at 17:46












  • $begingroup$
    @ZacharySelk Ok, but then, what meaning does the $x$ in the lower bound of the external integral have: $int_x^1int_0^1 A dxdy$ (if $A$ is the integrand)?
    $endgroup$
    – J. Doe
    Jan 6 at 17:52






  • 1




    $begingroup$
    Draw a picture of the region.
    $endgroup$
    – Zachary Selk
    Jan 6 at 17:53






  • 1




    $begingroup$
    Yeah now it's totally doable.
    $endgroup$
    – Zachary Selk
    Jan 6 at 18:12








1




1




$begingroup$
Maybe try reversing order of integration?
$endgroup$
– Zachary Selk
Jan 6 at 17:44




$begingroup$
Maybe try reversing order of integration?
$endgroup$
– Zachary Selk
Jan 6 at 17:44




2




2




$begingroup$
Obviously, $$x=u/v^2qquad v=y$$ hence the domain $$0<x<y<1$$ translates as $$0<u/v^2<v<1$$ that is, $$0<u<v^3<1$$
$endgroup$
– Did
Jan 6 at 17:46






$begingroup$
Obviously, $$x=u/v^2qquad v=y$$ hence the domain $$0<x<y<1$$ translates as $$0<u/v^2<v<1$$ that is, $$0<u<v^3<1$$
$endgroup$
– Did
Jan 6 at 17:46














$begingroup$
@ZacharySelk Ok, but then, what meaning does the $x$ in the lower bound of the external integral have: $int_x^1int_0^1 A dxdy$ (if $A$ is the integrand)?
$endgroup$
– J. Doe
Jan 6 at 17:52




$begingroup$
@ZacharySelk Ok, but then, what meaning does the $x$ in the lower bound of the external integral have: $int_x^1int_0^1 A dxdy$ (if $A$ is the integrand)?
$endgroup$
– J. Doe
Jan 6 at 17:52




1




1




$begingroup$
Draw a picture of the region.
$endgroup$
– Zachary Selk
Jan 6 at 17:53




$begingroup$
Draw a picture of the region.
$endgroup$
– Zachary Selk
Jan 6 at 17:53




1




1




$begingroup$
Yeah now it's totally doable.
$endgroup$
– Zachary Selk
Jan 6 at 18:12




$begingroup$
Yeah now it's totally doable.
$endgroup$
– Zachary Selk
Jan 6 at 18:12










2 Answers
2






active

oldest

votes


















5












$begingroup$

$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$




$ds{int_{0}^{1}int_{x}^{1}y^{4}expo{xy^{2}}dd y,dd x:
{LARGE ?}}$.




begin{align}
&bbox[10px,#ffd]{int_{0}^{1}int_{x}^{1}y^{4}expo{xy^{2}}dd y,dd x} =
int_{0}^{1}y^{4}int_{0}^{y}expo{xy^{2}}dd x,dd y =
int_{0}^{1}y^{4}{expo{y^{3}} - 1 over y^{2}},dd y
\[5mm] = &
int_{0}^{1}pars{y^{2}expo{y^{3}} - y^{2}}dd y =
left.{expo{y^{3}} - y^{3} over 3},rightvert_{0}^{1} =
bbx{expo{} - 2 over 3} approx 0.2394
end{align}






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Well, solving a much more general problem:



    $$mathcal{I}_text{n}left(alpharight):=int_0^alphaint_x^alphatext{y}^text{n}cdotexpleft(xcdottext{y}^{text{n}-2}right)spacetext{d}text{y}spacetext{d}xtag1$$



    Using that (for all $x$):



    $$expleft(xright)=sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}tag2$$



    We can write:



    $$mathcal{I}_text{n}left(alpharight)=int_0^alphaint_x^alphatext{y}^text{n}cdotsum_{text{k}=0}^inftyfrac{left(xcdottext{y}^{text{n}-2}right)^text{k}}{text{k}!}spacetext{d}text{y}spacetext{d}x=$$
    $$int_0^alphaint_x^alphatext{y}^text{n}cdotsum_{text{k}=0}^inftyfrac{x^text{k}cdottext{y}^{text{k}left(text{n}-2right)}}{text{k}!}spacetext{d}text{y}spacetext{d}x=$$
    $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}int_x^alphatext{y}^text{n}cdottext{y}^{text{k}left(text{n}-2right)}spacetext{d}text{y}right}spacetext{d}x=$$
    $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}int_x^alphatext{y}^{text{n}+text{k}left(text{n}-2right)}spacetext{d}text{y}right}spacetext{d}x=$$
    $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}cdotleft[frac{text{y}^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}right]_x^alpharight}spacetext{d}x=$$
    $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}cdotleft(frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}-frac{x^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}right)right}spacetext{d}x=$$
    $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotint_0^alpha x^text{k}spacetext{d}x-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotint_0^alpha x^{1+text{n}+text{k}left(text{n}-1right)}spacetext{d}xright}=$$
    $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{1+text{k}}}{1+text{k}}-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{1+1+text{n}+text{k}left(text{n}-1right)}}{1+1+text{n}+text{k}left(text{n}-1right)}right}=$$
    $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{1}{1+text{k}}-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{2+text{n}+text{k}left(text{n}-1right)}right}=$$
    $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotleft{frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{1+text{k}}-frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{2+text{n}+text{k}left(text{n}-1right)}right}tag3$$



    When $alpha=1$, we get:



    $$mathcal{I}_text{n}left(1right):=int_0^1int_x^1text{y}^text{n}cdotexpleft(xcdottext{y}^{text{n}-2}right)spacetext{d}text{y}spacetext{d}x=$$
    $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotleft{frac{1}{1+text{k}}-frac{1}{2+text{n}+text{k}left(text{n}-1right)}right}=$$
    $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{k}}cdotfrac{1}{2+text{n}+text{k}left(text{n}-1right)}tag4$$






    share|cite|improve this answer











    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3064163%2fsolve-the-integral-int-01-int1-xy4exy2dydx%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
      newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
      newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
      newcommand{dd}{mathrm{d}}
      newcommand{ds}[1]{displaystyle{#1}}
      newcommand{expo}[1]{,mathrm{e}^{#1},}
      newcommand{ic}{mathrm{i}}
      newcommand{mc}[1]{mathcal{#1}}
      newcommand{mrm}[1]{mathrm{#1}}
      newcommand{pars}[1]{left(,{#1},right)}
      newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
      newcommand{root}[2]{,sqrt[#1]{,{#2},},}
      newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
      newcommand{verts}[1]{leftvert,{#1},rightvert}$




      $ds{int_{0}^{1}int_{x}^{1}y^{4}expo{xy^{2}}dd y,dd x:
      {LARGE ?}}$      .




      begin{align}
      &bbox[10px,#ffd]{int_{0}^{1}int_{x}^{1}y^{4}expo{xy^{2}}dd y,dd x} =
      int_{0}^{1}y^{4}int_{0}^{y}expo{xy^{2}}dd x,dd y =
      int_{0}^{1}y^{4}{expo{y^{3}} - 1 over y^{2}},dd y
      \[5mm] = &
      int_{0}^{1}pars{y^{2}expo{y^{3}} - y^{2}}dd y =
      left.{expo{y^{3}} - y^{3} over 3},rightvert_{0}^{1} =
      bbx{expo{} - 2 over 3} approx 0.2394
      end{align}






      share|cite|improve this answer









      $endgroup$


















        5












        $begingroup$

        $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
        newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
        newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
        newcommand{dd}{mathrm{d}}
        newcommand{ds}[1]{displaystyle{#1}}
        newcommand{expo}[1]{,mathrm{e}^{#1},}
        newcommand{ic}{mathrm{i}}
        newcommand{mc}[1]{mathcal{#1}}
        newcommand{mrm}[1]{mathrm{#1}}
        newcommand{pars}[1]{left(,{#1},right)}
        newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
        newcommand{root}[2]{,sqrt[#1]{,{#2},},}
        newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
        newcommand{verts}[1]{leftvert,{#1},rightvert}$




        $ds{int_{0}^{1}int_{x}^{1}y^{4}expo{xy^{2}}dd y,dd x:
        {LARGE ?}}$        .




        begin{align}
        &bbox[10px,#ffd]{int_{0}^{1}int_{x}^{1}y^{4}expo{xy^{2}}dd y,dd x} =
        int_{0}^{1}y^{4}int_{0}^{y}expo{xy^{2}}dd x,dd y =
        int_{0}^{1}y^{4}{expo{y^{3}} - 1 over y^{2}},dd y
        \[5mm] = &
        int_{0}^{1}pars{y^{2}expo{y^{3}} - y^{2}}dd y =
        left.{expo{y^{3}} - y^{3} over 3},rightvert_{0}^{1} =
        bbx{expo{} - 2 over 3} approx 0.2394
        end{align}






        share|cite|improve this answer









        $endgroup$
















          5












          5








          5





          $begingroup$

          $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
          newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
          newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
          newcommand{dd}{mathrm{d}}
          newcommand{ds}[1]{displaystyle{#1}}
          newcommand{expo}[1]{,mathrm{e}^{#1},}
          newcommand{ic}{mathrm{i}}
          newcommand{mc}[1]{mathcal{#1}}
          newcommand{mrm}[1]{mathrm{#1}}
          newcommand{pars}[1]{left(,{#1},right)}
          newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
          newcommand{root}[2]{,sqrt[#1]{,{#2},},}
          newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
          newcommand{verts}[1]{leftvert,{#1},rightvert}$




          $ds{int_{0}^{1}int_{x}^{1}y^{4}expo{xy^{2}}dd y,dd x:
          {LARGE ?}}$          .




          begin{align}
          &bbox[10px,#ffd]{int_{0}^{1}int_{x}^{1}y^{4}expo{xy^{2}}dd y,dd x} =
          int_{0}^{1}y^{4}int_{0}^{y}expo{xy^{2}}dd x,dd y =
          int_{0}^{1}y^{4}{expo{y^{3}} - 1 over y^{2}},dd y
          \[5mm] = &
          int_{0}^{1}pars{y^{2}expo{y^{3}} - y^{2}}dd y =
          left.{expo{y^{3}} - y^{3} over 3},rightvert_{0}^{1} =
          bbx{expo{} - 2 over 3} approx 0.2394
          end{align}






          share|cite|improve this answer









          $endgroup$



          $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
          newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
          newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
          newcommand{dd}{mathrm{d}}
          newcommand{ds}[1]{displaystyle{#1}}
          newcommand{expo}[1]{,mathrm{e}^{#1},}
          newcommand{ic}{mathrm{i}}
          newcommand{mc}[1]{mathcal{#1}}
          newcommand{mrm}[1]{mathrm{#1}}
          newcommand{pars}[1]{left(,{#1},right)}
          newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
          newcommand{root}[2]{,sqrt[#1]{,{#2},},}
          newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
          newcommand{verts}[1]{leftvert,{#1},rightvert}$




          $ds{int_{0}^{1}int_{x}^{1}y^{4}expo{xy^{2}}dd y,dd x:
          {LARGE ?}}$          .




          begin{align}
          &bbox[10px,#ffd]{int_{0}^{1}int_{x}^{1}y^{4}expo{xy^{2}}dd y,dd x} =
          int_{0}^{1}y^{4}int_{0}^{y}expo{xy^{2}}dd x,dd y =
          int_{0}^{1}y^{4}{expo{y^{3}} - 1 over y^{2}},dd y
          \[5mm] = &
          int_{0}^{1}pars{y^{2}expo{y^{3}} - y^{2}}dd y =
          left.{expo{y^{3}} - y^{3} over 3},rightvert_{0}^{1} =
          bbx{expo{} - 2 over 3} approx 0.2394
          end{align}







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 6 at 21:48









          Felix MarinFelix Marin

          67.9k7107142




          67.9k7107142























              2












              $begingroup$

              Well, solving a much more general problem:



              $$mathcal{I}_text{n}left(alpharight):=int_0^alphaint_x^alphatext{y}^text{n}cdotexpleft(xcdottext{y}^{text{n}-2}right)spacetext{d}text{y}spacetext{d}xtag1$$



              Using that (for all $x$):



              $$expleft(xright)=sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}tag2$$



              We can write:



              $$mathcal{I}_text{n}left(alpharight)=int_0^alphaint_x^alphatext{y}^text{n}cdotsum_{text{k}=0}^inftyfrac{left(xcdottext{y}^{text{n}-2}right)^text{k}}{text{k}!}spacetext{d}text{y}spacetext{d}x=$$
              $$int_0^alphaint_x^alphatext{y}^text{n}cdotsum_{text{k}=0}^inftyfrac{x^text{k}cdottext{y}^{text{k}left(text{n}-2right)}}{text{k}!}spacetext{d}text{y}spacetext{d}x=$$
              $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}int_x^alphatext{y}^text{n}cdottext{y}^{text{k}left(text{n}-2right)}spacetext{d}text{y}right}spacetext{d}x=$$
              $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}int_x^alphatext{y}^{text{n}+text{k}left(text{n}-2right)}spacetext{d}text{y}right}spacetext{d}x=$$
              $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}cdotleft[frac{text{y}^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}right]_x^alpharight}spacetext{d}x=$$
              $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}cdotleft(frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}-frac{x^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}right)right}spacetext{d}x=$$
              $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotint_0^alpha x^text{k}spacetext{d}x-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotint_0^alpha x^{1+text{n}+text{k}left(text{n}-1right)}spacetext{d}xright}=$$
              $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{1+text{k}}}{1+text{k}}-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{1+1+text{n}+text{k}left(text{n}-1right)}}{1+1+text{n}+text{k}left(text{n}-1right)}right}=$$
              $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{1}{1+text{k}}-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{2+text{n}+text{k}left(text{n}-1right)}right}=$$
              $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotleft{frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{1+text{k}}-frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{2+text{n}+text{k}left(text{n}-1right)}right}tag3$$



              When $alpha=1$, we get:



              $$mathcal{I}_text{n}left(1right):=int_0^1int_x^1text{y}^text{n}cdotexpleft(xcdottext{y}^{text{n}-2}right)spacetext{d}text{y}spacetext{d}x=$$
              $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotleft{frac{1}{1+text{k}}-frac{1}{2+text{n}+text{k}left(text{n}-1right)}right}=$$
              $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{k}}cdotfrac{1}{2+text{n}+text{k}left(text{n}-1right)}tag4$$






              share|cite|improve this answer











              $endgroup$


















                2












                $begingroup$

                Well, solving a much more general problem:



                $$mathcal{I}_text{n}left(alpharight):=int_0^alphaint_x^alphatext{y}^text{n}cdotexpleft(xcdottext{y}^{text{n}-2}right)spacetext{d}text{y}spacetext{d}xtag1$$



                Using that (for all $x$):



                $$expleft(xright)=sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}tag2$$



                We can write:



                $$mathcal{I}_text{n}left(alpharight)=int_0^alphaint_x^alphatext{y}^text{n}cdotsum_{text{k}=0}^inftyfrac{left(xcdottext{y}^{text{n}-2}right)^text{k}}{text{k}!}spacetext{d}text{y}spacetext{d}x=$$
                $$int_0^alphaint_x^alphatext{y}^text{n}cdotsum_{text{k}=0}^inftyfrac{x^text{k}cdottext{y}^{text{k}left(text{n}-2right)}}{text{k}!}spacetext{d}text{y}spacetext{d}x=$$
                $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}int_x^alphatext{y}^text{n}cdottext{y}^{text{k}left(text{n}-2right)}spacetext{d}text{y}right}spacetext{d}x=$$
                $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}int_x^alphatext{y}^{text{n}+text{k}left(text{n}-2right)}spacetext{d}text{y}right}spacetext{d}x=$$
                $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}cdotleft[frac{text{y}^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}right]_x^alpharight}spacetext{d}x=$$
                $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}cdotleft(frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}-frac{x^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}right)right}spacetext{d}x=$$
                $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotint_0^alpha x^text{k}spacetext{d}x-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotint_0^alpha x^{1+text{n}+text{k}left(text{n}-1right)}spacetext{d}xright}=$$
                $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{1+text{k}}}{1+text{k}}-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{1+1+text{n}+text{k}left(text{n}-1right)}}{1+1+text{n}+text{k}left(text{n}-1right)}right}=$$
                $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{1}{1+text{k}}-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{2+text{n}+text{k}left(text{n}-1right)}right}=$$
                $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotleft{frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{1+text{k}}-frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{2+text{n}+text{k}left(text{n}-1right)}right}tag3$$



                When $alpha=1$, we get:



                $$mathcal{I}_text{n}left(1right):=int_0^1int_x^1text{y}^text{n}cdotexpleft(xcdottext{y}^{text{n}-2}right)spacetext{d}text{y}spacetext{d}x=$$
                $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotleft{frac{1}{1+text{k}}-frac{1}{2+text{n}+text{k}left(text{n}-1right)}right}=$$
                $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{k}}cdotfrac{1}{2+text{n}+text{k}left(text{n}-1right)}tag4$$






                share|cite|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Well, solving a much more general problem:



                  $$mathcal{I}_text{n}left(alpharight):=int_0^alphaint_x^alphatext{y}^text{n}cdotexpleft(xcdottext{y}^{text{n}-2}right)spacetext{d}text{y}spacetext{d}xtag1$$



                  Using that (for all $x$):



                  $$expleft(xright)=sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}tag2$$



                  We can write:



                  $$mathcal{I}_text{n}left(alpharight)=int_0^alphaint_x^alphatext{y}^text{n}cdotsum_{text{k}=0}^inftyfrac{left(xcdottext{y}^{text{n}-2}right)^text{k}}{text{k}!}spacetext{d}text{y}spacetext{d}x=$$
                  $$int_0^alphaint_x^alphatext{y}^text{n}cdotsum_{text{k}=0}^inftyfrac{x^text{k}cdottext{y}^{text{k}left(text{n}-2right)}}{text{k}!}spacetext{d}text{y}spacetext{d}x=$$
                  $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}int_x^alphatext{y}^text{n}cdottext{y}^{text{k}left(text{n}-2right)}spacetext{d}text{y}right}spacetext{d}x=$$
                  $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}int_x^alphatext{y}^{text{n}+text{k}left(text{n}-2right)}spacetext{d}text{y}right}spacetext{d}x=$$
                  $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}cdotleft[frac{text{y}^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}right]_x^alpharight}spacetext{d}x=$$
                  $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}cdotleft(frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}-frac{x^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}right)right}spacetext{d}x=$$
                  $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotint_0^alpha x^text{k}spacetext{d}x-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotint_0^alpha x^{1+text{n}+text{k}left(text{n}-1right)}spacetext{d}xright}=$$
                  $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{1+text{k}}}{1+text{k}}-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{1+1+text{n}+text{k}left(text{n}-1right)}}{1+1+text{n}+text{k}left(text{n}-1right)}right}=$$
                  $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{1}{1+text{k}}-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{2+text{n}+text{k}left(text{n}-1right)}right}=$$
                  $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotleft{frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{1+text{k}}-frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{2+text{n}+text{k}left(text{n}-1right)}right}tag3$$



                  When $alpha=1$, we get:



                  $$mathcal{I}_text{n}left(1right):=int_0^1int_x^1text{y}^text{n}cdotexpleft(xcdottext{y}^{text{n}-2}right)spacetext{d}text{y}spacetext{d}x=$$
                  $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotleft{frac{1}{1+text{k}}-frac{1}{2+text{n}+text{k}left(text{n}-1right)}right}=$$
                  $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{k}}cdotfrac{1}{2+text{n}+text{k}left(text{n}-1right)}tag4$$






                  share|cite|improve this answer











                  $endgroup$



                  Well, solving a much more general problem:



                  $$mathcal{I}_text{n}left(alpharight):=int_0^alphaint_x^alphatext{y}^text{n}cdotexpleft(xcdottext{y}^{text{n}-2}right)spacetext{d}text{y}spacetext{d}xtag1$$



                  Using that (for all $x$):



                  $$expleft(xright)=sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}tag2$$



                  We can write:



                  $$mathcal{I}_text{n}left(alpharight)=int_0^alphaint_x^alphatext{y}^text{n}cdotsum_{text{k}=0}^inftyfrac{left(xcdottext{y}^{text{n}-2}right)^text{k}}{text{k}!}spacetext{d}text{y}spacetext{d}x=$$
                  $$int_0^alphaint_x^alphatext{y}^text{n}cdotsum_{text{k}=0}^inftyfrac{x^text{k}cdottext{y}^{text{k}left(text{n}-2right)}}{text{k}!}spacetext{d}text{y}spacetext{d}x=$$
                  $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}int_x^alphatext{y}^text{n}cdottext{y}^{text{k}left(text{n}-2right)}spacetext{d}text{y}right}spacetext{d}x=$$
                  $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}int_x^alphatext{y}^{text{n}+text{k}left(text{n}-2right)}spacetext{d}text{y}right}spacetext{d}x=$$
                  $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}cdotleft[frac{text{y}^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}right]_x^alpharight}spacetext{d}x=$$
                  $$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}cdotleft(frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}-frac{x^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}right)right}spacetext{d}x=$$
                  $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotint_0^alpha x^text{k}spacetext{d}x-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotint_0^alpha x^{1+text{n}+text{k}left(text{n}-1right)}spacetext{d}xright}=$$
                  $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{1+text{k}}}{1+text{k}}-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{1+1+text{n}+text{k}left(text{n}-1right)}}{1+1+text{n}+text{k}left(text{n}-1right)}right}=$$
                  $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{1}{1+text{k}}-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{2+text{n}+text{k}left(text{n}-1right)}right}=$$
                  $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotleft{frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{1+text{k}}-frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{2+text{n}+text{k}left(text{n}-1right)}right}tag3$$



                  When $alpha=1$, we get:



                  $$mathcal{I}_text{n}left(1right):=int_0^1int_x^1text{y}^text{n}cdotexpleft(xcdottext{y}^{text{n}-2}right)spacetext{d}text{y}spacetext{d}x=$$
                  $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotleft{frac{1}{1+text{k}}-frac{1}{2+text{n}+text{k}left(text{n}-1right)}right}=$$
                  $$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{k}}cdotfrac{1}{2+text{n}+text{k}left(text{n}-1right)}tag4$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 23 at 8:19

























                  answered Jan 6 at 20:52









                  JanJan

                  21.8k31240




                  21.8k31240






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3064163%2fsolve-the-integral-int-01-int1-xy4exy2dydx%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Human spaceflight

                      Image
                      Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e...
                      Read more

                      Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

                      Image
                      0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
                      Read more

                      張江高科駅

                      Image
                      張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
                      Read more