Area of the region bounded by the curves $y=x^2$ & $y=x^4$












0














One of my last assignments for my first semester of calculus is a list of area, volume, and work problems using integrals, I worked one out, but can't find the answer anywhere, and just want to make sure I'm correct, here's my work:



Find the area of the region bounded by the curves $y=x^2$ and $y=x^4$



$x^4=x^2$



$x^4-x^2=0$



$x^2(x^2-1)=0$



$x=1, -1$



$int_1^1(x^4-x^2)dx$



$left[frac{x^5}{5}-frac{x^3}{3}right]$ from $-1$ to $1$



$left[frac{(1)^5}{5}-frac{(1)^3}{3}right] - left[frac{(-1)^5}{5}-frac{(-1)^3}{3}right]$



$left[frac{1}{5}-frac{1}{3}right] - left[frac{-1}{5}+frac{1^3}{3}right]$



$left[frac{3-5}{15}-frac{-3+5}{15}right]$



$left[frac{-2}{15}+frac{-2}{15}right] = frac{-4}{15}$



I double checked the definite integral online and it got $frac{-4}{15}$ as well, so I guess that means I"m correct I'm just confused as to how an area can be negative unless I'm missing something obvious.



My prof also left a note that says "*Make sure you have the entire region", where "entire" is underlined, is there more to this problem than what I've done?






calculus







share|cite|improve this question





























    0














    One of my last assignments for my first semester of calculus is a list of area, volume, and work problems using integrals, I worked one out, but can't find the answer anywhere, and just want to make sure I'm correct, here's my work:



    Find the area of the region bounded by the curves $y=x^2$ and $y=x^4$



    $x^4=x^2$



    $x^4-x^2=0$



    $x^2(x^2-1)=0$



    $x=1, -1$



    $int_1^1(x^4-x^2)dx$



    $left[frac{x^5}{5}-frac{x^3}{3}right]$ from $-1$ to $1$



    $left[frac{(1)^5}{5}-frac{(1)^3}{3}right] - left[frac{(-1)^5}{5}-frac{(-1)^3}{3}right]$



    $left[frac{1}{5}-frac{1}{3}right] - left[frac{-1}{5}+frac{1^3}{3}right]$



    $left[frac{3-5}{15}-frac{-3+5}{15}right]$



    $left[frac{-2}{15}+frac{-2}{15}right] = frac{-4}{15}$



    I double checked the definite integral online and it got $frac{-4}{15}$ as well, so I guess that means I"m correct I'm just confused as to how an area can be negative unless I'm missing something obvious.



    My prof also left a note that says "*Make sure you have the entire region", where "entire" is underlined, is there more to this problem than what I've done?






    calculus







    share|cite|improve this question



























      0












      0








      0







      One of my last assignments for my first semester of calculus is a list of area, volume, and work problems using integrals, I worked one out, but can't find the answer anywhere, and just want to make sure I'm correct, here's my work:



      Find the area of the region bounded by the curves $y=x^2$ and $y=x^4$



      $x^4=x^2$



      $x^4-x^2=0$



      $x^2(x^2-1)=0$



      $x=1, -1$



      $int_1^1(x^4-x^2)dx$



      $left[frac{x^5}{5}-frac{x^3}{3}right]$ from $-1$ to $1$



      $left[frac{(1)^5}{5}-frac{(1)^3}{3}right] - left[frac{(-1)^5}{5}-frac{(-1)^3}{3}right]$



      $left[frac{1}{5}-frac{1}{3}right] - left[frac{-1}{5}+frac{1^3}{3}right]$



      $left[frac{3-5}{15}-frac{-3+5}{15}right]$



      $left[frac{-2}{15}+frac{-2}{15}right] = frac{-4}{15}$



      I double checked the definite integral online and it got $frac{-4}{15}$ as well, so I guess that means I"m correct I'm just confused as to how an area can be negative unless I'm missing something obvious.



      My prof also left a note that says "*Make sure you have the entire region", where "entire" is underlined, is there more to this problem than what I've done?






      calculus







      share|cite|improve this question















      One of my last assignments for my first semester of calculus is a list of area, volume, and work problems using integrals, I worked one out, but can't find the answer anywhere, and just want to make sure I'm correct, here's my work:



      Find the area of the region bounded by the curves $y=x^2$ and $y=x^4$



      $x^4=x^2$



      $x^4-x^2=0$



      $x^2(x^2-1)=0$



      $x=1, -1$



      $int_1^1(x^4-x^2)dx$



      $left[frac{x^5}{5}-frac{x^3}{3}right]$ from $-1$ to $1$



      $left[frac{(1)^5}{5}-frac{(1)^3}{3}right] - left[frac{(-1)^5}{5}-frac{(-1)^3}{3}right]$



      $left[frac{1}{5}-frac{1}{3}right] - left[frac{-1}{5}+frac{1^3}{3}right]$



      $left[frac{3-5}{15}-frac{-3+5}{15}right]$



      $left[frac{-2}{15}+frac{-2}{15}right] = frac{-4}{15}$



      I double checked the definite integral online and it got $frac{-4}{15}$ as well, so I guess that means I"m correct I'm just confused as to how an area can be negative unless I'm missing something obvious.



      My prof also left a note that says "*Make sure you have the entire region", where "entire" is underlined, is there more to this problem than what I've done?






      calculus




      calculus






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 7 at 7:04









      Eevee Trainer

      3,412325




      3,412325










      asked Dec 7 at 6:55









      Ben Dreslinski

      31




      31






















          1 Answer
          1






          active

          oldest

          votes


















          1














          I think you do indeed have the entire region. The only problem is that your integral should have been



          $$int_{-1}^1(x^2-x^4)dx$$



          instead, because $x^2≥x^4$ in the region where $-1<x<1$.






          share|cite|improve this answer























          • Oh I see, that would make it positive which makes much more sense. I knew it was the integral where f(x) > g(x), I just figured that x^4 is greater than x^2, totally forgot it was over that interval. Thanks a lot!
            – Ben Dreslinski
            Dec 7 at 7:03










          • Fixed it. ()()()
            – glowstonetrees
            Dec 7 at 17:50











          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%2f3029584%2farea-of-the-region-bounded-by-the-curves-y-x2-y-x4%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1














          I think you do indeed have the entire region. The only problem is that your integral should have been



          $$int_{-1}^1(x^2-x^4)dx$$



          instead, because $x^2≥x^4$ in the region where $-1<x<1$.






          share|cite|improve this answer























          • Oh I see, that would make it positive which makes much more sense. I knew it was the integral where f(x) > g(x), I just figured that x^4 is greater than x^2, totally forgot it was over that interval. Thanks a lot!
            – Ben Dreslinski
            Dec 7 at 7:03










          • Fixed it. ()()()
            – glowstonetrees
            Dec 7 at 17:50
















          1














          I think you do indeed have the entire region. The only problem is that your integral should have been



          $$int_{-1}^1(x^2-x^4)dx$$



          instead, because $x^2≥x^4$ in the region where $-1<x<1$.






          share|cite|improve this answer























          • Oh I see, that would make it positive which makes much more sense. I knew it was the integral where f(x) > g(x), I just figured that x^4 is greater than x^2, totally forgot it was over that interval. Thanks a lot!
            – Ben Dreslinski
            Dec 7 at 7:03










          • Fixed it. ()()()
            – glowstonetrees
            Dec 7 at 17:50














          1












          1








          1






          I think you do indeed have the entire region. The only problem is that your integral should have been



          $$int_{-1}^1(x^2-x^4)dx$$



          instead, because $x^2≥x^4$ in the region where $-1<x<1$.






          share|cite|improve this answer














          I think you do indeed have the entire region. The only problem is that your integral should have been



          $$int_{-1}^1(x^2-x^4)dx$$



          instead, because $x^2≥x^4$ in the region where $-1<x<1$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 7 at 14:39

























          answered Dec 7 at 6:58









          glowstonetrees

          2,285317




          2,285317












          • Oh I see, that would make it positive which makes much more sense. I knew it was the integral where f(x) > g(x), I just figured that x^4 is greater than x^2, totally forgot it was over that interval. Thanks a lot!
            – Ben Dreslinski
            Dec 7 at 7:03










          • Fixed it. ()()()
            – glowstonetrees
            Dec 7 at 17:50


















          • Oh I see, that would make it positive which makes much more sense. I knew it was the integral where f(x) > g(x), I just figured that x^4 is greater than x^2, totally forgot it was over that interval. Thanks a lot!
            – Ben Dreslinski
            Dec 7 at 7:03










          • Fixed it. ()()()
            – glowstonetrees
            Dec 7 at 17:50
















          Oh I see, that would make it positive which makes much more sense. I knew it was the integral where f(x) > g(x), I just figured that x^4 is greater than x^2, totally forgot it was over that interval. Thanks a lot!
          – Ben Dreslinski
          Dec 7 at 7:03




          Oh I see, that would make it positive which makes much more sense. I knew it was the integral where f(x) > g(x), I just figured that x^4 is greater than x^2, totally forgot it was over that interval. Thanks a lot!
          – Ben Dreslinski
          Dec 7 at 7:03












          Fixed it. ()()()
          – glowstonetrees
          Dec 7 at 17:50




          Fixed it. ()()()
          – glowstonetrees
          Dec 7 at 17:50


















          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.





          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


          Please pay close attention to the following guidance:


          • 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.


          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%2f3029584%2farea-of-the-region-bounded-by-the-curves-y-x2-y-x4%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

          Måne

          Image
          För jordens måne, se Månen. Några månar i solsystemet i skala till vår måne En måne , naturlig satellit eller drabant är en himlakropp som kretsar kring en planet eller en asteroid i ett solsystem. Ordet i bestämd form – månen – avser planeten jordens måne. Det finns inte i något fall belagt att någon måne har egna månar. De flesta stora månar i solsystemet är bundna i synkron rotation med en och samma sida vänd mot planeten. Ett exempel på en måne som inte beter sig på detta vis, är Hyperion som kaotiskt roterar kring Saturnus. En asteroidmåne är en måne som kretsar kring en asteroid. En måne kan ha bildats samtidigt som moderplaneten eller senare infångats av planetens gravitationskraft. I vårt eget solsystem känner vi i nuläget till 173 månar runt sex av de åtta planeterna [ 1 ] : Jorden – månen Mars – Phobos och Deimos Jupiter – Ganymedes (som är solsystemets största måne), Io, Europa, Callisto och 59 andra kända månar Saturnus – Titan (den enda måne...
          Read more

          Storängen

          Image
          För stadsdelen i Mariehamn, se Storängen, Åland. För ett naturreservat i Norrtälje kommun, se Storängens naturreservat. Storängen från luften i september 1967, vy mot väst. Till vänster syns Järlasjön och Saltsjöbanans spår. Till höger löper Värmdövägen och längst ut till höger syns anläggningsarbetena för Värmdöleden. Storängen är ett villaområde i centrala Nacka i Stockholms län. Området är beläget strax sydost om Nacka Forum vid Saltsjöbanan mellan Lillängen och Saltsjö-Duvnäs. Storängens villasamhälle grundades 1904 av "Tjänstemännens Egnahemsförening vid Storängen" och de första villorna började byggas samma år. Redan 1909 var 107 tomter bebyggda av totalt runt 160. [ 1 ] År 2013 fanns drygt 150 villor med lika många familjer i området. Egnahemsföreningen existerar fortfarande och fungerar idag som intresseförening för de boende i Storängen. Storängen anses vara ett av Sveriges bäst bevarade villasamhällen från 1900-talets början, tack vare sina tidstypis...
          Read more

          VLT Carioca

          Image
          VLT Carioca Informações Proprietário Prefeitura do Rio de Janeiro Local [[Rio de Janeiro, RJ]] País {{country data Brasil| country flag2| name = Brasil| variant = | size = }} Tipo de transporte Bonde (Veículo leve sobre trilhos) Número de linhas 2 (+1 em construção) Número de estações 26 (+3 em construção) Tráfego 80 mil passageiros/dia [ 1 ] (dezembro/2018) Website www.vltrio.com.br Funcionamento Início de funcionamento 5 de junho de 2016 (2 anos) Operadora(s) Concessionária do VLT Carioca S.A. Número de veículos 32 Alstom Citadis Comprimento dos veículos 44 m (144 ft) Dados técnicos Extensão do sistema 28 km (17,4 mi) [ 2 ] Frequência 5–10 minutos Bitola Bitola padrão/internacional 1 435 mm (4,71 ft) Eletrificação APS, 750V DC Velocidade média 15 km/h (9,32 mph) [ 2 ] Velocidade máxima 50 km/h (31,1 mph) ...
          Read more