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How does one calculate the area of a rectangle using a single integral?



The Next CEO of Stack OverflowEvaluating an integral using the Fundamental Theorem of CalculusOptimizing the area of a rectangle with one side against a wall using the am-gm inequalityHow can I tell the difference if my integral is measuring arc length or area?Proof request for finding the area of a rectangle using an ellipse and an integral.Trouble in finding the area of the curve using IntegrationCalculate integral with help the Euler's integralsGaussian Integral using single integrationCan arc length be a straight line?Area of a Rectangle using a Double Integral in Polar CoordinatesHow fast is the area of rectangle increasing?










1












$begingroup$


I tried to ask this in a different way and did not correctly explain myself.



I am ok integrating the line $y = x$ , let us say from $0$ to $2$ using calculus.
If I want to get the square I can easily multiply by two and using calculus the dimensions work. The area is a square and when we integrate we have a square.



Here is my question. If I use a straight line above the x axis my equation becomes $y$ = ( some constant) . Now I use calculus and integrate from 0 to 2 , I also get the correct answer BUT I have to imagine it is a rectangle because when you ingtegrate you have one $x$ term and it is not a square. The answers match OK it's the dimensions that bother me.



I did not do a good job explaining this on my previous question. Sorry










share|cite|improve this question











$endgroup$
















    1












    $begingroup$


    I tried to ask this in a different way and did not correctly explain myself.



    I am ok integrating the line $y = x$ , let us say from $0$ to $2$ using calculus.
    If I want to get the square I can easily multiply by two and using calculus the dimensions work. The area is a square and when we integrate we have a square.



    Here is my question. If I use a straight line above the x axis my equation becomes $y$ = ( some constant) . Now I use calculus and integrate from 0 to 2 , I also get the correct answer BUT I have to imagine it is a rectangle because when you ingtegrate you have one $x$ term and it is not a square. The answers match OK it's the dimensions that bother me.



    I did not do a good job explaining this on my previous question. Sorry










    share|cite|improve this question











    $endgroup$














      1












      1








      1





      $begingroup$


      I tried to ask this in a different way and did not correctly explain myself.



      I am ok integrating the line $y = x$ , let us say from $0$ to $2$ using calculus.
      If I want to get the square I can easily multiply by two and using calculus the dimensions work. The area is a square and when we integrate we have a square.



      Here is my question. If I use a straight line above the x axis my equation becomes $y$ = ( some constant) . Now I use calculus and integrate from 0 to 2 , I also get the correct answer BUT I have to imagine it is a rectangle because when you ingtegrate you have one $x$ term and it is not a square. The answers match OK it's the dimensions that bother me.



      I did not do a good job explaining this on my previous question. Sorry










      share|cite|improve this question











      $endgroup$




      I tried to ask this in a different way and did not correctly explain myself.



      I am ok integrating the line $y = x$ , let us say from $0$ to $2$ using calculus.
      If I want to get the square I can easily multiply by two and using calculus the dimensions work. The area is a square and when we integrate we have a square.



      Here is my question. If I use a straight line above the x axis my equation becomes $y$ = ( some constant) . Now I use calculus and integrate from 0 to 2 , I also get the correct answer BUT I have to imagine it is a rectangle because when you ingtegrate you have one $x$ term and it is not a square. The answers match OK it's the dimensions that bother me.



      I did not do a good job explaining this on my previous question. Sorry







      calculus integration






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 19 at 19:48









      ADITYA PRAKASH

      365110




      365110










      asked Mar 19 at 16:00









      SedumjoySedumjoy

      661316




      661316




















          2 Answers
          2






          active

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          3












          $begingroup$

          The dimensions do work correctly in case of a rectangle as well. Let us say the height of the rectangle is $k$ and width is $b-a$.



          Recall that integration is basically summing up infinitely many small rectangles with infinitesimally small widths.



          $$beginalignedtextArea &=lim_nto inftysum_i=1^nunderbracek_textheightcdotunderbraceleft(dfracb-anright)_textwidth of each infinitesimal rectangle\&=int_a^bunderbracek_textunitsunderbracemathrm dx_textunits\&=kxbiggr|_a^b=k(b-a) text sq. unitsendaligned$$






          share|cite|improve this answer









          $endgroup$




















            1












            $begingroup$

            Consider a rectangle formed by the equation $y=K$ extending from $x=a$ to $x=b$.
            enter image description hereenter image description here



            Calculate the area of a small rectangular strip formed by the coordinates $(x,0)$, $(x+dx,0)$,$(x+dx,y)$ and $(x,y)$.



            The area of this rectangle will be, $ydx$. Convert the whole rectangular area into smaller rectangles of area $ydx$ and sum all of them from $x=a$ to $x=b$ using integration.



            The area under the curve $y=f(x)$ bounded by the X Axis and the lines $x=a$ and $x=b$ is given by,



            $$int_x=a^x=bydx=int_x=a^x=bf(x)dx=int_x=a^x=bKdx=K(b-a)$$
            Hope this helps...






            share|cite|improve this answer









            $endgroup$













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              2 Answers
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              2 Answers
              2






              active

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              active

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              active

              oldest

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              3












              $begingroup$

              The dimensions do work correctly in case of a rectangle as well. Let us say the height of the rectangle is $k$ and width is $b-a$.



              Recall that integration is basically summing up infinitely many small rectangles with infinitesimally small widths.



              $$beginalignedtextArea &=lim_nto inftysum_i=1^nunderbracek_textheightcdotunderbraceleft(dfracb-anright)_textwidth of each infinitesimal rectangle\&=int_a^bunderbracek_textunitsunderbracemathrm dx_textunits\&=kxbiggr|_a^b=k(b-a) text sq. unitsendaligned$$






              share|cite|improve this answer









              $endgroup$

















                3












                $begingroup$

                The dimensions do work correctly in case of a rectangle as well. Let us say the height of the rectangle is $k$ and width is $b-a$.



                Recall that integration is basically summing up infinitely many small rectangles with infinitesimally small widths.



                $$beginalignedtextArea &=lim_nto inftysum_i=1^nunderbracek_textheightcdotunderbraceleft(dfracb-anright)_textwidth of each infinitesimal rectangle\&=int_a^bunderbracek_textunitsunderbracemathrm dx_textunits\&=kxbiggr|_a^b=k(b-a) text sq. unitsendaligned$$






                share|cite|improve this answer









                $endgroup$















                  3












                  3








                  3





                  $begingroup$

                  The dimensions do work correctly in case of a rectangle as well. Let us say the height of the rectangle is $k$ and width is $b-a$.



                  Recall that integration is basically summing up infinitely many small rectangles with infinitesimally small widths.



                  $$beginalignedtextArea &=lim_nto inftysum_i=1^nunderbracek_textheightcdotunderbraceleft(dfracb-anright)_textwidth of each infinitesimal rectangle\&=int_a^bunderbracek_textunitsunderbracemathrm dx_textunits\&=kxbiggr|_a^b=k(b-a) text sq. unitsendaligned$$






                  share|cite|improve this answer









                  $endgroup$



                  The dimensions do work correctly in case of a rectangle as well. Let us say the height of the rectangle is $k$ and width is $b-a$.



                  Recall that integration is basically summing up infinitely many small rectangles with infinitesimally small widths.



                  $$beginalignedtextArea &=lim_nto inftysum_i=1^nunderbracek_textheightcdotunderbraceleft(dfracb-anright)_textwidth of each infinitesimal rectangle\&=int_a^bunderbracek_textunitsunderbracemathrm dx_textunits\&=kxbiggr|_a^b=k(b-a) text sq. unitsendaligned$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 19 at 16:18









                  Paras KhoslaParas Khosla

                  2,716423




                  2,716423





















                      1












                      $begingroup$

                      Consider a rectangle formed by the equation $y=K$ extending from $x=a$ to $x=b$.
                      enter image description hereenter image description here



                      Calculate the area of a small rectangular strip formed by the coordinates $(x,0)$, $(x+dx,0)$,$(x+dx,y)$ and $(x,y)$.



                      The area of this rectangle will be, $ydx$. Convert the whole rectangular area into smaller rectangles of area $ydx$ and sum all of them from $x=a$ to $x=b$ using integration.



                      The area under the curve $y=f(x)$ bounded by the X Axis and the lines $x=a$ and $x=b$ is given by,



                      $$int_x=a^x=bydx=int_x=a^x=bf(x)dx=int_x=a^x=bKdx=K(b-a)$$
                      Hope this helps...






                      share|cite|improve this answer









                      $endgroup$

















                        1












                        $begingroup$

                        Consider a rectangle formed by the equation $y=K$ extending from $x=a$ to $x=b$.
                        enter image description hereenter image description here



                        Calculate the area of a small rectangular strip formed by the coordinates $(x,0)$, $(x+dx,0)$,$(x+dx,y)$ and $(x,y)$.



                        The area of this rectangle will be, $ydx$. Convert the whole rectangular area into smaller rectangles of area $ydx$ and sum all of them from $x=a$ to $x=b$ using integration.



                        The area under the curve $y=f(x)$ bounded by the X Axis and the lines $x=a$ and $x=b$ is given by,



                        $$int_x=a^x=bydx=int_x=a^x=bf(x)dx=int_x=a^x=bKdx=K(b-a)$$
                        Hope this helps...






                        share|cite|improve this answer









                        $endgroup$















                          1












                          1








                          1





                          $begingroup$

                          Consider a rectangle formed by the equation $y=K$ extending from $x=a$ to $x=b$.
                          enter image description hereenter image description here



                          Calculate the area of a small rectangular strip formed by the coordinates $(x,0)$, $(x+dx,0)$,$(x+dx,y)$ and $(x,y)$.



                          The area of this rectangle will be, $ydx$. Convert the whole rectangular area into smaller rectangles of area $ydx$ and sum all of them from $x=a$ to $x=b$ using integration.



                          The area under the curve $y=f(x)$ bounded by the X Axis and the lines $x=a$ and $x=b$ is given by,



                          $$int_x=a^x=bydx=int_x=a^x=bf(x)dx=int_x=a^x=bKdx=K(b-a)$$
                          Hope this helps...






                          share|cite|improve this answer









                          $endgroup$



                          Consider a rectangle formed by the equation $y=K$ extending from $x=a$ to $x=b$.
                          enter image description hereenter image description here



                          Calculate the area of a small rectangular strip formed by the coordinates $(x,0)$, $(x+dx,0)$,$(x+dx,y)$ and $(x,y)$.



                          The area of this rectangle will be, $ydx$. Convert the whole rectangular area into smaller rectangles of area $ydx$ and sum all of them from $x=a$ to $x=b$ using integration.



                          The area under the curve $y=f(x)$ bounded by the X Axis and the lines $x=a$ and $x=b$ is given by,



                          $$int_x=a^x=bydx=int_x=a^x=bf(x)dx=int_x=a^x=bKdx=K(b-a)$$
                          Hope this helps...







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Mar 19 at 17:34









                          SNEHIL SANYALSNEHIL SANYAL

                          656110




                          656110



























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