Area between the curve $y$, the $x$-axis and the lines

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Find the total area between the curve $y=x^2-4x+3$, the $x$-axis and the lines $x=0$ and $x=3$.




I have drawn the graph and concluded that:



$$int_0^1 0 - (x^2-4x+3), dx + int_1^3 0 - (x^2-4x+3), dx = -frac-263$$ Which differs from the answer key: $4$. What is the mistake here? Is it feasiable to solve this problem without relying on the graph?










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  • the first integral should be positive, because the upper graphic is $x^2-4x+3$
    – haqnatural
    Aug 31 at 4:04














up vote
1
down vote

favorite













Find the total area between the curve $y=x^2-4x+3$, the $x$-axis and the lines $x=0$ and $x=3$.




I have drawn the graph and concluded that:



$$int_0^1 0 - (x^2-4x+3), dx + int_1^3 0 - (x^2-4x+3), dx = -frac-263$$ Which differs from the answer key: $4$. What is the mistake here? Is it feasiable to solve this problem without relying on the graph?










share|cite|improve this question





















  • the first integral should be positive, because the upper graphic is $x^2-4x+3$
    – haqnatural
    Aug 31 at 4:04












up vote
1
down vote

favorite









up vote
1
down vote

favorite












Find the total area between the curve $y=x^2-4x+3$, the $x$-axis and the lines $x=0$ and $x=3$.




I have drawn the graph and concluded that:



$$int_0^1 0 - (x^2-4x+3), dx + int_1^3 0 - (x^2-4x+3), dx = -frac-263$$ Which differs from the answer key: $4$. What is the mistake here? Is it feasiable to solve this problem without relying on the graph?










share|cite|improve this question














Find the total area between the curve $y=x^2-4x+3$, the $x$-axis and the lines $x=0$ and $x=3$.




I have drawn the graph and concluded that:



$$int_0^1 0 - (x^2-4x+3), dx + int_1^3 0 - (x^2-4x+3), dx = -frac-263$$ Which differs from the answer key: $4$. What is the mistake here? Is it feasiable to solve this problem without relying on the graph?







calculus integration definite-integrals area






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asked Aug 31 at 3:56









Mauricio Mendes

968




968











  • the first integral should be positive, because the upper graphic is $x^2-4x+3$
    – haqnatural
    Aug 31 at 4:04
















  • the first integral should be positive, because the upper graphic is $x^2-4x+3$
    – haqnatural
    Aug 31 at 4:04















the first integral should be positive, because the upper graphic is $x^2-4x+3$
– haqnatural
Aug 31 at 4:04




the first integral should be positive, because the upper graphic is $x^2-4x+3$
– haqnatural
Aug 31 at 4:04










3 Answers
3






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up vote
1
down vote



accepted










The required area is the green area:



$hspace3cm$enter image description here



The proper set up of integrals:
$$int_0^1 (x^2-4x+3)-0, dx + int_1^3 0 - (x^2-4x+3), dx = frac83.$$






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    1
    down vote













    $int_0^1 (x^2-4x+3)mathbb dx-int_1^3(x^2-4x+3)mathbb dx=[(fracx^33-2x^2+3x)]_0^1-[fracx^33-2x^2+3x]_1^3=frac43-[0-frac43]=frac83$






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      0
      down vote













      Hint:



      Area under curve of $y=f(x)$ in the domain of $a < x < b$ is given by
      beginalign
      mathrmArea ; = ; int_a^b Big|, f(x) , Big| ; dx
      endalign
      Take absolute value of your integrand.



      For your case, $f(x) = x^2 - 4x + 3$.
      Factorization gives $f(x) = (x-3)(x-1)$.
      Now you can easily see when $f(x) > 0$ or $<0$ without using a graph.






      share|cite|improve this answer




















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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        1
        down vote



        accepted










        The required area is the green area:



        $hspace3cm$enter image description here



        The proper set up of integrals:
        $$int_0^1 (x^2-4x+3)-0, dx + int_1^3 0 - (x^2-4x+3), dx = frac83.$$






        share|cite|improve this answer
























          up vote
          1
          down vote



          accepted










          The required area is the green area:



          $hspace3cm$enter image description here



          The proper set up of integrals:
          $$int_0^1 (x^2-4x+3)-0, dx + int_1^3 0 - (x^2-4x+3), dx = frac83.$$






          share|cite|improve this answer






















            up vote
            1
            down vote



            accepted







            up vote
            1
            down vote



            accepted






            The required area is the green area:



            $hspace3cm$enter image description here



            The proper set up of integrals:
            $$int_0^1 (x^2-4x+3)-0, dx + int_1^3 0 - (x^2-4x+3), dx = frac83.$$






            share|cite|improve this answer












            The required area is the green area:



            $hspace3cm$enter image description here



            The proper set up of integrals:
            $$int_0^1 (x^2-4x+3)-0, dx + int_1^3 0 - (x^2-4x+3), dx = frac83.$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Aug 31 at 4:37









            farruhota

            15.2k2734




            15.2k2734




















                up vote
                1
                down vote













                $int_0^1 (x^2-4x+3)mathbb dx-int_1^3(x^2-4x+3)mathbb dx=[(fracx^33-2x^2+3x)]_0^1-[fracx^33-2x^2+3x]_1^3=frac43-[0-frac43]=frac83$






                share|cite|improve this answer
























                  up vote
                  1
                  down vote













                  $int_0^1 (x^2-4x+3)mathbb dx-int_1^3(x^2-4x+3)mathbb dx=[(fracx^33-2x^2+3x)]_0^1-[fracx^33-2x^2+3x]_1^3=frac43-[0-frac43]=frac83$






                  share|cite|improve this answer






















                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    $int_0^1 (x^2-4x+3)mathbb dx-int_1^3(x^2-4x+3)mathbb dx=[(fracx^33-2x^2+3x)]_0^1-[fracx^33-2x^2+3x]_1^3=frac43-[0-frac43]=frac83$






                    share|cite|improve this answer












                    $int_0^1 (x^2-4x+3)mathbb dx-int_1^3(x^2-4x+3)mathbb dx=[(fracx^33-2x^2+3x)]_0^1-[fracx^33-2x^2+3x]_1^3=frac43-[0-frac43]=frac83$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Aug 31 at 4:22









                    Chris Custer

                    6,3472622




                    6,3472622




















                        up vote
                        0
                        down vote













                        Hint:



                        Area under curve of $y=f(x)$ in the domain of $a < x < b$ is given by
                        beginalign
                        mathrmArea ; = ; int_a^b Big|, f(x) , Big| ; dx
                        endalign
                        Take absolute value of your integrand.



                        For your case, $f(x) = x^2 - 4x + 3$.
                        Factorization gives $f(x) = (x-3)(x-1)$.
                        Now you can easily see when $f(x) > 0$ or $<0$ without using a graph.






                        share|cite|improve this answer
























                          up vote
                          0
                          down vote













                          Hint:



                          Area under curve of $y=f(x)$ in the domain of $a < x < b$ is given by
                          beginalign
                          mathrmArea ; = ; int_a^b Big|, f(x) , Big| ; dx
                          endalign
                          Take absolute value of your integrand.



                          For your case, $f(x) = x^2 - 4x + 3$.
                          Factorization gives $f(x) = (x-3)(x-1)$.
                          Now you can easily see when $f(x) > 0$ or $<0$ without using a graph.






                          share|cite|improve this answer






















                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            Hint:



                            Area under curve of $y=f(x)$ in the domain of $a < x < b$ is given by
                            beginalign
                            mathrmArea ; = ; int_a^b Big|, f(x) , Big| ; dx
                            endalign
                            Take absolute value of your integrand.



                            For your case, $f(x) = x^2 - 4x + 3$.
                            Factorization gives $f(x) = (x-3)(x-1)$.
                            Now you can easily see when $f(x) > 0$ or $<0$ without using a graph.






                            share|cite|improve this answer












                            Hint:



                            Area under curve of $y=f(x)$ in the domain of $a < x < b$ is given by
                            beginalign
                            mathrmArea ; = ; int_a^b Big|, f(x) , Big| ; dx
                            endalign
                            Take absolute value of your integrand.



                            For your case, $f(x) = x^2 - 4x + 3$.
                            Factorization gives $f(x) = (x-3)(x-1)$.
                            Now you can easily see when $f(x) > 0$ or $<0$ without using a graph.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Aug 31 at 4:06









                            K_inverse

                            22829




                            22829



























                                 

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