Calculate the double integral. [closed]












-1












$begingroup$


First of all I would like to ask you if you know a very good material that could help me with range of integration.



$$int_{-1}^ {1} int_{0}^{x+2}y,dy,dx$$



How do I solve this question?










share|cite|improve this question











$endgroup$



closed as off-topic by Saad, Nosrati, RRL, Eric Wofsey, Leucippus Dec 22 '18 at 7:08


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Nosrati, RRL, Eric Wofsey, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    Is it $$int_{-1}^1int_0^{x+2}y~dydx~~?$$
    $endgroup$
    – Dave
    Dec 11 '18 at 0:36








  • 2




    $begingroup$
    tutorial.math.lamar.edu/Classes/CalcIII/DoubleIntegrals.aspx This is a good start may be for double integrals.
    $endgroup$
    – mm-crj
    Dec 11 '18 at 0:39


















-1












$begingroup$


First of all I would like to ask you if you know a very good material that could help me with range of integration.



$$int_{-1}^ {1} int_{0}^{x+2}y,dy,dx$$



How do I solve this question?










share|cite|improve this question











$endgroup$



closed as off-topic by Saad, Nosrati, RRL, Eric Wofsey, Leucippus Dec 22 '18 at 7:08


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Nosrati, RRL, Eric Wofsey, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    Is it $$int_{-1}^1int_0^{x+2}y~dydx~~?$$
    $endgroup$
    – Dave
    Dec 11 '18 at 0:36








  • 2




    $begingroup$
    tutorial.math.lamar.edu/Classes/CalcIII/DoubleIntegrals.aspx This is a good start may be for double integrals.
    $endgroup$
    – mm-crj
    Dec 11 '18 at 0:39
















-1












-1








-1





$begingroup$


First of all I would like to ask you if you know a very good material that could help me with range of integration.



$$int_{-1}^ {1} int_{0}^{x+2}y,dy,dx$$



How do I solve this question?










share|cite|improve this question











$endgroup$




First of all I would like to ask you if you know a very good material that could help me with range of integration.



$$int_{-1}^ {1} int_{0}^{x+2}y,dy,dx$$



How do I solve this question?







integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 3:47









Key Flex

8,58261233




8,58261233










asked Dec 11 '18 at 0:32









user2860452user2860452

588




588




closed as off-topic by Saad, Nosrati, RRL, Eric Wofsey, Leucippus Dec 22 '18 at 7:08


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Nosrati, RRL, Eric Wofsey, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Saad, Nosrati, RRL, Eric Wofsey, Leucippus Dec 22 '18 at 7:08


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Nosrati, RRL, Eric Wofsey, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    Is it $$int_{-1}^1int_0^{x+2}y~dydx~~?$$
    $endgroup$
    – Dave
    Dec 11 '18 at 0:36








  • 2




    $begingroup$
    tutorial.math.lamar.edu/Classes/CalcIII/DoubleIntegrals.aspx This is a good start may be for double integrals.
    $endgroup$
    – mm-crj
    Dec 11 '18 at 0:39




















  • $begingroup$
    Is it $$int_{-1}^1int_0^{x+2}y~dydx~~?$$
    $endgroup$
    – Dave
    Dec 11 '18 at 0:36








  • 2




    $begingroup$
    tutorial.math.lamar.edu/Classes/CalcIII/DoubleIntegrals.aspx This is a good start may be for double integrals.
    $endgroup$
    – mm-crj
    Dec 11 '18 at 0:39


















$begingroup$
Is it $$int_{-1}^1int_0^{x+2}y~dydx~~?$$
$endgroup$
– Dave
Dec 11 '18 at 0:36






$begingroup$
Is it $$int_{-1}^1int_0^{x+2}y~dydx~~?$$
$endgroup$
– Dave
Dec 11 '18 at 0:36






2




2




$begingroup$
tutorial.math.lamar.edu/Classes/CalcIII/DoubleIntegrals.aspx This is a good start may be for double integrals.
$endgroup$
– mm-crj
Dec 11 '18 at 0:39






$begingroup$
tutorial.math.lamar.edu/Classes/CalcIII/DoubleIntegrals.aspx This is a good start may be for double integrals.
$endgroup$
– mm-crj
Dec 11 '18 at 0:39












1 Answer
1






active

oldest

votes


















3












$begingroup$

The given bounds for $x$ are from $-1$ to $1$ and the given bounds for $y$ are from $0$ to $x+2$
$$int_{-1}^1int_0^{x+2} y dy dx$$



Now first take the inner integral $$int_0^{x+2} y dy=dfrac{(x+2)^2}{2}$$



Now we get $$int_{-1}^1dfrac{(x+2)^2}{2} dx=dfrac{13}{3}$$






share|cite|improve this answer









$endgroup$




















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    The given bounds for $x$ are from $-1$ to $1$ and the given bounds for $y$ are from $0$ to $x+2$
    $$int_{-1}^1int_0^{x+2} y dy dx$$



    Now first take the inner integral $$int_0^{x+2} y dy=dfrac{(x+2)^2}{2}$$



    Now we get $$int_{-1}^1dfrac{(x+2)^2}{2} dx=dfrac{13}{3}$$






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      The given bounds for $x$ are from $-1$ to $1$ and the given bounds for $y$ are from $0$ to $x+2$
      $$int_{-1}^1int_0^{x+2} y dy dx$$



      Now first take the inner integral $$int_0^{x+2} y dy=dfrac{(x+2)^2}{2}$$



      Now we get $$int_{-1}^1dfrac{(x+2)^2}{2} dx=dfrac{13}{3}$$






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        The given bounds for $x$ are from $-1$ to $1$ and the given bounds for $y$ are from $0$ to $x+2$
        $$int_{-1}^1int_0^{x+2} y dy dx$$



        Now first take the inner integral $$int_0^{x+2} y dy=dfrac{(x+2)^2}{2}$$



        Now we get $$int_{-1}^1dfrac{(x+2)^2}{2} dx=dfrac{13}{3}$$






        share|cite|improve this answer









        $endgroup$



        The given bounds for $x$ are from $-1$ to $1$ and the given bounds for $y$ are from $0$ to $x+2$
        $$int_{-1}^1int_0^{x+2} y dy dx$$



        Now first take the inner integral $$int_0^{x+2} y dy=dfrac{(x+2)^2}{2}$$



        Now we get $$int_{-1}^1dfrac{(x+2)^2}{2} dx=dfrac{13}{3}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 11 '18 at 0:36









        Key FlexKey Flex

        8,58261233




        8,58261233















            Popular posts from this blog

            Biblatex bibliography style without URLs when DOI exists (in Overleaf with Zotero bibliography)

            ComboBox Display Member on multiple fields

            Is it possible to collect Nectar points via Trainline?