A double integral

(5 pts)

The first assignment from our online presentations page asks you to compute a double integral. You should choose your double integral by forum login name and answer the question by responding here. For example, the web page says that my double integral is:

$$\int _0^1\int _0^3\left(-3 x^2-4 x y+9 y^2\right)dydx.$$

Thus, if I were going to answer the question, then that's the integral I should work out.

audreybenjamin
«1

Comments

  • The webpage says that my integral is:
    %% int_0^1 int_0^3 (-3x^2 + 6xy - 3y^2) dx dy.%%

    I can evaluate this as follows:

    %% int_0^1 int_0^3 (-3x^2 + 6xy - 3y^2) dx dy = int_0^1 (-3x^2y+3xy^2-y^3)|_0^3 dx =%%

    %% int_0^1 (-9x^2+27x-27) dx = (-3x^3 + 27/2 x^2 - 27x)|_0^1 = -3 + 27/2 - 27 = -33/2.%%

    mark
  • edited March 26

    Okay then, I guess I'm first.


    My double integral is:
    $$A=\int_0^3 \int_0^2 (-3x^2+2xy+3y^2)dydx.$$

    The double integral can be split into the addition of three separate double integrals:

    $$A=\int_0^3 \int_0^2 -3x^2 \ dydx + \int_0^3 \int_0^2 2xy \ dydx + \int_0^3 \int_0^2 3y^2 \ dydx.$$

    We then solve the inner integral of each and evaluate over the limits of integration (in my case starting with integrating with respect to %%y%% and from %%y=0%% to %%y=2%%).

    After taking the integral and evaluating %%y%% we get:
    $$A=\int_0^3 -3yx^2 \ dx \Big|_0^2 + \int_0^3 y^2x \ dx \Big|_0^2 + \int_0^3 y^3 \ dx \Big|_0^2.$$
    $$A=\int_0^3 -6x^2 \ dx + \int_0^3 4x \ dx + \int_0^3 8 \ dx.$$

    We repeat with respect to %%x%% now to get:

    $$A=-54+18+24=-12$$

    mark
  • edited March 26

    My assignment gives me the integral:
    $$A = \int_0^3 \int_0^3 (-3x^2 + 2xy + 9y^2) dydx.$$
    We can first integrate the inside, and we get
    $$A = \int_0^3 (-3x^2y + xy^2 + 3y^3)\Big|_0^3 dx = \int_0^3 (-9x^2 + 9x +81) dx$$
    Next, we integrate as normal and evaluate as a normal integral.
    $$= (-3x^3 +\frac{9x}{2}^2 + 81x)\Big|_0^3 = -81 + \frac{729}{4} + 243 = \frac{1377}{4}$$

    mark
  • edited March 26

    The integral given to me is $$A=\int _0^1 \int _0^1 (6x^2-2xy-6y^2)dydx.$$

    We can start off on the inner integral which would be the y integral so the x-values stay constant. $$A=\int _0^1(6x^2-xy^2-2y^3|^1_0)dx= \int^1_0 (6x^2-x-2)dx.$$

    After we get to one derivative, we just solve that and ta-da!
    $$A= \int^1_0 (6x^2-x-2)dx= 3x^3- \frac{1}{2} x^2-2x |^1_0 = 2- \frac{1}{2}-2=-\frac{1}{2}.$$

    mark
  • edited April 14

    My integral was:
    $$\int_0^2 \int_0^2 6x^2+6xy+9y^2 \,dy \,dx$$

    First, integrate with respect to y, start with the inner integral:
    $$\int_0^2 6x^2+6xy+9y^2 \,dy$$

    $$= (6x^2y+3xy^2+3y^3)\Big|_0^2 12x^2+12x+24.$$

    Now, integrate this with respect to x, do the outer integral:
    $$\int_0^2 12x^2+12x+24 \,dx$$

    $$=(4x^3+6x^2+24x)\Big|_0^2 = 4(8)+6(4)+24(2) = 104.$$

    mark
  • edited March 26

    My integral is :
    %%int_0^3int_0^2(-3x^2-2xy+3y^2)dydx%%
    So
    %%int_0^3int_0^2(-3x^2-2xy+3y^2)dydx=int_0^3(-3x^2y-xy^2+y^3|_0^2)dx=%%
    %%int_0^3(-6x^2-4x+8)dx=-2x^3-2x^2+8x|_0^3=-48%%

    mark
  • edited March 26

    my integral is %%int_0^3int_0^3(9x^2-2xy+3y^2)dydx%%

    first I integrated with respect to y,
    %%int_0^3int_0^3(9x^2-2xy+3y^2)dydx=9x^2y-xy^2+y^3|_0^3=27x^2-9x+27%%
    then I integrated with respect to x,
    %%int_0^3(27x^2-9x+27)dx=9x^3-9x^2+27x|_0^3=
    9(27)-(81/2)+27(3)=(567/2)%%

    mark
  • edited March 31

    My integral is %%int_0^1int_0^2 (9x^2-4xy-3y^2)dydx%%

    I integrated with respect to y,
    %%int_0^1int_0^2(9x^2-4xy-3y^2)dydx =int_0^1(9x^2y-2xy^2-y^3)dx|_0^2%%
    %%=int_0^1(9x^2(2)-2x(2)^2-2^3)-(0-0-0)dx%%
    %%=int_0^1(18x^2-8x-8)dx%%

    Then I integrated with respect to x,
    %%int_0^2(18x^2-8x-8)dy%%
    %%=int_0^2(6x^3-4x^2-8x)dy|_0^1%%
    %%=(6(1)^3-4(1)^2-8(1))%%
    %%=(6-4-8)%%
    %%=-6%%

  • gusgus
    edited March 26

    My integral is $$\int^2_0\int^1_0 (-6x^2-2xy+9y^2)dydx$$
    The first step would be to integrate with respect to y:
    $$\int ^2_0\int^1_0 (-6x^2-2xy+9y^2)dydx=\int^2_0(-6x^2y-2/2xy^2+9/3y^3)\Big|^1_0dx$$
    $$=\int^2_0(-6x^2(1)-x(1)^2+3(1)^3)-(0-0+0)dx=\int^2_0(-6x^2-x+3)dx$$
    The next step would be to integrate with respect to x:
    $$\int^2_0(-6x^2-x+3)dx=-6/3x^3-x^2/2+3x\Big|^2_0=(-2(2)^3-(2)^2/2+3(2))-(0-0+0)=-16-2+6=-12$$

    mark
  • edited March 26

    My integral is;
    $\int_0^1 \int_0^1 (-6x^2-2xy-6y^2) dydx$.

    First you take the integral of the middle part "$\int_0^1 (-6x^2-2xy-6y^2)dy$" which equals:
    $(-6yx^2-xy^2-2y^3)|_0^1=(-6(1)x^2-x(1)^2-2(1)^3)-(-6(0)x^2-x(0)^2-2(0)^3)$

    $=-6x^2-x-2$

    Then you plug that back into for the middle intergral:

    $\int_0^1 (-6x^2-x-2) dx =(-2x^3- \frac{1}{2}x^2-2x)|_0^1$

    $=(-2(1)^3-\frac{1}{2}(1)^2-2(1))-(-2(0)^3-\frac{1}{2}(0)^2-2(0))$

    $-2-\frac{1}{2}-2=\frac{-9}{2}$

    mark
  • edited March 26

    My integral is:
    $$A=\int_{0}^{1}\int_{0}^{1}(9x^2 - 4xy - 6y^2)dydx$$

    take the integral w/ respect to y

    $$\int_{0}^{1}[9x^2y-4x(\frac{1}{2}y^2)-6(\frac{}1{3}y^3)]|_{0}^{1}dx$$

    $$\int_{0}^{1}(9x^2y-2xy^2-2y^3)|_{0}^{1}dx$$

    evaluate the integral from 0 to 1

    $$\int_{0}^{1}[9x^2(1)-2x(1)^2-2(1)^3]-(0)dx$$

    $$\int_{0}^{1}(9x^2-2x-2)dx$$

    take the integral w/ respect to x

    $$[9(\frac{1}{3}x^3)-2(\frac{1}{2}x^2)-(2x)]|_{0}^{1}$$

    $$(3x^3-x^2-2x)|_{0}^{1}$$

    evaluate the integral from 0 to 1

    $$[3(1)^3-(1)^2-2(1)]-(0)$$

    $$3-1-2$$

    $$0$$

    mark
  • edited March 26

    My integral is:
    $$ \int_{0}^{3} \int_{0}^{1} (3x^2-4xy-3y^2)dydx$$

    Firstly we integrate the inside with respect to $y$:

    $$ \int_{0}^{3} \int_{0}^{1} (3x^2-4xy-3y^2)dydx $$

    $$= \int_{0}^{3}(3x^2y-2xy^2-y^3) \Big|_0^1 dx$$

    $$ = \int_{0}^{3} ((3x^2(1)-2x(1)^2-(1)^3) dx$$

    $$ = \int_{0}^{3} (3x^2-2x-1) dx $$

    Next we integrate with respect to $x$:
    $$ \int_{0}^{3}(3x^2-2x-1) dx $$

    $$= (x^3-x^2-x) \Big|_0^3 $$

    $$= 3^3-3^2-3-0 $$

    $$= 15$$

    mark
  • edited March 26

    My integral was:
    %%\int_{0}^{3}\int_{0}^{3} (9x^2+6xy-3y^2)dydx%%
    Integrate with respect to y:
    %%\int_{0}^{3}\int_{0}^{3}(9x^2+6xy-3y^2)dydx%%

    %% =\int_{0}^{3}(9x^2y+3xy^2-y^3)|_{0}^{3}dx%%

    %% =\int_{0}^{3}(27x^2+27x-27)dx%%

    Integrate with respect to x:
    %% =\int_{0}^{3}(27x^2+27x-27)dx%%

    %% =(9x^3+\frac{27}{2}x^2-27x)|_0^3%%

    %% =9(3)^3+\frac{27}{2}(3)^2-27(3)%%

    %% =243+\frac{243}{2}-81%%

    %% =\frac{567}{2}%%

    mark
  • My given integral is:
    $$ \int_0^3 \int_0^2 (-3x^2 + 2xy + 3y^2) dydx $$

    First integrate the inside:
    $$ = \int_0^3 (-3x^2y + xy^2 + y^3) |_0^2 dx $$
    $$ = \int_0^3 (-6x^2 + 4x + 8) dx $$

    Now integrate and evaluate like regular:
    $$ = -2x^3 + 2x^2 + 8x |_0^3 $$
    $$ = -2(3)^3 + 2(3)^2 + 8(3) $$
    $$ = -12 $$

    mark
  • @zachary Looks good! I did edit your post for formatting a bit. For example, to get something like
    $$(x^2+y)\Big|_0^2 = 4y,$$
    rather than
    $x^2+y$ from 0 to 2.

    You can enter edit mode or just right-click on the math to see what I typed.

  • @veronica I think you made a little mistake here. Here's the first couple of steps where I'm using color and parentheses to really highlight the order in which things happen:

    $$\begin{align}
    \int _0^1\color{red}{\left(\int _0^2\left(9 x^2-4 x y-3 y^2\right)dy\right)}dx &=
    \int_0^1 \color{red}{\left(9 x^2 y-2 x y^2-y^3\right)\Big|_0^2} dx \\
    &= \int_0^1 \color{red}{\left(18 x^2-8 x-8\right)} dx.
    \end{align}$$

    Note that the inner integral (which is red and in parentheses) is with respect to $y$ and that the bounds for $y$ are specified to be from 0 to 2. You integrated correctly with respect to $y$ but then evaluated from 0 to 1.

    I'll let you take it from there!

  • edited March 27

    My given integral is:
    %%int_0^2int_0^3(9x^2+4xy+6y^2)dydx%%

    First integrate the inside with respect to y holding x constant:
    %%=int_0^2(9x^2y+2xy^2+2y^3)|_0^3dx%%
    %%=int_0^2(27x^2+18x+54)dx%%

    Then integrate and evaluate as usual:
    %%=9x^3+9x^2+54x|_0^2%%
    %%=9(2)^3+9(2)^2+54(2)%%
    %%=216%%

    mark
  • edited April 6

    My given integral...

    %%int_0^2int_0^1(-3x^2+6xy-6y^2)dydx%%

    After integrating with respect to Y...

    %%int_0^2(-3x^2+3xy^2-2y^3)dx%%

    After plugging in 1 and 0...

    %%int_0^2(-3x^2+3x-2)dx%%

    After integrating with respect to X...

    %%(-x^3+(3/2)x^2-2x)%%

    Evaluating from 0 to 2

    %%-(2)^3+(3/2)(2)^2-(2)%%

    %%-8+6-4=-6%%

    mark
  • @Mitchell Don't indent your asciimath input! Doing so tells the software to interpret your input as code, rather than math. I edited your post so that the first math section is aligned correctly - and look how pretty it looks!

  • edited March 30

    The integral assigned to me was:

    %%int_0^3int_0^2(9x^2+2xy+9y^2)dydx%%

    After integrating the inside with respect to Y, and hold the X value constant:

    %%int_0^3 (9x^2y+2xy^2/2+9y^3/3)|_0^2dx%%

    %%int_0^3 (18x^2+4x+24)dx%%

    Now integrate with respect to X:

    %%(18x^3/3+4x^2/2+24x)|_0^3%%

    %%162+18+72%%

    %%=252%%

    mark
  • edited March 30

    My integral is

    %%int_0^2int_0^1 (-6x^2+4xy-6y^2)dydx%%

    First, integrate with respect to y...

    %%int_0^2 (-6x^2y+2xy^2-2y^3)|_0^1dx%%

    %%int_0^2 (-6x^2+2x-2)dx%%

    Then integrate with respect to x...

    %%(-2x^3+x^2-2x)|_0^2%%

    %%(-2(2)^3+(2)^2-2(2))-(0+0-0)%%

    %%=-16%%

    mark
  • edited March 30

    My given integral is,

    %%A= int_0^1int_0^2 (9x^2+2xy+9y^2)dy dx%%

    We start by integrating dy, which is the inner integral,

    %%int_0^1(9x^2y+xy^2+9y^2)|_0^2 dx%%

    Plugging in our values for the limits of integration,

    %%int_0^1(9(2)x^2+(2)^2x+9/3(2)^2)-0dx%%

    Our integral with respect to dx now looks like,

    %%int_0^1(18x^2+4x+12)dx%%

    By conducting the same process as above, we obtain,

    %%6x^3+2x^2+12|_0^1%%

    Evaluating,

    %%6(1)^3+2(1)^2+12(1) - 0%%

    Our final answer is 20.

  • edited March 30

    My Integral was

    $$\int_{0}^{2}\int_{0}^{2}(3x^2+6xy-3y^2)dydx$$

    First Integrate with respect to y

    $$\int_{0}^{2}(3yx^2+3xy^2-y^3)|_{0}^{2}dx$$

    $$\int_{0}^{2}(6x^2+12x-8)-(0)dx$$

    $$\int_{0}^{2}(6x^2+12x-8)dx$$

    Now Integrate with respect to x

    $$\int_{0}^{2}(6x^2+12x-8)dx$$

    $$(2x^3+6x^2-8x)|_{0}^{2}$$

    $$(16+24-16)-(0)$$

    And the final answer is :

    $$24$$

    mark
  • edited March 30

    The integral I got was:

    %%int_0^3int_0^3 (-6x^2+6xy+6y^2)dydx%%

    First, we integrate the inner integral with respect to y and hold x constant:

    %%int_0^3 (-6x^2y+3xy^2-2y^3)|_0^3dx%%
    %%=int_0^3 (-18x^2+27x+54)dx%%

    Next, we integrate with respect to x:

    %%=(-6x^3+27/2x^2+54x)|_0^3%%
    %%=-162+(243/2)+162%%
    %%=(243/2)%%
    %%=121.5%%

    mark
  • edited March 30

    My integral is
    %%int_0^3int_0^2(3x^2-4xy-6y^2)dydx%%
    First integrate inner integral
    %%int_0^3(3x^2y-2xy^2-2y^3)|_0^2dx%%
    %%int_0^3(3(2)x^2-2(2)^2x-2(2)^3)-(0)dx%%
    %%int_0^3(6x^2-8x-16)dx%%
    Then integrate outer integral
    %%(2x^3-4x^2-16x)|_0^3%%
    %%(2(3)^3-4(3)^2-16(3))-(0)%%
    %%54-36-48=-30%%

    davidmark
  • edited March 30

    My integral is:

    $$A = \int_{0}^{3}\int_{0}^{3}6x^2-4xy+9y^2dydx.$$

    First we integrate with respect to y
    $$\int_{0}^{3}\int_{0}^{3}6x^2-4xy+9y^2dydx = \int_{0}^{3}(6x^2-2xy^2+3y^3)\Big|^3_0dx$$
    $$= \int_{0}^{3}\Big(6x^2(3)-2x(3)^2+3(3)^3\Big)dx$$
    $$ = \int_{0}^{3}(18x^2-18x+81)dx.$$

    Next, we integrate with respect to x
    $$\int_{0}^{3}(18x^2-18x+81)dx = (6x^3-9x^2+81x)\Big|^3_0$$
    $$=162-81+243$$
    $$A=324.$$

    mark
  • edited March 30

    @david Please use the typesetting tools that are described in Meta. There are a number of good reasons for this, not the least of which is the fact that I can (to some extent) automate the grading process when the answers are input as text, rather than images as you've done.

    I might also point out this forum post for advice on how you might typeset an integral.

  • edited April 14

    My double integral is:
    %%A = int_0^3 int_0^3 (6x^2+2xy+9y^2) dy dx%%
    Integrating with respect to y first results in the following integral:
    %% int_0^3 (6x^2y+xy^2+3y^3)|_0^3dx = 9 int_0^3(2x^2+x+9)dx%%
    Integrating with respect to x results in:
    %% 9(2/3x^3+1/2x^2+9x)|_0^3 = 9(18+4.5+27)=445.5%%
    So,
    %%A = int_0^3 int_0^3 (6x^2+2xy+9y^2) dy dx = 445.5%%

    mark
  • edited March 30

    My integral is:
    %%A=int_0^3int_0^1(-3x^2+2xy +6y^2) dy dx%%
    First to derive with respect to y:
    %%int_0^3 (-3x^2+xy^2+2y^3)|_0^1%%
    If we plug in our terms for y, we get:
    %%int_0^3(-3x^2+x(1)^2+2(1)^3)-(-3x^2+x(0)^2+2(0)^3)%%
    Which simplifies to:
    %%int_0^3 (x^2+8)dx%%
    Now we integrate with respect to x:
    %%1/3x^3+8x|_0^3%%
    %%=1/3(3)^3+8(3)%%
    %%A=33%%

    mark
  • edited March 31

    My assigned integral was:
    %%A=int_0^2\0^2(9x^2+2xy-3y^2)dydx%%

    I started by integrating with respect to y:
    %%int_0^2(9x^2+xy^2-y^3)\right|_0^2dx%%

    When plugging in the terms for y, I got:
    %%int_0^2(9x^2+x(2)^2-(2)^3)-(9x^2+x(0)^2-(0)^3)dx%%

    Which I simplified to:
    %%int_0^2(4x^2-8)dx%%

    Which I then integrated with respect to x:
    %%((4/3)x^3-8x)\right|_0^2%%

    Then I plugged in my x values:
    %%((4/3)(2)^3-8(2))-((4/3)(0)^3-8(0))%%

    And got a final answer of:
    %%A=(32/3)-16=(-16/3)%%

    mark
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