# A double integral

in Assignments

(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.

## 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.%%

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:

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

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.

$$= (-3x^3 +\frac{9x}{2}^2 + 81x)\Big|_0^3 = -81 + \frac{729}{4} + 243 = \frac{1377}{4}$$

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!

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

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:

$$\int_0^2 12x^2+12x+24 \,dx$$

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

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%%

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)%%

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%%

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:

$$\int^2_0(-6x^2-x+3)dx=-6/3x^3-x^2/2+3x\Big|^2_0=(-2(2)^3-(2)^2/2+3(2))-(0-0+0)=-16-2+6=-12$$

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:

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

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

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

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

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

$$3-1-2$$

$$0$$

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 $x$:

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

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

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

$$= 15$$

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}%%

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:

$$ = -2x^3 + 2x^2 + 8x |_0^3 $$

$$ = -2(3)^3 + 2(3)^2 + 8(3) $$

$$ = -12 $$

@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

innerintegral (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!

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%%

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%%

@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!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%%

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%%

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.

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 :

$$24$$

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%%

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%%

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

$$\int_{0}^{3}(18x^2-18x+81)dx = (6x^3-9x^2+81x)\Big|^3_0$$

$$=162-81+243$$

$$A=324.$$

@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.

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%%

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%%

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)%%