Calculating a Double Integral in the First Quadrant

[Petrus]

Calculate the double integral

,
where D is the set of all points in the first quadrant which satisfies the inequality

.
I am confused how to calculate the a,b $$\int_a^b$$ (I don't know what you call that in english)
Shall I do like this
$$y=4-x$$ and put it in the function so we get $$x^2(4-x)^2$$ and then?

 

[tkhunny]

Re: multiple integrate

You're defining limits, not making a substitution.

\int\limits_{0}^{4}\int\limits_{0}^{4-y}x^{2}y^{2}dxdy

\int\limits_{0}^{4}y^{2}\left(\int\limits_{0}^{4-y}x^{2}dx\right) dy

 

[MarkFL]

...
I am confused how to calculate the a,b $$\int_a^b$$ (I don't know what you call that in english)...

Hello Petrus,

Those values, $a$ and $b$, are referred to as the limits of integration. $a$ is the lower limit of integration, and $b$ is the upper limit of integration.

 

[Petrus]

I am having problem integrate $$\frac{1}{3}\int_0^4(4-y)^2y^2 dy$$
and I did not understand how we got this limits of integration.

 

[MarkFL]

I am having problem integrate $$\frac{1}{3}\int_0^4(4-y)^2y^2 dy$$
and I did not understand how we got this limits of integration.

You have integrated the inner integral incorrectly. What is:

$$\int x^2\,dx$$ ?

As far as the limits go, can you describe the region over which you are integrating?

 

[Petrus]

You have integrated the inner integral incorrectly. What is:

$$\int x^2\,dx$$ ?

As far as the limits go, can you describe the region over which you are integrating?

$$\frac{x^3}{3}$$

- - - Updated - - -

ops I did not notice $$\frac{1}{3}\int_0^4(4-y)^3y^2 dy$$

 

[MarkFL]

Yes, less the constant of integration. Do you see that you need $$(4-y)^3$$ in your remaining integrand?

edit: Yes, you caught the minor error! (Yes) Now, do you know how to proceed?

 

[Petrus]

Yes, less the constant of integration. Do you see that you need $$(4-y)^3$$ in your remaining integrand?

edit: Yes, you caught the minor error! (Yes) Now, do you know how to proceed?

Actually not. If I am honest I got so much problem with integration, I was trying integration by part but did not work well

 

[MarkFL]

I think the most straightforward method here would be to expand the cubed binomial, then distribute the factor $y^2$. What is your integrand now?

 

[Ackbach]

I think the most straightforward method here would be to expand the cubed binomial, then distribute the factor $y^2$. What is your integrand now?

Or $u=4-y$ would make the algebra a little easier, as you'd only have to expand a quadratic instead of a cubic.

 

[Petrus]

I think the most straightforward method here would be to expand the cubed binomial, then distribute the factor $y^2$. What is your integrand now?

Do you mean like this?
$$\frac{1}{3}\int_0^4y^2(-y^3+12y^2-48y+64)$$

 

[MarkFL]

Or $u=4-y$ would make the algebra a little easier, as you'd only have to expand a quadratic instead of a cubic.

Yes, great suggestion!

If you use this substitution Petrus, be mindful to rewrite the limits and differential in terms of the new variable, and use the rule:

$$\int_a^b f(x)\,dx=-\int_b^a f(x)\,dx$$

 

[MarkFL]

Do you mean like this?
$$\frac{1}{3}\int_0^4y^2(-y^3+12y^2-48y+64)\,dy$$

Yes...I have added the missing differential. (Wink)

Now distribute the $y^2$, and integrate term by term to get the anti-derivative for use in the FTOC.

 

[Petrus]

Yes...I have added the missing differential. (Wink)

Now distribute the $y^2$, and integrate term by term to get the anti-derivative for use in the FTOC.

Is this what you mean $$\frac{1}{3}\int_0^4y^2 dy+\frac{1}{3}\int_0^4-y^3+12y^2-48y+64 dy$$

(Sorry for forgeting dy, this is something I always forget cause I think in my brain that I derivate dy

Edit: after I posted I start to notice that Is wrong.. you mean $$\frac{1}{3}\int_0^4y^2(-y^3)+\frac{1}{3}\int_0^4y^212y^2+\frac{1}{3}\int_0^4y^2(-48y)+\frac{1}{3}\int_0^4y^2*64$$

 

[MarkFL]

No, that is an illegal move. You want to distribute the $y^2$ factor:

$$\frac{1}{3}\int_0^4(-y^5+12y^4-48y^3+64y^2)\,dy$$

 

[Petrus]

No, that is an illegal move. You want to distribute the $y^2$ factor:

$$\frac{1}{3}\int_0^4(-y^5+12y^4-48y^3+64y^2)\,dy$$

Yeah I notice when I submited, I just felt stupid that I posted it after I did submit because I obvious did not think correct.

 

[MarkFL]

Yeah I notice when I submited, I just felt stupid that I posted it after I did submit because I obvious did not think correct.

Hey, we all make mistakes...some in haste. I've made my share for sure! (Happy)

Now, what is the anti-derivative you want to use in your application of the FTOC?

 

[Petrus]

Hey, we all make mistakes...some in haste. I've made my share for sure! (Happy)

Now, what is the anti-derivative you want to use in your application of the FTOC?

$$\frac{1}{3}\int_0^4(-y^5+12y^4-48y^3+64y^2)\,dy$$
I think you call it 'Second part' the one $$\int_a^bf(x)=F(b)-F(a)$$
our anti derivate is $$\frac{-y^6}{6}+\frac{12y^5}{5}-\frac{48y^4}{4}+\frac{64y^3}{3}$$
So now we got $$\frac{1}{3}[\frac{-y^6}{6}+\frac{12y^5}{5}-\frac{48y^4}{4}+\frac{64y^3}{3}]_0^4 =\frac{1}{3}(\frac{1024}{15}-0)=\frac{1024}{45}$$

 

[MarkFL]

Yes, good work! (Cool)

I was taught to call it the anti-derivative form of the fundamental theorem of calculus. I notice many now call it the second theorem. Whatever you call it though, it gives us a powerful method for evaluating definite integrals. (Smile)

 

[Petrus]

Yes, great suggestion!

If you use this substitution Petrus, be mindful to rewrite the limits and differential in terms of the new variable, and use the rule:

$$\int_a^b f(x)\,dx=-\int_b^a f(x)\,dx$$

if $$u=4-y$$ then is $$y=4-u$$
so we got $$\frac{1}{3}\int_4^0u^3(4-u)^2 du$$ and with that rule we got $$-\frac{1}{3}\int_0^4u^3(4-u)^2 du$$ Is this correct?Edit: forgot to add 'du'

 

[MarkFL]

if $$u=4-y$$ then is $$y=4-u$$
so we got $$\frac{1}{3}\int_4^0u^3(4-u)^2$$ and with that rule we got $$-\frac{1}{3}\int_0^4u^3(4-u)^2$$ Is this correct?

Not quite...with:

$$u=4-y\,\therefore\,du=-dy$$

Notice this negative sign on the differential gets brought out front (this is one reason it is important to always include the differential in your integrations :D), and so rewriting everything in terms of the new variable $u$, we get:

$$-\frac{1}{3}\int_4^0u^3(4-u)^2\,du=\frac{1}{3}\int_0^4u^3(4-u)^2\,du$$

 

[Petrus]

Not quite...with:

$$u=4-y\,\therefore\,du=-dy$$

Notice this negative sign on the differential gets brought out front (this is one reason it is important to always include the differential in your integrations :D), and so rewriting everything in terms of the new variable $u$, we get:

$$-\frac{1}{3}\int_4^0u^3(4-u)^2\,du=\frac{1}{3}\int_0^4u^3(4-u)^2\,du$$

I still don't understand why 1/3 got negative.

 

[MarkFL]

I still don't understand why 1/3 got negative.

Because $dy=-du$ and the negative sign represents the constant $-1$ and can be brought out in front of the integral as a factor.

 

[Petrus]

Because $dy=-du$ and the negative sign represents the constant $-1$ and can be brought out in front of the integral as a factor.

Thanks Mark and Ackback!:) I now got same answer answer and integrade in two way:)
The only thing I did not understand is how we got this limits of integration. for x we got from 0 to 4-y and for y we got from 0 to 4 I did not understand how you get that.

 

[Ackbach]

Thanks Mark and Ackback!:) I now got same answer answer and integrade in two way:)
The only thing I did not understand is how we got this limits of integration. for x we got from 0 to 4-y and for y we got from 0 to 4 I did not understand how you get that.

Because the original limits are $x$-values. When you do a $u$-substitution, you can either write the limits as $u$-values (by substituting into the $u=4-x$, in your case), or you can write the new limits as $u(0)$ to $u(4)$. Either way, the rule is as follows for $u$-substitutions:

1. For definite integrals, you must translate the integrand, the differential, and the limits. You can perform the entire integral in the $u$-domain, if you like.
2. For indefinite integrals, you must translate the integrand and the differential into the $u$-domain. However, once you have an antiderivative computed, you must go back to $x$'s, since that is what you were given.

 

[MarkFL]

Thanks Mark and Ackbach!:) I now got same answer answer and integrade in two way:)
The only thing I did not understand is how we got this limits of integration. for x we got from 0 to 4-y and for y we got from 0 to 4 I did not understand how you get that.

We were told to consider the first quadrant points satisfying the inequality:

$$x+y\le4$$

So for $x$, this implies:

$$0\le x\le 4-y$$

And for $y$, this implies:

$$0\le y\le 4$$

 

[Petrus]

We were told to consider the first quadrant points satisfying the inequality:

$$x+y\le4$$

So for $x$, this implies:

$$0\le x\le 4-y$$

And for $y$, this implies:

$$0\le y\le 4$$

this is the part i strugle:
$$0\le y\le 4$$
why don't it become
$$0\le y\le 4-x$$

 

[MarkFL]

We only want to use the hypotenuse boundary in the inner integral, which we chose to be in terms of $x$. We could have integrated in the other order, and written:

$$I=\int_0^4\int_0^{4-x}x^2y^2\,dy\,dx$$

I suggest reading about Fubini's theorem, and the distinction between Type I and Type II regions.