# Repeated Integrals

1. Mar 14, 2007

### Bucky

1. The problem statement, all variables and given/known data
Evaluate the following repeated integral
$$\int^2_1 \int^4_2 \sqrt{x-y} dx dy$$

2. Relevant equations

3. The attempt at a solution

$$\int^2_1 \int^4_2 x^{1/2} - y^{1/2} dx dy$$
$$\int^2_1 [ \frac {2x^{3/2}}{3} - xy^{1/2} ]^4_2 dy$$
$$\int^2_1 [ \frac {2(4)^{3/2}}{3} - (4)y^{1/2} ] - [ \frac {2(2)^{3/2}}{3} - (2)y^{1/2} ] dy$$

$$\int^2_1 [ \frac {2\sqrt{64}}{3} - (4)y^{1/2} ] - [ \frac {2\sqrt{8}}{3} - (2)\sqrt{y} ] dy$$

$$\int^2_1 [ \frac {16}{3} - (4)\sqrt{y}] - [ \frac {2(2)^{3/2}}{3} - (4)\sqrt{2} ] dy$$

$$\int^2_1 [ \frac {16 - 4 \sqrt{2}}{3} - 2 y^{1/2}] dy$$

$$\frac {16 - 4 \sqrt{2}}{3} \int^2_1 [ - 2 y^{1/2}] dy$$
(can i take it out since it's a constant?)

$$\frac {16 - 4 \sqrt{2}}{3} \int^2_1 [ - 2 y^{3/2}]^2_1$$

$$\frac {16 - 4 \sqrt{2}}{3} \int^2_1 [ \frac{- 2 y^{3/2}}{\frac{3}{2}}]^2_1$$

$$\frac {16 - 4 \sqrt{2}}{3} \int^2_1 [ \frac{- 4 y^{3/2}}{3}]^2_1$$

...and so on. basically it doesn't work out. answer i'm meant to get is

$$\frac{4(9\sqrt{3} - 4\sqrt{2} - 1)}{15}$$

can anyone see what i've done wrong? Invairably it's sloppy algebra -.-

2. Mar 14, 2007

### mattmns

edit... Woops, you made a huge mistake on the first step!

It is not always true that $\sqrt{x-y} = \sqrt{x}-\sqrt{y}$

Consider x = 2, y =1. Then $\sqrt{2-1} = \sqrt{1} = 1, \ \text{but} \ \sqrt{2}-\sqrt{1} = \sqrt{2} - 1 \neq 1$

So you cannot just switch things out like that.
-----

Assuming that was OK, you made a mistake when you pulled out the "constant".

It is not really a constant in the way you used it.

Remember the properties of integrals:

$$\int (f-g) = \int f - \int g$$

$$\int cf = c\int f$$

So you should have had:

\begin{align*} \int_1^2 \left(\frac {16 - 4 \sqrt{2}}{3} - 2 y^{1/2}\right)dy & = \int_1^2 \frac{\left(16- 4\sqrt{2}}{3}\right)dy - \int_1^2 2y^{1/2}dy \\ & = \left(\frac{16- 4\sqrt{2}}{3}\right) \int_1^2 dy - 2\int_1^2 y^{1/2}dy \end{align*}

Note that I posted this last part before I noticed the first mistake, so the idea behind my post is correct, but you should not get the same thing (since the first step was wrong).

edit.. Hmm, my code is not being updated, and what is cmyk0000 lol?

Last edited: Mar 14, 2007
3. Mar 14, 2007

### Bucky

ok so i looked at this again and it may in fact be a standard integral?

for the sake of my hands, i won't type the whole thing out in latex...but can someone confirm if i've started this properly?

$$\int^2_1 \int^4_2 \sqrt{x-y} dx dy$$

using standard integral

$$\int (ax+b)^n = \frac {(ax+b)^{n+1}}{a (n+1)}$$

$$\frac {(x-y)^{3/2}}{\frac{3}{2}}$$

$$\frac {2(x-y)^{3/2}}{3}$$

4. Mar 14, 2007

### mattmns

Looks good.

5. Mar 14, 2007

### Bucky

ok so i've mucked it up...

$$\int^2_1 [\frac {2(x-y)^{3/2}}{3} ]^4_2 dy$$

$$[\frac {2((4)-y)^{3/2}}{3} ] - [\frac {2((2)-y)^{3/2}}{3} ]$$

i get to this point. my first thought was "oh it's a letter and a number in a bracket to a power ( (4-y)^3/2 )...can't i just solve it out?", but that seems very akward and messy...

$$2\frac {\sqrt{-2y^3 - 8y^2 - 96y + 128}}{3} - 2\frac {\sqrt{y^3 + 6y^2 + 12y + 8}}{3}$$

6. Mar 14, 2007

### mattmns

What you wrote first is good, but you left out the integral in the second line, so I am not sure if you are just not writing it out, or if you are misunderstanding something.

You should have:

\begin{align*} \int^2_1 \frac {2(x-y)^{3/2}}{3} \Big |_2^4 dy & = \int_2^1 \left(\frac {2((4)-y)^{3/2}}{3} - \frac {2((2)-y)^{3/2}}{3} \right)dy \\ &= \int^2_1 \frac {2((4)-y)^{3/2}}{3} dy - \int_1^2 \frac {2((2)-y)^{3/2}}{3} dy \end{align*}

From here $y$ is now a variable, so you have a new integral to evaluate. Try to evaluate this integral now.

Last edited: Mar 14, 2007
7. Mar 14, 2007

### Bucky

ok this is all mathcrafting, may be lies and methods that'd make no sense...

$$\int^2_1 \frac {2((4)-y)^{3/2}}{3} dy - \int_1^2 \frac {2((2)-y)^{3/2}}{3} dy$$

if i take 2/3 out as a fraction in both equations, i am left with a bracket to integrate.

$$\int^2_1 \frac {2}{3} * ((4)-y)^{3/2} dy - \int_1^2 \frac {2}{3} * ((2)-y)^{3/2} dy$$

from earlier, the rule for integrating the bracket is..

$$\int (ax+b)^n = \frac {(ax+b)^{n+1}}{a (n+1)}$$

giving

$$\frac{2}{3} x \frac{(4-y)^{5/2}}{-1(\frac{5}{2})} - \frac{2}{3} x \frac{(2-y)^{5/2}}{-1(\frac{5}{2})}$$

$$[\frac{4(4-y)^{5/2}}{25}]^2_1 - [\frac{4(2-y)^{5/2}}{25}^2_1]$$

somethings wrong i'm sure, but that may just be complete lack of confidence in my ability to do maths.

8. Mar 14, 2007

### mattmns

Everything looks good up to the very last step, I think you just made a silly mistake with the fractions.

$$\frac{2}{3} \times \frac{(4-y)^{5/2}}{-1(\frac{5}{2})} - \frac{2}{3} \times \frac{(2-y)^{5/2}}{-1(\frac{5}{2})} \Big |_1^2$$

Note: If you want to use x to means times then you should use \times in your LaTeX code.

From there you get (factor out the 2/3 from both parts, and "flip" the fraction in the denominator):

$$\frac{2}{3}\left( -\frac{2}{5}(4-y)^{5/2} - -\frac{2}{5}(2-y)^{5/2} \Big |_1^2 \right)$$

From there we can factor the 2/5 out and we get:

$$\frac{2}{3}\times \frac{2}{5}\left( -(4-y)^{5/2} + (2-y)^{5/2} \Big |_1^2 \right)$$

Now you just "plug and chug," and you should get the correct answer.

Last edited: Mar 14, 2007
9. Mar 14, 2007

### Bucky

ok got it, thanks a lot for the help!

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