# Solve Repeated Integral: \int^2_1 \int^4_2 \sqrt{x-y} dx dy

• Bucky
In summary, the repeated integral given is evaluated using standard integral rules and properties of integrals. A mistake in algebra is corrected and the integral is solved step by step to reach the final answer.
Bucky

## Homework Statement

Evaluate the following repeated integral
$$\int^2_1 \int^4_2 \sqrt{x-y} dx dy$$

## 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 -.-

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:

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

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

\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?

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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

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

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

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

Looks good.

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...

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

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:
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.

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:
ok got it, thanks a lot for the help!

## What is the repeated integral in this problem?

The repeated integral in this problem is the integration of the function √(x-y) with respect to both x and y. This means that we will first integrate the function with respect to x, and then integrate the resulting expression with respect to y.

## What is the purpose of solving a repeated integral?

The purpose of solving a repeated integral is to find the total area under a given function within a specific region. In this case, we are finding the area under the curve of √(x-y) within the rectangular region bounded by x=1, x=2, y=2, and y=4.

## What are the steps to solve a repeated integral?

1. Begin by integrating the inner function with respect to x, treating y as a constant.
2. Evaluate this expression at the upper and lower bounds of y, resulting in a new function of y.
3. Integrate this new function with respect to y, treating x as a constant.
4. Evaluate this expression at the upper and lower bounds of x, resulting in a final numerical value.

## What is the meaning of the upper and lower bounds in this problem?

The upper and lower bounds in this problem represent the limits of integration for both x and y. These values define the rectangular region in which we are finding the area under the curve of √(x-y).

## What are some real-world applications of solving a repeated integral?

Repeated integrals have various applications in physics, engineering, and economics. For example, they can be used to find the center of mass of an object, calculate the volume of a three-dimensional shape, or determine the average value of a function over a given region.

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