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Order of integration

  • Thread starter boneill3
  • Start date
  • #1
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Homework Statement



Solve the double integration by changing the order of integration

Homework Equations




\int_{0}^{2} \int_{y/2}^{1} cos(x^2)dxdy


The Attempt at a Solution



I've been trying to sketch the graph with

y/2<= x <= 1

and

0 <= y <= 2

Is the problem trying to get the {y/2} in the limit of \int_{y/2}^{1} cos(x^2)dx to be the equivalent value of x ?

I'm not if my graph that I sketched should be a triangle from the origin out to x= 1 y=2

or should it be a curve from (0,1) to (1,cos(1^2))

regards
Brendan
 

Answers and Replies

  • #2
djeitnstine
Gold Member
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if x= y/2, y = 2x
 
  • #3
127
0
I see that you have made the integral.

\int_{0}^{2} \int_{y/2}^{1} cos(x^2)dxdy

into

\int_{0}^{2} \int_{2x}^{1} cos(x^2)dxdy

by solving y/2 for x = 2x.

Can you use the same rational for

\int_{1}^{3} \int_{0}^{ln(x)} x dydx

solving y = ln(x) for x therefore x = e^(y()

than the integral would be:

int_{1}^{3} \int_{0}^{e^(y)} x dydx


regards
Brendan
 
  • #4
djeitnstine
Gold Member
614
0
I see that you have made the integral.

\int_{0}^{2} \int_{y/2}^{1} cos(x^2)dxdy

into

\int_{0}^{2} \int_{2x}^{1} cos(x^2)dxdy

__________________________________

This is incorrect. When solving for y in y/2 = x (y=2x) It is necessary to sketch the curves so that you can see the boundaries clearly. It seems to me you have a triangle with boundaries y=2x, x=1, and y=0. If you solve for 'y' like that its necessary to change the order of integration and write new limits.

Now think about what you did when you exchanged y for x in your limits...when you try to integrate the function you can't...the whole reason why you change order in this case is so that you may perform the double integrals.

In the second case, it is again similar to the triangle but with the upper boundary ln{x}
 
  • #5
127
0
I've sketched the graph of
\int_{1}^{3} \int_{0}^{ln(x)} x dydx

And the boundaries are the:

x axis <= y <= ln(x)
1 <= x <=3

Is the integral now

\int_{x}^{ln(x)} \int_{1}^{3} x dxdy ?

regards
 
  • #6
djeitnstine
Gold Member
614
0
Is the integral now

\int_{x}^{ln(x)} \int_{1}^{3} x dxdy ?
You should restudy double integration. That double integral there makes no sense. Just a hint that allows you to know whether what you're doing is right or wrong, you should change the function in terms of your second integrating variable.

\int_{1}^{3} \int_{0}^{ln(x)} x dydx


this double integral goes from the lower y limit to the upper, then from the left x limit to the right. In order to change that order you have to change the function y=ln{x} in terms of x....e^y = x

so when doing dxdy we start from dx and are going from the left limit to the right limit, as you see on your graph the limits now become \int_{e^y}^{3} dx and for dy its from lower to upper... \int_{0}^{ln(3)} dy

putting it all together \int_{0}^{ln(3)}\int_{e^y}^{3} x dxdy (note that I said before the function should be in terms of the second integrating variable...)
 
  • #7
127
0
Thanks for your reply,

\int_{0}^{ln(3)}\int_{e^y}^{3} x dxdy


with the X limits \int_{e^y}^{3} when I graphed the original function those limits were 1 <= x <=3

Why do we change 1 to be e^y and not change 3
 

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