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Multiple Integral

Is this sketch of the region correct?
(edited 6 years ago)
Nope. The "2nd region" clearly should have y ranging from 1 to e (and x ranging from ln y to 1).
Reply 2
Avatar for Zacken
Zacken
22
Original post by ChrisLorient
I can read. Just not sketch.

No idea what the upper limit being e actually means, other than drawing an e^x curve. Ln(y) I thought cut the axis at 1. But if it also finishes at 1...


ln(x) cuts the x-axis at 1, ln(y) is not ln(x).

y represents the values on the y-axis, so your 2nd region should be contained in the region 1 < y < e.
Original post by Zacken
ln(x) cuts the x-axis at 1, ln(y) is not ln(x).

y represents the values on the y-axis, so your 2nd region should be contained in the region 1 < y < e.


Correct? Or back to the drawing board?
(edited 6 years ago)
IMAG0893[1].jpg175.0KB
I feel it should be greater than 1, as surely the value for the first region alone is 1?
Original post by ChrisLorient
I feel it should be greater than 1, as surely the value for the first region alone is 1?


If you integrate, you can differentiate the result to see if you get back to the original function - as a check. Try it on the function outlined in red.

Also, please don't delete the original question - no one else will be able to benefit from the thread, and it then becomes difficult, to say the least, to assist/help further.

IMAG0893[1].jpg
Original post by ghostwalker
If you integrate, you can differentiate the result to see if you get back to the original function - as a check. Try it on the function outlined in red.

Also, please don't delete the original question - no one else will be able to benefit from the thread, and it then becomes difficult, to say the least, to assist/help further.

IMAG0893[1].jpg


First of all, the sketch is correct in the above post? Just want to make sure.

I used x = ln(y) to get y = e^x
Original post by ChrisLorient
First of all, the sketch is correct in the above post? Just want to make sure.

I used x = ln(y) to get y = e^x


The sketch is consistent with the formula for I you've given in your 2nd post's image.
(edited 6 years ago)
Original post by ghostwalker
The sketch is consistent with the formula for I you've given in your 2nd post's image.


Are the limits correct then? The ones I changed to based on my new integral.

It is that I am getting xe^x, which integrated with respect to x is xe^x - e^x.

This is the same as e^x(x-1), which between 1 and 0 gives me 1 for the final answer.
Original post by ghostwalker
If you integrate, you can differentiate the result to see if you get back to the original function - as a check. Try it on the function outlined in red.

Also, please don't delete the original question - no one else will be able to benefit from the thread, and it then becomes difficult, to say the least, to assist/help further.

IMAG0893[1].jpg


Differentiating xe^x - e^x gives me xe^x, which is consistent.

Does that mean 1 is correct then?
Original post by ChrisLorient
Are the limits correct then? The ones I changed to based on my new integral.

It is that I am getting xe^x, which integrated with respect to x is xe^x - e^x.

This is the same as e^x(x-1), which between 1 and 0 gives me 1 for the final answer.


Limits look fine.

According to your post, you're integrating xe^x-x, not just xe^x

Edit: Actually, looking further back, that -x shouldn't be there, so your integral would have been correct.
(edited 6 years ago)
Original post by ghostwalker
Limits look fine.

According to your post, you're integrating xe^x-x, not just xe^x

Edit: Actually, looking further back, that -x shouldn't be there, so your integral would have been correct.


Yeah, I corrected it. -x was rogue.

So it should be fine, no?

But isn't 1 the value for the first region of the integral alone? Shouldn't I therefore sort of expect a larger value?

I will write it out in neat and post it.
Original post by ghostwalker
Limits look fine.

According to your post, you're integrating xe^x-x, not just xe^x

Edit: Actually, looking further back, that -x shouldn't be there, so your integral would have been correct.


Working.
IMAG0901[1].jpg152.2KB
Original post by ChrisLorient
Working.


OK, working looks fine.

If by the "first region" you mean, the square at the bottom, then no, the integral over that square is actually 1/2. Since the area of the square is 1 unit^2, this is the centre of gravity in the x direction, which you would expect to be 1/2 for the square alone.
Original post by ghostwalker
OK, working looks fine.

If by the "first region" you mean, the square at the bottom, then no, the integral over that square is actually 1/2. Since the area of the square is 1 unit^2, this is the centre of gravity in the x direction, which you would expect to be 1/2 for the square alone.


Ah, OK. So that all seems correct then?
Original post by ChrisLorient
Ah, OK. So that all seems correct then?


If that's all there was to the original question, then yes.
Original post by ghostwalker
If that's all there was to the original question, then yes.


It asked to reverse the order of the integral and hence write it as one integral, and then find the value of I.

Thanks for the help.
Reply 17
Avatar for simon0
simon0
13
As a check, you could evaluate using the first set of definite integrals (the first line on your document) and it seems to agree.
Original post by simon0
As a check, you could evaluate using the first set of definite integrals (the first line on your document) and it seems to agree.


How would I state the region bounded in the new integral?

Let's assume the region of the double integral formed is A.

{A: x = 0, x = 1, y =0 , y = e^x}

What is the proper way of doing that for this double integral?

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