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July 25th, 2019, 07:02 AM   #1
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double integral

Hello ,anyone got an idea about getting the area of the place surrounded by
y=-x^2+x and y=-x by double integrals .
really thanks by the way
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 July 25th, 2019, 07:29 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,585 Thanks: 1430 $I = \displaystyle \int_0^2 \int_{-x}^{x-x^2}~dy~dx$ Thanks from topsquark and shadow dancer
July 25th, 2019, 10:18 PM   #3
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Quote:
 Originally Posted by romsek $I = \displaystyle \int_0^2 \int_{-x}^{x-x^2}~dy~dx$
You've got any idea about the notation

July 25th, 2019, 11:01 PM   #4
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Quote:
 Originally Posted by shadow dancer You've got any idea about the notation
what do you mean?

July 26th, 2019, 12:14 AM   #5
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Quote:
 Originally Posted by romsek what do you mean?
Notation like f(x,y) kind of thing. Or it's not needed?

Last edited by skipjack; August 1st, 2019 at 09:48 AM.

July 26th, 2019, 06:40 AM   #6
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Quote:
 Originally Posted by shadow dancer Notation like f(x,y) kind of thing. Or it's not needed?
It's a definite integral. It's not a function of $x$ and $y$. Those just serve as dummy variables for the integral.

Last edited by skipjack; August 1st, 2019 at 09:49 AM.

 August 1st, 2019, 08:49 AM #7 Newbie   Joined: Nov 2018 From: Iran Posts: 12 Thanks: 0 Math Focus: calculus So the integral is wrong? I mean, is there no way of calculating that area with double integral? Last edited by skipjack; August 1st, 2019 at 09:47 AM.
August 1st, 2019, 08:54 AM   #8
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Quote:
 Originally Posted by shadow dancer So the integral is wrong? I mean, is there no way of calculating that area with double integral?
Where did you get that idea?

I just said that the integral as given is a definite number.

It's not a function of x or y.

Last edited by skipjack; August 1st, 2019 at 09:47 AM.

August 1st, 2019, 09:51 AM   #9
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Quote:
 Originally Posted by shadow dancer Notation like f(x,y) kind of thing.
The two functions mentioned in the question were used as integration limits.

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