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November 01, 2025, 03:21:11 pm

Author Topic: Area under a curve -by symmetry-?  (Read 614 times)  Share 

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Phenotype

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Area under a curve -by symmetry-?
« on: May 25, 2011, 11:20:16 am »
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I'm just getting so bad at spesh...
What does it mean -by symmetry- when finding areas under a curve? In my Essentials specialist book, there are examples which give a graph which positive and negative x values and it asks you to find the area of both of them...

But their examples, when working it out, say By Symmetry.

Here's one example:

Find the area

Answer:
Area

-By Symmetry-

Area <---what did they do there?
0 on bottom, 1 on top (forgot how to do this in LaTeX)

=
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/0

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Re: Area under a curve -by symmetry-?
« Reply #1 on: May 25, 2011, 02:09:46 pm »
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Have you drawn the graph? :P
The function is even (you can tell by the x to the power of 2)!
So if you take the graph for , and flip it around the y-axis, you get the graph for .
In particular, you can pick up the enclosed area from [0,1], flip it, and fit it exactly over the area from [-1,0], so they're the same area, and the integral over [-1,1] is twice the integral over [0,1].
« Last Edit: May 25, 2011, 02:15:22 pm by /0 »

soopertaco

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Re: Area under a curve -by symmetry-?
« Reply #2 on: May 29, 2011, 11:38:34 am »
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Okay so i can't give you a visual since i dont know how to use latex or any of those programs...

BUT when it says by symmetry from what i understand it to be is that you only have to find area and multiply that by two because BY SYMMETRY there are two of the same area.
for example say you have a function sin(x) x element of: [-pi,pi] and the question asks you to find the shaded region under the graph in the 1st quadrant and the area bounded by the graph and the x-axis in the 3rd quadrant. i hope you you're getting a solid enough visual from my explanation so far.

From this you are able to see that the area in the first quadrant is the SAME as the area in the 3rd quadrant and thus all you need to do i realise this and take the integral of sin(x) from [0,pi] and once you've found this area you can simply multiply the area by 2 as it is SYMMETRICAL about the y-axis.
essentially what they've done with the integral in your question is instead of finding the area and then multiplying it by two they take the liberty in placing the two in front of the integral straight away.

These symmetry questions will normally (well at least i've found it to be this way) have symmetry about the y-axis making it easier for you by making one of the limits zero.

:)
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