Integration involving two curves
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From: Mike Terry on 24 Jul 2010 10:53 "Fred Nurk" <albert.xtheunknown0(a)gmail.com> wrote in message news:Dny2o.1535$FH2.583(a)viwinnwfe02.internal.bigpond.com... > Greg Neill wrote: > > <snip> > > Plot the curves for the given range (on the same set of axes). Then ask > > yourself what it means to find the area under the curves. > > The textbook sketches http://sites.google.com/site/xtheunknown0/maths/ > antidifferentiation > > The line is higher than the parabola from 0 to 3 and the parabola is > higher than the line for the rest. ....so you want to add A + B A = area under parabola from 0 to 3 B = area under line for the rest Your original formula seems to be trying to calculate the area *between* the two curves, not the area *under* the curves. (I don't like the wording of the original question though!) Mike. > > Fred
From: Ray Vickson on 24 Jul 2010 13:46 On Jul 24, 2:29 am, Fred Nurk <albert.xtheunkno...(a)gmail.com> wrote: > Greg Neill wrote: > > <snip> > > Plot the curves for the given range (on the same set of axes). Then ask > > yourself what it means to find the area under the curves. > > The textbook sketcheshttp://sites.google.com/site/xtheunknown0/maths/ > antidifferentiation > > The line is higher than the parabola from 0 to 3 and the parabola is > higher than the line for the rest. That means that the wording of the problem is ambiguous. For example, the area "under" the curve y = x from x = -1 to x = +1 is zero (because the negative area for x < 0 cancels the positive one for x > 0). However, the area "between" the curves y = x and y = 0 from x = -1 to x = +1 is 1: it consists of two parts having (absolute) areas of 1/2. You need to decide what, exactly, is wanted here: is it like a computation of a long-run monetary average, where negative returns can reduce your profit, or is it like the computation a painter would perform when deciding how much paint to buy to cover the region? R.G. Vickson > > Fred
From: Noone on 24 Jul 2010 17:05 On Sat, 24 Jul 2010 09:29:07 GMT, Fred Nurk <albert.xtheunknown0(a)gmail.com> wrote: [snip] >The textbook sketches http://sites.google.com/site/xtheunknown0/maths/ >antidifferentiation > >The line is higher than the parabola from 0 to 3 and the parabola is >higher than the line for the rest. > >Fred From your original post: y = x^2 and 2x + y = 15 OK, so I'll presume the problem is asking for the area of the "yellow" part. If you imagine a line perpendicular to the x-axis at x = 3, you can compute that area by adding the part under the parabola from x=0 to x=3, and the part under the straight line from x=3 to x=15/2.
From: Ray Vickson on 24 Jul 2010 19:30 On Jul 24, 2:05 pm, Noone <no...(a)domain.invalid> wrote: > On Sat, 24 Jul 2010 09:29:07 GMT, Fred Nurk > > <albert.xtheunkno...(a)gmail.com> wrote: > > [snip] > > >The textbook sketcheshttp://sites.google.com/site/xtheunknown0/maths/ > >antidifferentiation > > >The line is higher than the parabola from 0 to 3 and the parabola is > >higher than the line for the rest. > > >Fred > > From your original post: y = x^2 and 2x + y = 15 > > OK, so I'll presume the problem is asking for the area of the "yellow" > part. If you imagine a line perpendicular to the x-axis at x = 3, > you can compute that area by adding the part under the parabola from > x=0 to x=3, and the part under the straight line from x=3 to x=15/2. Alternatively, he can compute the area as A = integral{L(y) dy, y=0..y1}, where L(y) = x-distance between the curves y = x^2 and 2x + y = 15 (at any given y-value), and y1 = abscissa at which the two curves cross. Note that L(y) = (15 - y)/2 - sqrt(y) and y1 is obtained from L(y1) = 0. R.G. Vickson
From: Fred Nurk on 25 Jul 2010 07:22
On Sat, 24 Jul 2010 15:53:28 +0100, Mike Terry wrote: > "Fred Nurk" <albert.xtheunknown0(a)gmail.com> wrote in message > news:Dny2o.1535$FH2.583(a)viwinnwfe02.internal.bigpond.com... >> <snip> >> The line is higher than the parabola from 0 to 3 and the parabola is >> higher than the line for the rest. > > ...so you want to add A + B > > A = area under parabola from 0 to 3 > > B = area under line for the rest Ahh!!! I've now got the right answer - and not just to this problem but two others similar to this! Thanks so much! > Your original formula seems to be trying to calculate the area *between* > the two curves, not the area *under* the curves. Yes - because my older self assumed that all questions involving two curves in the chapter titled integration meant 'find the area between two curves'. > (I don't like the wording of the original question though!) > <snip> If I found myself in your shoes I'd say the same - after all, I was avoiding the posting of the entire textbook question (I might do it next time :)) so I only put enough thought into phrasing a question that conveyed my difficulty *moderately* well. |