[maths] Re: more of calculus
- From: "Nelson Blachman" <blachman@xxxxxxx>
- To: <maths@xxxxxxxxxxxxx>
- Date: Thu, 14 Dec 2006 07:55:11 -0800
Ned,
In regard to integrating 1 dx from x=-1 to 5, notice that this just means
adding up the widths dx=6/n of all the intervals dx from the first to the
nth. So the result is n*6/n=6.
To integrate 3x from x=-1 to 5. you have to muultiply the width dx=6/n by
the value of x associated with each one and add them all up.
The first has x=-1,+ 3/n in its middle,and the
last has 5 - 3/n in its middle.
You can figure out what the mth has in its middle and add up dx times 3x
for each of these, from m=1 to m=m.
But these summands average out to 6 (the sum of the first plus the last
divided by 2) because the first and last have this average and the second
and second-last have this average, etc.
So you get 6*6 for the second term 3x of the integrand, making the total
42.
I know of no such nice way to deal with the limit in the second problem.
--Nelson
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