Question 292749: Hi
How exactly do I do b + c of this question?
a) Using a large scale, sketch the curve y =1/x between the values of x = 1 and x=2.
b) Divide the interval into n equal parts and by comparing the area under the curve, with the areas of suitable rectangles, show that 1/n * (n/(n+1) + n/(n+2) + ... + n/2n)< log2 < 1/n * (1 + n/(n+1) + n/(n+2) + ... + n/(2n-1)
c) lim (n -> infinity) [n/(n+1)+ n/(n+2)+ ... + n/2n] = log2
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! a) Using a large scale, sketch the curve y =1/x between the values of x = 1 and x=2.
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Do as directed. Draw the curve over that interval. Make it large enough
so you can see what you are doing.
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Comment: Note that the width of each base interval is 1/n.
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b) Divide the interval into n equal parts and by comparing the area under the curve, with the areas of suitable rectangles, show that 1/n * [n/(n+1) + n/(n+2) + ... + n/2n])< log2 < 1/n * [1 + n/(n+1) + n/(n+2) + ... + n/(2n-1)]
Note: The 1st base interval goes from 1 to 1+1/n = (n+1)/n
The height of the curve at x = 1 is 1
The height of the curve at x = 1 + 1/n is n/(n+1)
The height of the curve at x = 1 + 2/n is n/(n+2) etc
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Draw a few rectangles UNDER the curve, with 1/n as base and height equal
to the height of the curve at x = 1 then 1+1/n then 1+2/n etc.
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The sum of those areas is less than the area under the curve from [1,2].\
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Now draw a few rectangles that extend ABOVE the curve with 1/n as base
and height equal to 1+1/n then 1+2/n then 1+3/n etc
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The sum of those areas is greater than the area under the curve from [1,2]
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c) lim (n -> infinity) [n/(n+1)+ n/(n+2)+ ... + n/2n] = log2
If you go thru the process with a few values of "n" you will see
that, as n gets larger, the sums get closer to log(2)
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As you increase the number of rectangles by increasing "n" the area
sum of the areas under the curve and sum of the areas over the curve
get closer to the true area under the curve, which is log(2)
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Cheers,
Stan H.
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