document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=292749>Question 292749</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Hi\r<BR>\n" );
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document.write( "How exactly do I do b + c of this question?\r<BR>\n" );
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document.write( "a) Using a large scale, sketch the curve y =1/x between the values of x = 1 and x=2. \r<BR>\n" );
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document.write( "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)\r<BR>\n" );
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document.write( "c) lim (n -> infinity) [n/(n+1)+ n/(n+2)+ ... + n/2n] = log2 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #211403 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=211403\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> a) Using a large scale, sketch the curve y =1/x between the values of x = 1 and x=2.<BR>\n" );
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document.write( "Do as directed.  Draw the curve over that interval.  Make it large enough<BR>\n" );
document.write( "so you can see what you are doing.<BR>\n" );
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document.write( "Comment: Note that the width of each base interval is 1/n.<BR>\n" );
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document.write( "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)]<BR>\n" );
document.write( "Note: The 1st base interval goes from 1 to 1+1/n = (n+1)/n<BR>\n" );
document.write( "The height of the curve at x = 1 is 1<BR>\n" );
document.write( "The height of the curve at x = 1 + 1/n is n/(n+1)<BR>\n" );
document.write( "The height of the curve at x = 1 + 2/n is n/(n+2) etc<BR>\n" );
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document.write( "Draw a few rectangles UNDER the curve, with 1/n as base and height equal<BR>\n" );
document.write( "to the height of the curve at x = 1 then 1+1/n then 1+2/n etc.<BR>\n" );
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document.write( "The sum of those areas is less than the area under the curve from [1,2].\<BR>\n" );
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document.write( "Now draw a few rectangles that extend ABOVE the curve with 1/n as base <BR>\n" );
document.write( "and height equal to 1+1/n then 1+2/n then 1+3/n etc <BR>\n" );
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document.write( "The sum of those areas is greater than the area under the curve from [1,2]<BR>\n" );
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document.write( "c) lim (n -> infinity) [n/(n+1)+ n/(n+2)+ ... + n/2n] = log2<BR>\n" );
document.write( "If you go thru the process with a few values of \"n\" you will see <BR>\n" );
document.write( "that, as n gets larger, the sums get closer to log(2)<BR>\n" );
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document.write( "As you increase the number of rectangles by increasing \"n\" the area<BR>\n" );
document.write( "sum of the areas under the curve and sum of the areas over the curve<BR>\n" );
document.write( "get closer to the true area under the curve, which is log(2)<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H.<BR>\n" );
document.write( "==================== </SPAN>\n" );
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