{"id":1152,"date":"2021-06-30T17:02:00","date_gmt":"2021-06-30T17:02:00","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/problem-set-approximating-areas\/"},"modified":"2021-11-17T01:50:07","modified_gmt":"2021-11-17T01:50:07","slug":"problem-set-approximating-areas","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/problem-set-approximating-areas\/","title":{"raw":"Problem Set: Approximating Areas","rendered":"Problem Set: Approximating Areas"},"content":{"raw":"
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1.\u00a0<\/strong>State whether the given sums are equal or unequal.<\/p>\r\n\r\n

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  1. [latex]\\underset{i=1}{\\overset{10}{\\Sigma}} i[\/latex] and [latex]\\underset{k=1}{\\overset{10}{\\Sigma}} k[\/latex]<\/li>\r\n \t
  2. [latex]\\underset{i=1}{\\overset{10}{\\Sigma}} i[\/latex] and [latex]\\underset{i=6}{\\overset{15}{\\Sigma}} (i-5)[\/latex]<\/li>\r\n \t
  3. [latex]\\underset{i=1}{\\overset{10}{\\Sigma}} i(i-1)[\/latex] and [latex]\\underset{j=0}{\\overset{9}{\\Sigma}} (j+1)j[\/latex]<\/li>\r\n \t
  4. [latex]\\underset{i=1}{\\overset{10}{\\Sigma}} i(i-1)[\/latex] and [latex]\\underset{k=1}{\\overset{10}{\\Sigma}}(k^2-k)[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1170572274949\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572274949\"]\r\n

    a. They are equal; both represent the sum of the first 10 whole numbers. b. They are equal; both represent the sum of the first 10 whole numbers. c. They are equal by substituting [latex]j=i-1[\/latex]. d. They are equal; the first sum factors the terms of the second.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

    In the following exercises (2-3), use the rules for sums of powers of integers to compute the sums.<\/p>\r\n\r\n

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    2.\u00a0<\/strong>[latex]\\displaystyle\\sum_{i=5}^{10} i[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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    3.\u00a0<\/strong>[latex]\\displaystyle\\sum_{i=5}^{10} i^2[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572510074\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572510074\"]\r\n

    [latex]385-30=355[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

    Suppose that [latex]\\underset{i=1}{\\overset{100}{\\Sigma}} a_i=15[\/latex] and [latex]\\underset{i=1}{\\overset{100}{\\Sigma}} b_i=-12[\/latex]. In the following exercises (4-7), compute the sums.<\/p>\r\n\r\n

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    4.\u00a0<\/strong>[latex]\\displaystyle\\sum_{i=1}^{100} (a_i+b_i)[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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    5.\u00a0<\/strong>[latex]\\displaystyle\\sum_{i=1}^{100} (a_i-b_i)[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572419248\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572419248\"]\r\n

    [latex]15-(-12)=27[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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    6.\u00a0<\/strong>[latex]\\displaystyle\\sum_{i=1}^{100} (3a_i-4b_i)[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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    7.\u00a0<\/strong>[latex]\\displaystyle\\sum_{i=1}^{100} (5a_i+4b_i)[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572601242\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572601242\"]\r\n

    [latex]5(15)+4(-12)=27[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

    In the following exercises (8-11), use summation properties and formulas to rewrite and evaluate the sums.<\/p>\r\n\r\n

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    \r\n\r\n8.\u00a0<\/strong>[latex]\\displaystyle\\sum_{k=1}^{20} 100(k^2-5k+1)[\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\n
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    9.\u00a0<\/strong>[latex]\\displaystyle\\sum_{j=1}^{50} (j^2-2j)[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170571638237\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571638237\"]\r\n

    [latex]\\underset{j=1}{\\overset{50}{\\Sigma}} j^2-2\\underset{j=1}{\\overset{50}{\\Sigma}} j=\\frac{(50)(51)(101)}{6}-\\frac{2(50)(51)}{2}=40,375[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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    10.\u00a0<\/strong>[latex]\\displaystyle\\sum_{j=11}^{20} (j^2-10j)[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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    11.\u00a0<\/strong>[latex]\\displaystyle\\sum_{k=1}^{25} [(2k)^2-100k][\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170571539210\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571539210\"]\r\n

    [latex]4\\underset{k=1}{\\overset{25}{\\Sigma}} k^2-100\\underset{k=1}{\\overset{25}{\\Sigma}} k=\\frac{4(25)(26)(51)}{9}-50(25)(26)=-10,400[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

    Let [latex]L_n[\/latex] denote the left-endpoint sum using [latex]n[\/latex] subintervals and let [latex]R_n[\/latex] denote the corresponding right-endpoint sum. In the following exercises (12-19), compute the indicated left and right sums for the given functions on the indicated interval.<\/p>\r\n\r\n

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    12<\/strong>. <\/strong>[latex]L_4[\/latex]\u00a0for [latex]f(x)=\\dfrac{1}{x-1}[\/latex] on [latex][2,3][\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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    13. <\/strong>[latex]R_4[\/latex]\u00a0for [latex]g(x)= \\cos (\\pi x)[\/latex] on [latex][0,1][\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170571562422\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571562422\"]\r\n

    [latex]R_4=0.25[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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    14.\u00a0<\/strong>[latex]L_6[\/latex]\u00a0for [latex]f(x)=\\dfrac{1}{x(x-1)}[\/latex] on [latex][2,5][\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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    \r\n\r\n15.\u00a0<\/strong>[latex]R_6[\/latex]\u00a0for [latex]f(x)=\\dfrac{1}{x(x-1)}[\/latex] on [latex][2,5][\/latex]\r\n\r\n[reveal-answer q=\"fs-id1170572218645\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572218645\"]\r\n

    [latex]R_6=0.372[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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    16.\u00a0\u00a0<\/strong>[latex]R_4[\/latex]\u00a0for [latex]\\dfrac{1}{x^2+1}[\/latex] on [latex][-2,2][\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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    17.<\/strong>\u00a0<\/em>[latex]L_4[\/latex]\u00a0for [latex]\\dfrac{1}{x^2+1}[\/latex] on [latex][-2,2][\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572643187\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572643187\"]\r\n

    [latex]L_4=2.20[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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    18. <\/strong>[latex]R_4[\/latex]\u00a0for [latex]x^2-2x+1[\/latex] on [latex][0,2][\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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    19.\u00a0<\/strong>[latex]L_8[\/latex]\u00a0for [latex]x^2-2x+1[\/latex] on [latex][0,2][\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572373696\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572373696\"]\r\n

    [latex]L_8=0.6875[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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    20.\u00a0<\/strong>Compute the left and right Riemann sums\u2014[latex]L_4[\/latex]\u00a0and [latex]R_4[\/latex], respectively\u2014for [latex]f(x)=(2-|x|)[\/latex] on [latex][-2,2][\/latex]. Compute their average value and compare it with the area under the graph of [latex]f[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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    21.\u00a0<\/strong>Compute the left and right Riemann sums\u2014[latex]L_6[\/latex]\u00a0and [latex]R_6[\/latex], respectively\u2014for [latex]f(x)=(3-|3-x|)[\/latex] on [latex][0,6][\/latex]. Compute their average value and compare it with the area under the graph of [latex]f[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1170572399037\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572399037\"]\r\n

    [latex]L_6=9.000=R_6[\/latex]. The graph of [latex]f[\/latex] is a triangle with area 9.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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    22.\u00a0<\/strong>Compute the left and right Riemann sums\u2014[latex]L_4[\/latex]\u00a0and [latex]R_4[\/latex], respectively\u2014for [latex]f(x)=\\sqrt{4-x^2}[\/latex] on [latex][-2,2][\/latex] and compare their values.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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    23.\u00a0<\/strong>Compute the left and right Riemann sums\u2014[latex]L_6[\/latex]\u00a0and [latex]R_6[\/latex], respectively\u2014for [latex]f(x)=\\sqrt{9-(x-3)^2}[\/latex] on [latex][0,6][\/latex] and compare their values.<\/p>\r\n[reveal-answer q=\"fs-id1170572629236\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572629236\"]\r\n

    [latex]L_6=13.12899=R_6[\/latex]. They are equal.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

    Express the following endpoint sums in sigma notation but do not evaluate them (24-27).<\/p>\r\n\r\n

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    24. <\/strong>[latex]L_{30}[\/latex]\u00a0for [latex]f(x)=x^2[\/latex] on [latex][1,2][\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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    25. <\/strong>[latex]L_{10}[\/latex]\u00a0for [latex]f(x)=\\sqrt{4-x^2}[\/latex] on [latex][-2,2][\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170571614670\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571614670\"]\r\n

    [latex]L_{10}=\\frac{4}{10}\\underset{i=1}{\\overset{10}{\\Sigma}} \\sqrt{4-(-2+4\\frac{(i-1)}{10})}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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    26.\u00a0<\/strong>[latex]R_{20}[\/latex]\u00a0for [latex]f(x)= \\sin x[\/latex] on [latex][0,\\pi][\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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    27.\u00a0<\/strong>[latex]R_{100}[\/latex]\u00a0for [latex]\\ln x[\/latex] on [latex][1,e][\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170571777861\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571777861\"]\r\n

    [latex]R_{100}=\\frac{e-1}{100}\\underset{i=1}{\\overset{100}{\\Sigma}} \\ln (1+(e-1)\\frac{i}{100})[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

    In the following exercises (28-33), graph the function then use a calculator or a computer program to evaluate the following left and right endpoint sums. Is the area under the curve between the left and right endpoint sums?<\/p>\r\n\r\n

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    28. [T] <\/strong>[latex]L_{100}[\/latex]\u00a0and [latex]R_{100}[\/latex]\u00a0for [latex]y=x^2-3x+1[\/latex] on the interval [latex][-1,1][\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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    29. [T]\u00a0<\/strong>[latex]L_{100}[\/latex]\u00a0and [latex]R_{100}[\/latex]\u00a0for [latex]y=x^2[\/latex] on the interval [latex][0,1][\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1167794173599\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794173599\"]\r\n

    \"A<\/span><\/p>\r\n[latex]R_{100}=0.33835, \\, L_{100}=0.32835[\/latex]. The plot shows that the left Riemann sum is an underestimate because the function is increasing. Similarly, the right Riemann sum is an overestimate. The area lies between the left and right Riemann sums. Ten rectangles are shown for visual clarity. This behavior persists for more rectangles.[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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    30. [T]\u00a0<\/strong>[latex]L_{50}[\/latex]\u00a0and [latex]R_{50}[\/latex]\u00a0for [latex]y=\\dfrac{x+1}{x^2-1}[\/latex] on the interval [latex][2,4][\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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    31. [T]\u00a0<\/strong>[latex]L_{100}[\/latex]\u00a0and [latex]R_{100}[\/latex]\u00a0for [latex]y=x^3[\/latex] on the interval [latex][-1,1][\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1167794094128\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794094128\"]\r\n

    \"A<\/span><\/p>\r\n[latex]L_{100}=-0.02, \\, R_{100}=0.02[\/latex]. The left endpoint sum is an underestimate because the function is increasing. Similarly, a right endpoint approximation is an overestimate. The area lies between the left and right endpoint estimates.[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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    32. [T]\u00a0<\/strong>[latex]L_{50}[\/latex]\u00a0and [latex]R_{50}[\/latex]\u00a0for [latex]y= \\tan x[\/latex] on the interval [latex]\\left[0,\\frac{\\pi}{4}\\right][\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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    33. [T]\u00a0<\/strong>[latex]L_{100}[\/latex]\u00a0and [latex]R_{100}[\/latex]\u00a0for [latex]y=e^{2x}[\/latex] on the interval [latex][-1,1][\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1167794047771\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794047771\"]\r\n

    \"A<\/span><\/p>\r\n[latex]L_{100}=3.555, \\, R_{100}=3.670[\/latex]. The plot shows that the left Riemann sum is an underestimate because the function is increasing. Ten rectangles are shown for visual clarity. This behavior persists for more rectangles.[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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    34.\u00a0<\/strong>Let [latex]t_j[\/latex]\u00a0denote the time that it took Tejay van Garteren to ride the [latex]j[\/latex]th stage of the Tour de France in 2014. If there were a total of 21 stages, interpret [latex]\\displaystyle\\sum_{j=1}^{21} t_j[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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    35.\u00a0<\/strong>Let [latex]r_j[\/latex] denote the total rainfall in Portland on the [latex]j[\/latex]th day of the year in 2009. Interpret [latex]\\displaystyle\\sum_{j=1}^{31} r_j[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1170571613663\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571613663\"]\r\n

    The sum represents the cumulative rainfall in January 2009.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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    36.\u00a0<\/strong>Let [latex]d_j[\/latex] denote the hours of daylight and [latex]\\delta_j[\/latex] denote the increase in the hours of daylight from day [latex]j-1[\/latex] to day [latex]j[\/latex] in Fargo, North Dakota, on the [latex]j[\/latex]th day of the year. Interpret [latex]d_1+\\underset{j=2}{\\overset{365}{\\Sigma}} \\delta_j[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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    37.\u00a0<\/strong>To help get in shape, Joe gets a new pair of running shoes. If Joe runs 1 mi each day in week 1 and adds [latex]\\frac{1}{10}[\/latex] mi to his daily routine each week, what is the total mileage on Joe\u2019s shoes after 25 weeks?<\/p>\r\n[reveal-answer q=\"fs-id1170572434973\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572434973\"]\r\n

    The total mileage is [latex]7 \\times \\underset{i=1}{\\overset{25}{\\Sigma}} (1+\\frac{(i-1)}{10})=7 \\times 25+\\frac{7}{10} \\times 12 \\times 25=385[\/latex] mi.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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    38.\u00a0<\/strong>The following table gives approximate values of the average annual atmospheric rate of increase in carbon dioxide (CO2<\/sub>) each decade since 1960, in parts per million (ppm). Estimate the total increase in atmospheric CO2<\/sub> between 1964 and 2013.<\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    Average Annual Atmospheric CO2<\/sub> Increase, 1964\u20132013Source<\/em>: http:\/\/www.esrl.noaa.gov\/gmd\/ccgg\/trends\/.<\/caption>\r\n
    Decade<\/th>\r\nPpm\/y<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    1964\u20131973<\/td>\r\n1.07<\/td>\r\n<\/tr>\r\n
    1974\u20131983<\/td>\r\n1.34<\/td>\r\n<\/tr>\r\n
    1984\u20131993<\/td>\r\n1.40<\/td>\r\n<\/tr>\r\n
    1994\u20132003<\/td>\r\n1.87<\/td>\r\n<\/tr>\r\n
    2004\u20132013<\/td>\r\n2.07<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<\/div>\r\n
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    39.\u00a0<\/strong>The following table gives the approximate increase in sea level in inches over 20 years starting in the given year. Estimate the net change in mean sea level from 1870 to 2010.<\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    Approximate 20-Year Sea Level Increases, 1870\u20131990Source<\/em>: http:\/\/link.springer.com\/article\/10.1007%2Fs10712-011-9119-1<\/caption>\r\n
    Starting Year<\/th>\r\n20-Year Change<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    1870<\/td>\r\n0.3<\/td>\r\n<\/tr>\r\n
    1890<\/td>\r\n1.5<\/td>\r\n<\/tr>\r\n
    1910<\/td>\r\n0.2<\/td>\r\n<\/tr>\r\n
    1930<\/td>\r\n2.8<\/td>\r\n<\/tr>\r\n
    1950<\/td>\r\n0.7<\/td>\r\n<\/tr>\r\n
    1970<\/td>\r\n1.1<\/td>\r\n<\/tr>\r\n
    1990<\/td>\r\n1.5<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[reveal-answer q=\"fs-id1170572572337\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572572337\"]\r\n

    Add the numbers to get 8.1-in. net increase.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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    40.\u00a0<\/strong>The following table gives the approximate increase in dollars in the average price of a gallon of gas per decade since 1950. If the average price of a gallon of gas in 2010 was $2.60, what was the average price of a gallon of gas in 1950?<\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    Approximate 10-Year Gas Price Increases, 1950\u20132000Source<\/em>: http:\/\/epb.lbl.gov\/homepages\/Rick_Diamond\/docs\/lbnl55011-trends.pdf.<\/caption>\r\n
    Starting Year<\/th>\r\n10-Year Change<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    1950<\/td>\r\n0.03<\/td>\r\n<\/tr>\r\n
    1960<\/td>\r\n0.05<\/td>\r\n<\/tr>\r\n
    1970<\/td>\r\n0.86<\/td>\r\n<\/tr>\r\n
    1980<\/td>\r\n\u22120.03<\/td>\r\n<\/tr>\r\n
    1990<\/td>\r\n0.29<\/td>\r\n<\/tr>\r\n
    2000<\/td>\r\n1.12<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<\/div>\r\n
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    41.\u00a0<\/strong>The following table gives the percent growth of the U.S. population beginning in July of the year indicated. If the U.S. population was 281,421,906 in July 2000, estimate the U.S. population in July 2010.<\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    Annual Percentage Growth of U.S. Population, 2000\u20132009Source<\/em>: http:\/\/www.census.gov\/popest\/data.<\/caption>\r\n
    Year<\/th>\r\n% Change\/Year<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    2000<\/td>\r\n1.12<\/td>\r\n<\/tr>\r\n
    2001<\/td>\r\n0.99<\/td>\r\n<\/tr>\r\n
    2002<\/td>\r\n0.93<\/td>\r\n<\/tr>\r\n
    2003<\/td>\r\n0.86<\/td>\r\n<\/tr>\r\n
    2004<\/td>\r\n0.93<\/td>\r\n<\/tr>\r\n
    2005<\/td>\r\n0.93<\/td>\r\n<\/tr>\r\n
    2006<\/td>\r\n0.97<\/td>\r\n<\/tr>\r\n
    2007<\/td>\r\n0.96<\/td>\r\n<\/tr>\r\n
    2008<\/td>\r\n0.95<\/td>\r\n<\/tr>\r\n
    2009<\/td>\r\n0.88<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[reveal-answer q=\"4049667\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"4049667\"]\r\n\r\nTo obtain the population in July 2001, multiply the population in July 2000 by 1.0112 to get 284,573,831\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170572330275\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572330275\"]\r\n

    309,389,957<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

    In the following exercises (42-45), estimate the areas under the curves by computing the left Riemann sums, [latex]L_8[\/latex].<\/p>\r\n\r\n

    \r\n
    42.\u00a0<\/strong>\"A<\/span><\/div>\r\n<\/div>\r\n
    \r\n
    43.\u00a0<\/strong>\"The\r\n[reveal-answer q=\"fs-id1170572129132\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572129132\"][latex]L_8=3+2+1+2+3+4+5+4=24[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n
    \r\n
    44.\u00a0<\/strong>\"The<\/span><\/div>\r\n<\/div>\r\n
    \r\n
    45.\u00a0<\/strong>\"The\r\n[reveal-answer q=\"fs-id1170572309714\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572309714\"][latex]L_8=3+5+7+6+8+6+5+4=44[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n
    \r\n
    \r\n

    46. [T]<\/strong> Use a computer algebra system to compute the Riemann sum, [latex]L_N[\/latex], for [latex]N=10,30,50[\/latex] for [latex]f(x)=\\sqrt{1-x^2}[\/latex] on [latex][-1,1][\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    \r\n

    47. [T]<\/strong> Use a computer algebra system to compute the Riemann sum, [latex]L_N[\/latex], for [latex]N=10,30,50[\/latex] for [latex]f(x)=\\dfrac{1}{\\sqrt{1+x^2}}[\/latex] on [latex][-1,1][\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1170571628959\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571628959\"]\r\n

    [latex]L_{10} \\approx 1.7604, \\, L_{30} \\approx 1.7625, \\, L_{50} \\approx 1.76265[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    \r\n

    48. [T]<\/strong> Use a computer algebra system to compute the Riemann sum, [latex]L_N[\/latex], for [latex]N=10,30,50[\/latex] for [latex]f(x)= \\sin^2 x[\/latex] on [latex][0,2\\pi][\/latex]. Compare these estimates with [latex]\\pi[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

    In the following exercises (49-50), use a calculator or a computer program to evaluate the endpoint sums [latex]R_N[\/latex]\u00a0and [latex]L_N[\/latex]\u00a0for [latex]N=1,10,100[\/latex]. How do these estimates compare with the exact answers, which you can find via geometry?<\/p>\r\n\r\n

    \r\n
    \r\n

    49. [T]\u00a0<\/strong>[latex]y= \\cos (\\pi x)[\/latex] on the interval [latex][0,1][\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572351589\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572351589\"]\r\n

    [latex]R_1=-1, \\, L_1=1, \\, R_{10}=-0.1, \\, L_{10}=0.1, \\, L_{100}=0.01[\/latex], and [latex]R_{100}=-0.1[\/latex]. By symmetry of the graph, the exact area is zero.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    \r\n

    50. [T]\u00a0<\/strong>[latex]y=3x+2[\/latex] on the interval [latex][3,5][\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

    In the following exercises (51-52), use a calculator or a computer program to evaluate the endpoint sums [latex]R_N[\/latex]\u00a0and [latex]L_N[\/latex]\u00a0for [latex]N=1,10,100[\/latex].<\/p>\r\n\r\n

    \r\n
    \r\n

    51. [T]\u00a0<\/strong>[latex]y=x^4-5x^2+4[\/latex] on the interval [latex][-2,2][\/latex], which has an exact area of [latex]\\frac{32}{15}[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572448495\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572448495\"]\r\n

    [latex]R_1=0, \\, L_1=0, \\, R_{10}=2.4499, \\, L_{10}=2.4499, \\, R_{100}=2.1365, \\, L_{100}=2.1365[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    \r\n

    52. [T]\u00a0<\/strong>[latex]y=\\ln x[\/latex] on the interval [latex][1,2][\/latex], which has an exact area of [latex]2\\ln (2)-1[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    \r\n

    53.\u00a0<\/strong>Explain why, if [latex]f(a)\\ge 0[\/latex] and [latex]f[\/latex] is increasing on [latex][a,b][\/latex], that the left endpoint estimate is a lower bound for the area below the graph of [latex]f[\/latex] on [latex][a,b][\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1170572503294\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572503294\"]\r\n

    If [latex][c,d][\/latex] is a subinterval of [latex][a,b][\/latex] under one of the left-endpoint sum rectangles, then the area of the rectangle contributing to the left-endpoint estimate is [latex]f(c)(d-c)[\/latex]. But, [latex]f(c)\\le f(x)[\/latex] for [latex]c\\le x\\le d[\/latex], so the area under the graph of [latex]f[\/latex] between [latex]c[\/latex] and [latex]d[\/latex] is [latex]f(c)(d-c)[\/latex] plus the area below the graph of [latex]f[\/latex] but above the horizontal line segment at height [latex]f(c)[\/latex], which is positive. As this is true for each left-endpoint sum interval, it follows that the left Riemann sum is less than or equal to the area below the graph of [latex]f[\/latex] on [latex][a,b][\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    \r\n

    54.\u00a0<\/strong>Explain why, if [latex]f(b)\\ge 0[\/latex] and [latex]f[\/latex] is decreasing on [latex][a,b][\/latex], that the left endpoint estimate is an upper bound for the area below the graph of [latex]f[\/latex] on [latex][a,b][\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    \r\n

    55.\u00a0<\/strong>Show that, in general, [latex]R_N-L_N=(b-a) \\times \\frac{f(b)-f(a)}{N}[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1170572624706\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572624706\"]\r\n

    [latex]L_N=\\frac{b-a}{N}\\underset{i=1}{\\overset{n}{\\Sigma}}f(a+(b-a)\\frac{i-1}{N})=\\frac{b-a}{N}\\underset{i=0}{\\overset{N-1}{\\Sigma}} f(a+(b-a)\\frac{i}{N})[\/latex] and [latex]R_N=\\frac{b-a}{N}\\underset{i=1}{\\overset{n}{\\Sigma}}f(a+(b-a)\\frac{i}{N})[\/latex]. The left sum has a term corresponding to [latex]i=0[\/latex] and the right sum has a term corresponding to [latex]i=N[\/latex]. In [latex]R_N-L_N[\/latex], any term corresponding to [latex]i=1,2,\\cdots,N-1[\/latex] occurs once with a plus sign and once with a minus sign, so each such term cancels and one is left with [latex]R_N-L_N=\\frac{b-a}{N}(f(a+(b-a))\\frac{N}{N})-(f(a)+(b-a)\\frac{0}{N})=\\frac{b-a}{N}(f(b)-f(a))[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    \r\n

    56.\u00a0<\/strong>Explain why, if [latex]f[\/latex] is increasing on [latex][a,b][\/latex], the error between either [latex]L_N[\/latex]\u00a0or [latex]R_N[\/latex]\u00a0and the area [latex]A[\/latex] below the graph of [latex]f[\/latex] is at most [latex](b-a)\\frac{f(b)-f(a)}{N}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    \r\n

    57.\u00a0<\/strong>For each of the three graphs:<\/p>\r\n\r\n

      \r\n \t
    1. Obtain a lower bound [latex]L(A)[\/latex] for the area enclosed by the curve by adding the areas of the squares enclosed completely<\/em> by the curve.<\/li>\r\n \t
    2. Obtain an upper bound [latex]U(A)[\/latex] for the area by adding to [latex]L(A)[\/latex] the areas [latex]B(A)[\/latex] of the squares enclosed partially<\/em> by the curve.\r\n\"Three<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1170571698240\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571698240\"]\r\n\r\nGraph 1: a. [latex]L(A)=0, \\, B(A)=20[\/latex]; b. [latex]U(A)=20[\/latex]\r\n\r\nGraph 2: a. [latex]L(A)=9[\/latex]; b. [latex]B(A)=11, \\, U(A)=20[\/latex]\r\n\r\nGraph 3: a. [latex]L(A)=11.0[\/latex]; b. [latex]B(A)=4.5, \\, U(A)=15.5[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n
      \r\n
      \r\n

      58.\u00a0<\/strong>In the previous exercise, explain why [latex]L(A)[\/latex] gets no smaller while [latex]U(A)[\/latex] gets no larger as the squares are subdivided into four boxes of equal area.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

      \r\n
      \r\n

      59.\u00a0<\/strong>A unit circle is made up of [latex]n[\/latex] wedges equivalent to the inner wedge in the figure. The base of the inner triangle is 1 unit and its height is [latex] \\sin \\left(\\frac{\\pi }{n}\\right)[\/latex]. The base of the outer triangle is [latex]B= \\cos \\left(\\frac{\\pi }{n}\\right)+ \\sin \\left(\\frac{\\pi }{n}\\right) \\tan \\left(\\frac{\\pi }{n}\\right)[\/latex] and the height is [latex]H=B \\sin \\left(\\frac{2\\pi }{n}\\right)[\/latex]. Use this information to argue that the area of a unit circle is equal to [latex]\\pi[\/latex].<\/p>\r\n\"A\r\n[reveal-answer q=\"fs-id1170572554300\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572554300\"]<\/span>\r\n\r\nLet [latex]A[\/latex] be the area of the unit circle. The circle encloses [latex]n[\/latex] congruent triangles each of area [latex]\\frac{ \\sin (\\frac{2\\pi }{n})}{2}[\/latex], so [latex]\\frac{n}{2} \\sin (\\frac{2\\pi }{n})\\le A[\/latex]. Similarly, the circle is contained inside [latex]n[\/latex] congruent triangles each of area [latex]\\frac{BH}{2}=\\frac{1}{2}( \\cos (\\frac{\\pi }{n})+ \\sin (\\frac{\\pi }{n}) \\tan (\\frac{\\pi }{n})) \\sin (\\frac{2\\pi }{n})[\/latex], so [latex]A\\le \\frac{n}{2} \\sin (\\frac{2\\pi }{n})( \\cos (\\frac{\\pi }{n}))+ \\sin (\\frac{\\pi }{n}) \\tan (\\frac{\\pi }{n})[\/latex]. As [latex]n\\to \\infty, \\, \\frac{n}{2} \\sin (\\frac{2\\pi }{n})=\\frac{\\pi \\sin \\left(\\frac{2\\pi }{n}\\right)}{(\\frac{2\\pi }{n})}\\to \\pi[\/latex], so we conclude [latex]\\pi \\le A[\/latex]. Also, as [latex]n\\to \\infty, \\, \\cos \\left(\\frac{\\pi }{n}\\right)+ \\sin \\left(\\frac{\\pi }{n}\\right) \\tan (\\frac{\\pi }{n})\\to 1[\/latex], so we also have [latex]A\\le \\pi[\/latex]. By the squeeze theorem for limits, we conclude that [latex]A=\\pi[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>","rendered":"

      \n
      \n

      1.\u00a0<\/strong>State whether the given sums are equal or unequal.<\/p>\n

        \n
      1. [latex]\\underset{i=1}{\\overset{10}{\\Sigma}} i[\/latex] and [latex]\\underset{k=1}{\\overset{10}{\\Sigma}} k[\/latex]<\/li>\n
      2. [latex]\\underset{i=1}{\\overset{10}{\\Sigma}} i[\/latex] and [latex]\\underset{i=6}{\\overset{15}{\\Sigma}} (i-5)[\/latex]<\/li>\n
      3. [latex]\\underset{i=1}{\\overset{10}{\\Sigma}} i(i-1)[\/latex] and [latex]\\underset{j=0}{\\overset{9}{\\Sigma}} (j+1)j[\/latex]<\/li>\n
      4. [latex]\\underset{i=1}{\\overset{10}{\\Sigma}} i(i-1)[\/latex] and [latex]\\underset{k=1}{\\overset{10}{\\Sigma}}(k^2-k)[\/latex]<\/li>\n<\/ol>\n
        Show Solution<\/span><\/p>\n
        \n

        a. They are equal; both represent the sum of the first 10 whole numbers. b. They are equal; both represent the sum of the first 10 whole numbers. c. They are equal by substituting [latex]j=i-1[\/latex]. d. They are equal; the first sum factors the terms of the second.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

        In the following exercises (2-3), use the rules for sums of powers of integers to compute the sums.<\/p>\n

        \n
        \n

        2.\u00a0<\/strong>[latex]\\displaystyle\\sum_{i=5}^{10} i[\/latex]<\/p>\n<\/div>\n<\/div>\n

        \n
        \n

        3.\u00a0<\/strong>[latex]\\displaystyle\\sum_{i=5}^{10} i^2[\/latex]<\/p>\n

        Show Solution<\/span><\/p>\n
        \n

        [latex]385-30=355[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

        Suppose that [latex]\\underset{i=1}{\\overset{100}{\\Sigma}} a_i=15[\/latex] and [latex]\\underset{i=1}{\\overset{100}{\\Sigma}} b_i=-12[\/latex]. In the following exercises (4-7), compute the sums.<\/p>\n

        \n
        \n

        4.\u00a0<\/strong>[latex]\\displaystyle\\sum_{i=1}^{100} (a_i+b_i)[\/latex]<\/p>\n<\/div>\n<\/div>\n

        \n
        \n

        5.\u00a0<\/strong>[latex]\\displaystyle\\sum_{i=1}^{100} (a_i-b_i)[\/latex]<\/p>\n

        Show Solution<\/span><\/p>\n
        \n

        [latex]15-(-12)=27[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

        \n
        \n

        6.\u00a0<\/strong>[latex]\\displaystyle\\sum_{i=1}^{100} (3a_i-4b_i)[\/latex]<\/p>\n<\/div>\n<\/div>\n

        \n
        \n

        7.\u00a0<\/strong>[latex]\\displaystyle\\sum_{i=1}^{100} (5a_i+4b_i)[\/latex]<\/p>\n

        Show Solution<\/span><\/p>\n
        \n

        [latex]5(15)+4(-12)=27[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

        In the following exercises (8-11), use summation properties and formulas to rewrite and evaluate the sums.<\/p>\n

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        8.\u00a0<\/strong>[latex]\\displaystyle\\sum_{k=1}^{20} 100(k^2-5k+1)[\/latex]<\/p>\n<\/div>\n<\/div>\n

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        9.\u00a0<\/strong>[latex]\\displaystyle\\sum_{j=1}^{50} (j^2-2j)[\/latex]<\/p>\n

        Show Solution<\/span><\/p>\n
        \n

        [latex]\\underset{j=1}{\\overset{50}{\\Sigma}} j^2-2\\underset{j=1}{\\overset{50}{\\Sigma}} j=\\frac{(50)(51)(101)}{6}-\\frac{2(50)(51)}{2}=40,375[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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        10.\u00a0<\/strong>[latex]\\displaystyle\\sum_{j=11}^{20} (j^2-10j)[\/latex]<\/p>\n<\/div>\n<\/div>\n

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        11.\u00a0<\/strong>[latex]\\displaystyle\\sum_{k=1}^{25} [(2k)^2-100k][\/latex]<\/p>\n

        Show Solution<\/span><\/p>\n
        \n

        [latex]4\\underset{k=1}{\\overset{25}{\\Sigma}} k^2-100\\underset{k=1}{\\overset{25}{\\Sigma}} k=\\frac{4(25)(26)(51)}{9}-50(25)(26)=-10,400[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

        Let [latex]L_n[\/latex] denote the left-endpoint sum using [latex]n[\/latex] subintervals and let [latex]R_n[\/latex] denote the corresponding right-endpoint sum. In the following exercises (12-19), compute the indicated left and right sums for the given functions on the indicated interval.<\/p>\n

        \n
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        12<\/strong>. <\/strong>[latex]L_4[\/latex]\u00a0for [latex]f(x)=\\dfrac{1}{x-1}[\/latex] on [latex][2,3][\/latex]<\/p>\n<\/div>\n<\/div>\n

        \n
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        13. <\/strong>[latex]R_4[\/latex]\u00a0for [latex]g(x)= \\cos (\\pi x)[\/latex] on [latex][0,1][\/latex]<\/p>\n

        Show Solution<\/span><\/p>\n
        \n

        [latex]R_4=0.25[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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        14.\u00a0<\/strong>[latex]L_6[\/latex]\u00a0for [latex]f(x)=\\dfrac{1}{x(x-1)}[\/latex] on [latex][2,5][\/latex]<\/p>\n<\/div>\n<\/div>\n

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        15.\u00a0<\/strong>[latex]R_6[\/latex]\u00a0for [latex]f(x)=\\dfrac{1}{x(x-1)}[\/latex] on [latex][2,5][\/latex]<\/p>\n

        Show Solution<\/span><\/p>\n
        \n

        [latex]R_6=0.372[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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        16.\u00a0\u00a0<\/strong>[latex]R_4[\/latex]\u00a0for [latex]\\dfrac{1}{x^2+1}[\/latex] on [latex][-2,2][\/latex]<\/p>\n<\/div>\n<\/div>\n

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        17.<\/strong>\u00a0<\/em>[latex]L_4[\/latex]\u00a0for [latex]\\dfrac{1}{x^2+1}[\/latex] on [latex][-2,2][\/latex]<\/p>\n

        Show Solution<\/span><\/p>\n
        \n

        [latex]L_4=2.20[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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        18. <\/strong>[latex]R_4[\/latex]\u00a0for [latex]x^2-2x+1[\/latex] on [latex][0,2][\/latex]<\/p>\n<\/div>\n<\/div>\n

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        19.\u00a0<\/strong>[latex]L_8[\/latex]\u00a0for [latex]x^2-2x+1[\/latex] on [latex][0,2][\/latex]<\/p>\n

        Show Solution<\/span><\/p>\n
        \n

        [latex]L_8=0.6875[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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        20.\u00a0<\/strong>Compute the left and right Riemann sums\u2014[latex]L_4[\/latex]\u00a0and [latex]R_4[\/latex], respectively\u2014for [latex]f(x)=(2-|x|)[\/latex] on [latex][-2,2][\/latex]. Compute their average value and compare it with the area under the graph of [latex]f[\/latex].<\/p>\n<\/div>\n<\/div>\n

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        \n

        21.\u00a0<\/strong>Compute the left and right Riemann sums\u2014[latex]L_6[\/latex]\u00a0and [latex]R_6[\/latex], respectively\u2014for [latex]f(x)=(3-|3-x|)[\/latex] on [latex][0,6][\/latex]. Compute their average value and compare it with the area under the graph of [latex]f[\/latex].<\/p>\n

        Show Solution<\/span><\/p>\n
        \n

        [latex]L_6=9.000=R_6[\/latex]. The graph of [latex]f[\/latex] is a triangle with area 9.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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        22.\u00a0<\/strong>Compute the left and right Riemann sums\u2014[latex]L_4[\/latex]\u00a0and [latex]R_4[\/latex], respectively\u2014for [latex]f(x)=\\sqrt{4-x^2}[\/latex] on [latex][-2,2][\/latex] and compare their values.<\/p>\n<\/div>\n<\/div>\n

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        \n

        23.\u00a0<\/strong>Compute the left and right Riemann sums\u2014[latex]L_6[\/latex]\u00a0and [latex]R_6[\/latex], respectively\u2014for [latex]f(x)=\\sqrt{9-(x-3)^2}[\/latex] on [latex][0,6][\/latex] and compare their values.<\/p>\n

        Show Solution<\/span><\/p>\n
        \n

        [latex]L_6=13.12899=R_6[\/latex]. They are equal.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

        Express the following endpoint sums in sigma notation but do not evaluate them (24-27).<\/p>\n

        \n
        \n

        24. <\/strong>[latex]L_{30}[\/latex]\u00a0for [latex]f(x)=x^2[\/latex] on [latex][1,2][\/latex]<\/p>\n<\/div>\n<\/div>\n

        \n
        \n

        25. <\/strong>[latex]L_{10}[\/latex]\u00a0for [latex]f(x)=\\sqrt{4-x^2}[\/latex] on [latex][-2,2][\/latex]<\/p>\n

        Show Solution<\/span><\/p>\n
        \n

        [latex]L_{10}=\\frac{4}{10}\\underset{i=1}{\\overset{10}{\\Sigma}} \\sqrt{4-(-2+4\\frac{(i-1)}{10})}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

        \n
        \n

        26.\u00a0<\/strong>[latex]R_{20}[\/latex]\u00a0for [latex]f(x)= \\sin x[\/latex] on [latex][0,\\pi][\/latex]<\/p>\n<\/div>\n<\/div>\n

        \n
        \n

        27.\u00a0<\/strong>[latex]R_{100}[\/latex]\u00a0for [latex]\\ln x[\/latex] on [latex][1,e][\/latex]<\/p>\n

        Show Solution<\/span><\/p>\n
        \n

        [latex]R_{100}=\\frac{e-1}{100}\\underset{i=1}{\\overset{100}{\\Sigma}} \\ln (1+(e-1)\\frac{i}{100})[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

        In the following exercises (28-33), graph the function then use a calculator or a computer program to evaluate the following left and right endpoint sums. Is the area under the curve between the left and right endpoint sums?<\/p>\n

        \n
        \n

        28. [T] <\/strong>[latex]L_{100}[\/latex]\u00a0and [latex]R_{100}[\/latex]\u00a0for [latex]y=x^2-3x+1[\/latex] on the interval [latex][-1,1][\/latex]<\/p>\n<\/div>\n<\/div>\n

        \n
        \n

        29. [T]\u00a0<\/strong>[latex]L_{100}[\/latex]\u00a0and [latex]R_{100}[\/latex]\u00a0for [latex]y=x^2[\/latex] on the interval [latex][0,1][\/latex]<\/p>\n

        Show Solution<\/span><\/p>\n
        \n

        \"A<\/span><\/p>\n

        [latex]R_{100}=0.33835, \\, L_{100}=0.32835[\/latex]. The plot shows that the left Riemann sum is an underestimate because the function is increasing. Similarly, the right Riemann sum is an overestimate. The area lies between the left and right Riemann sums. Ten rectangles are shown for visual clarity. This behavior persists for more rectangles.<\/p><\/div>\n<\/div>\n<\/div>\n<\/div>\n

        \n
        \n

        30. [T]\u00a0<\/strong>[latex]L_{50}[\/latex]\u00a0and [latex]R_{50}[\/latex]\u00a0for [latex]y=\\dfrac{x+1}{x^2-1}[\/latex] on the interval [latex][2,4][\/latex]<\/p>\n<\/div>\n<\/div>\n

        \n
        \n

        31. [T]\u00a0<\/strong>[latex]L_{100}[\/latex]\u00a0and [latex]R_{100}[\/latex]\u00a0for [latex]y=x^3[\/latex] on the interval [latex][-1,1][\/latex]<\/p>\n

        Show Solution<\/span><\/p>\n
        \n

        \"A<\/span><\/p>\n

        [latex]L_{100}=-0.02, \\, R_{100}=0.02[\/latex]. The left endpoint sum is an underestimate because the function is increasing. Similarly, a right endpoint approximation is an overestimate. The area lies between the left and right endpoint estimates.<\/p><\/div>\n<\/div>\n<\/div>\n<\/div>\n

        \n
        \n

        32. [T]\u00a0<\/strong>[latex]L_{50}[\/latex]\u00a0and [latex]R_{50}[\/latex]\u00a0for [latex]y= \\tan x[\/latex] on the interval [latex]\\left[0,\\frac{\\pi}{4}\\right][\/latex]<\/p>\n<\/div>\n<\/div>\n

        \n
        \n

        33. [T]\u00a0<\/strong>[latex]L_{100}[\/latex]\u00a0and [latex]R_{100}[\/latex]\u00a0for [latex]y=e^{2x}[\/latex] on the interval [latex][-1,1][\/latex]<\/p>\n

        Show Solution<\/span><\/p>\n
        \n

        \"A<\/span><\/p>\n

        [latex]L_{100}=3.555, \\, R_{100}=3.670[\/latex]. The plot shows that the left Riemann sum is an underestimate because the function is increasing. Ten rectangles are shown for visual clarity. This behavior persists for more rectangles.<\/p><\/div>\n<\/div>\n<\/div>\n<\/div>\n

        \n
        \n

        34.\u00a0<\/strong>Let [latex]t_j[\/latex]\u00a0denote the time that it took Tejay van Garteren to ride the [latex]j[\/latex]th stage of the Tour de France in 2014. If there were a total of 21 stages, interpret [latex]\\displaystyle\\sum_{j=1}^{21} t_j[\/latex].<\/p>\n<\/div>\n<\/div>\n

        \n
        \n

        35.\u00a0<\/strong>Let [latex]r_j[\/latex] denote the total rainfall in Portland on the [latex]j[\/latex]th day of the year in 2009. Interpret [latex]\\displaystyle\\sum_{j=1}^{31} r_j[\/latex].<\/p>\n

        Show Solution<\/span><\/p>\n
        \n

        The sum represents the cumulative rainfall in January 2009.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

        \n
        \n

        36.\u00a0<\/strong>Let [latex]d_j[\/latex] denote the hours of daylight and [latex]\\delta_j[\/latex] denote the increase in the hours of daylight from day [latex]j-1[\/latex] to day [latex]j[\/latex] in Fargo, North Dakota, on the [latex]j[\/latex]th day of the year. Interpret [latex]d_1+\\underset{j=2}{\\overset{365}{\\Sigma}} \\delta_j[\/latex].<\/p>\n<\/div>\n<\/div>\n

        \n
        \n

        37.\u00a0<\/strong>To help get in shape, Joe gets a new pair of running shoes. If Joe runs 1 mi each day in week 1 and adds [latex]\\frac{1}{10}[\/latex] mi to his daily routine each week, what is the total mileage on Joe\u2019s shoes after 25 weeks?<\/p>\n

        Show Solution<\/span><\/p>\n
        \n

        The total mileage is [latex]7 \\times \\underset{i=1}{\\overset{25}{\\Sigma}} (1+\\frac{(i-1)}{10})=7 \\times 25+\\frac{7}{10} \\times 12 \\times 25=385[\/latex] mi.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

        \n
        \n

        38.\u00a0<\/strong>The following table gives approximate values of the average annual atmospheric rate of increase in carbon dioxide (CO2<\/sub>) each decade since 1960, in parts per million (ppm). Estimate the total increase in atmospheric CO2<\/sub> between 1964 and 2013.<\/p>\n\n\n\n\n\n\n\n\n\n
        Average Annual Atmospheric CO2<\/sub> Increase, 1964\u20132013Source<\/em>: http:\/\/www.esrl.noaa.gov\/gmd\/ccgg\/trends\/.<\/caption>\n
        Decade<\/th>\nPpm\/y<\/th>\n<\/tr>\n<\/thead>\n
        1964\u20131973<\/td>\n1.07<\/td>\n<\/tr>\n
        1974\u20131983<\/td>\n1.34<\/td>\n<\/tr>\n
        1984\u20131993<\/td>\n1.40<\/td>\n<\/tr>\n
        1994\u20132003<\/td>\n1.87<\/td>\n<\/tr>\n
        2004\u20132013<\/td>\n2.07<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n
        \n
        \n

        39.\u00a0<\/strong>The following table gives the approximate increase in sea level in inches over 20 years starting in the given year. Estimate the net change in mean sea level from 1870 to 2010.<\/p>\n\n\n\n\n\n\n\n\n\n\n\n
        Approximate 20-Year Sea Level Increases, 1870\u20131990Source<\/em>: http:\/\/link.springer.com\/article\/10.1007%2Fs10712-011-9119-1<\/caption>\n
        Starting Year<\/th>\n20-Year Change<\/th>\n<\/tr>\n<\/thead>\n
        1870<\/td>\n0.3<\/td>\n<\/tr>\n
        1890<\/td>\n1.5<\/td>\n<\/tr>\n
        1910<\/td>\n0.2<\/td>\n<\/tr>\n
        1930<\/td>\n2.8<\/td>\n<\/tr>\n
        1950<\/td>\n0.7<\/td>\n<\/tr>\n
        1970<\/td>\n1.1<\/td>\n<\/tr>\n
        1990<\/td>\n1.5<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
        Show Solution<\/span><\/p>\n
        \n

        Add the numbers to get 8.1-in. net increase.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

        \n
        \n

        40.\u00a0<\/strong>The following table gives the approximate increase in dollars in the average price of a gallon of gas per decade since 1950. If the average price of a gallon of gas in 2010 was $2.60, what was the average price of a gallon of gas in 1950?<\/p>\n\n\n\n\n\n\n\n\n\n\n
        Approximate 10-Year Gas Price Increases, 1950\u20132000Source<\/em>: http:\/\/epb.lbl.gov\/homepages\/Rick_Diamond\/docs\/lbnl55011-trends.pdf.<\/caption>\n
        Starting Year<\/th>\n10-Year Change<\/th>\n<\/tr>\n<\/thead>\n
        1950<\/td>\n0.03<\/td>\n<\/tr>\n
        1960<\/td>\n0.05<\/td>\n<\/tr>\n
        1970<\/td>\n0.86<\/td>\n<\/tr>\n
        1980<\/td>\n\u22120.03<\/td>\n<\/tr>\n
        1990<\/td>\n0.29<\/td>\n<\/tr>\n
        2000<\/td>\n1.12<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n
        \n
        \n

        41.\u00a0<\/strong>The following table gives the percent growth of the U.S. population beginning in July of the year indicated. If the U.S. population was 281,421,906 in July 2000, estimate the U.S. population in July 2010.<\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
        Annual Percentage Growth of U.S. Population, 2000\u20132009Source<\/em>: http:\/\/www.census.gov\/popest\/data.<\/caption>\n
        Year<\/th>\n% Change\/Year<\/th>\n<\/tr>\n<\/thead>\n
        2000<\/td>\n1.12<\/td>\n<\/tr>\n
        2001<\/td>\n0.99<\/td>\n<\/tr>\n
        2002<\/td>\n0.93<\/td>\n<\/tr>\n
        2003<\/td>\n0.86<\/td>\n<\/tr>\n
        2004<\/td>\n0.93<\/td>\n<\/tr>\n
        2005<\/td>\n0.93<\/td>\n<\/tr>\n
        2006<\/td>\n0.97<\/td>\n<\/tr>\n
        2007<\/td>\n0.96<\/td>\n<\/tr>\n
        2008<\/td>\n0.95<\/td>\n<\/tr>\n
        2009<\/td>\n0.88<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
        Hint<\/span><\/p>\n
        \n

        To obtain the population in July 2001, multiply the population in July 2000 by 1.0112 to get 284,573,831<\/p>\n<\/div>\n<\/div>\n

        Show Solution<\/span><\/p>\n
        \n

        309,389,957<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

        In the following exercises (42-45), estimate the areas under the curves by computing the left Riemann sums, [latex]L_8[\/latex].<\/p>\n

        \n
        42.\u00a0<\/strong>\"A<\/span><\/div>\n<\/div>\n
        \n
        43.\u00a0<\/strong>\"The<\/p>\n
        Show Solution<\/span><\/p>\n
        [latex]L_8=3+2+1+2+3+4+5+4=24[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n
        \n
        44.\u00a0<\/strong>\"The<\/span><\/div>\n<\/div>\n
        \n
        45.\u00a0<\/strong>\"The<\/p>\n
        Show Solution<\/span><\/p>\n
        [latex]L_8=3+5+7+6+8+6+5+4=44[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n
        \n
        \n

        46. [T]<\/strong> Use a computer algebra system to compute the Riemann sum, [latex]L_N[\/latex], for [latex]N=10,30,50[\/latex] for [latex]f(x)=\\sqrt{1-x^2}[\/latex] on [latex][-1,1][\/latex].<\/p>\n<\/div>\n<\/div>\n

        \n
        \n

        47. [T]<\/strong> Use a computer algebra system to compute the Riemann sum, [latex]L_N[\/latex], for [latex]N=10,30,50[\/latex] for [latex]f(x)=\\dfrac{1}{\\sqrt{1+x^2}}[\/latex] on [latex][-1,1][\/latex].<\/p>\n

        Show Solution<\/span><\/p>\n
        \n

        [latex]L_{10} \\approx 1.7604, \\, L_{30} \\approx 1.7625, \\, L_{50} \\approx 1.76265[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

        \n
        \n

        48. [T]<\/strong> Use a computer algebra system to compute the Riemann sum, [latex]L_N[\/latex], for [latex]N=10,30,50[\/latex] for [latex]f(x)= \\sin^2 x[\/latex] on [latex][0,2\\pi][\/latex]. Compare these estimates with [latex]\\pi[\/latex].<\/p>\n<\/div>\n<\/div>\n

        In the following exercises (49-50), use a calculator or a computer program to evaluate the endpoint sums [latex]R_N[\/latex]\u00a0and [latex]L_N[\/latex]\u00a0for [latex]N=1,10,100[\/latex]. How do these estimates compare with the exact answers, which you can find via geometry?<\/p>\n

        \n
        \n

        49. [T]\u00a0<\/strong>[latex]y= \\cos (\\pi x)[\/latex] on the interval [latex][0,1][\/latex]<\/p>\n

        Show Solution<\/span><\/p>\n
        \n

        [latex]R_1=-1, \\, L_1=1, \\, R_{10}=-0.1, \\, L_{10}=0.1, \\, L_{100}=0.01[\/latex], and [latex]R_{100}=-0.1[\/latex]. By symmetry of the graph, the exact area is zero.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

        \n
        \n

        50. [T]\u00a0<\/strong>[latex]y=3x+2[\/latex] on the interval [latex][3,5][\/latex]<\/p>\n<\/div>\n<\/div>\n

        In the following exercises (51-52), use a calculator or a computer program to evaluate the endpoint sums [latex]R_N[\/latex]\u00a0and [latex]L_N[\/latex]\u00a0for [latex]N=1,10,100[\/latex].<\/p>\n

        \n
        \n

        51. [T]\u00a0<\/strong>[latex]y=x^4-5x^2+4[\/latex] on the interval [latex][-2,2][\/latex], which has an exact area of [latex]\\frac{32}{15}[\/latex]<\/p>\n

        Show Solution<\/span><\/p>\n
        \n

        [latex]R_1=0, \\, L_1=0, \\, R_{10}=2.4499, \\, L_{10}=2.4499, \\, R_{100}=2.1365, \\, L_{100}=2.1365[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

        \n
        \n

        52. [T]\u00a0<\/strong>[latex]y=\\ln x[\/latex] on the interval [latex][1,2][\/latex], which has an exact area of [latex]2\\ln (2)-1[\/latex]<\/p>\n<\/div>\n<\/div>\n

        \n
        \n

        53.\u00a0<\/strong>Explain why, if [latex]f(a)\\ge 0[\/latex] and [latex]f[\/latex] is increasing on [latex][a,b][\/latex], that the left endpoint estimate is a lower bound for the area below the graph of [latex]f[\/latex] on [latex][a,b][\/latex].<\/p>\n

        Show Solution<\/span><\/p>\n
        \n

        If [latex][c,d][\/latex] is a subinterval of [latex][a,b][\/latex] under one of the left-endpoint sum rectangles, then the area of the rectangle contributing to the left-endpoint estimate is [latex]f(c)(d-c)[\/latex]. But, [latex]f(c)\\le f(x)[\/latex] for [latex]c\\le x\\le d[\/latex], so the area under the graph of [latex]f[\/latex] between [latex]c[\/latex] and [latex]d[\/latex] is [latex]f(c)(d-c)[\/latex] plus the area below the graph of [latex]f[\/latex] but above the horizontal line segment at height [latex]f(c)[\/latex], which is positive. As this is true for each left-endpoint sum interval, it follows that the left Riemann sum is less than or equal to the area below the graph of [latex]f[\/latex] on [latex][a,b][\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

        \n
        \n

        54.\u00a0<\/strong>Explain why, if [latex]f(b)\\ge 0[\/latex] and [latex]f[\/latex] is decreasing on [latex][a,b][\/latex], that the left endpoint estimate is an upper bound for the area below the graph of [latex]f[\/latex] on [latex][a,b][\/latex].<\/p>\n<\/div>\n<\/div>\n

        \n
        \n

        55.\u00a0<\/strong>Show that, in general, [latex]R_N-L_N=(b-a) \\times \\frac{f(b)-f(a)}{N}[\/latex].<\/p>\n

        Show Solution<\/span><\/p>\n
        \n

        [latex]L_N=\\frac{b-a}{N}\\underset{i=1}{\\overset{n}{\\Sigma}}f(a+(b-a)\\frac{i-1}{N})=\\frac{b-a}{N}\\underset{i=0}{\\overset{N-1}{\\Sigma}} f(a+(b-a)\\frac{i}{N})[\/latex] and [latex]R_N=\\frac{b-a}{N}\\underset{i=1}{\\overset{n}{\\Sigma}}f(a+(b-a)\\frac{i}{N})[\/latex]. The left sum has a term corresponding to [latex]i=0[\/latex] and the right sum has a term corresponding to [latex]i=N[\/latex]. In [latex]R_N-L_N[\/latex], any term corresponding to [latex]i=1,2,\\cdots,N-1[\/latex] occurs once with a plus sign and once with a minus sign, so each such term cancels and one is left with [latex]R_N-L_N=\\frac{b-a}{N}(f(a+(b-a))\\frac{N}{N})-(f(a)+(b-a)\\frac{0}{N})=\\frac{b-a}{N}(f(b)-f(a))[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

        \n
        \n

        56.\u00a0<\/strong>Explain why, if [latex]f[\/latex] is increasing on [latex][a,b][\/latex], the error between either [latex]L_N[\/latex]\u00a0or [latex]R_N[\/latex]\u00a0and the area [latex]A[\/latex] below the graph of [latex]f[\/latex] is at most [latex](b-a)\\frac{f(b)-f(a)}{N}[\/latex].<\/p>\n<\/div>\n<\/div>\n

        \n
        \n

        57.\u00a0<\/strong>For each of the three graphs:<\/p>\n

          \n
        1. Obtain a lower bound [latex]L(A)[\/latex] for the area enclosed by the curve by adding the areas of the squares enclosed completely<\/em> by the curve.<\/li>\n
        2. Obtain an upper bound [latex]U(A)[\/latex] for the area by adding to [latex]L(A)[\/latex] the areas [latex]B(A)[\/latex] of the squares enclosed partially<\/em> by the curve.
          \n\"Three<\/li>\n<\/ol>\n
          Show Solution<\/span><\/p>\n
          \n

          Graph 1: a. [latex]L(A)=0, \\, B(A)=20[\/latex]; b. [latex]U(A)=20[\/latex]<\/p>\n

          Graph 2: a. [latex]L(A)=9[\/latex]; b. [latex]B(A)=11, \\, U(A)=20[\/latex]<\/p>\n

          Graph 3: a. [latex]L(A)=11.0[\/latex]; b. [latex]B(A)=4.5, \\, U(A)=15.5[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

          \n
          \n

          58.\u00a0<\/strong>In the previous exercise, explain why [latex]L(A)[\/latex] gets no smaller while [latex]U(A)[\/latex] gets no larger as the squares are subdivided into four boxes of equal area.<\/p>\n<\/div>\n<\/div>\n

          \n
          \n

          59.\u00a0<\/strong>A unit circle is made up of [latex]n[\/latex] wedges equivalent to the inner wedge in the figure. The base of the inner triangle is 1 unit and its height is [latex]\\sin \\left(\\frac{\\pi }{n}\\right)[\/latex]. The base of the outer triangle is [latex]B= \\cos \\left(\\frac{\\pi }{n}\\right)+ \\sin \\left(\\frac{\\pi }{n}\\right) \\tan \\left(\\frac{\\pi }{n}\\right)[\/latex] and the height is [latex]H=B \\sin \\left(\\frac{2\\pi }{n}\\right)[\/latex]. Use this information to argue that the area of a unit circle is equal to [latex]\\pi[\/latex].<\/p>\n

          \"A<\/p>\n

          Show Solution<\/span><\/p>\n
          <\/span><\/p>\n

          Let [latex]A[\/latex] be the area of the unit circle. The circle encloses [latex]n[\/latex] congruent triangles each of area [latex]\\frac{ \\sin (\\frac{2\\pi }{n})}{2}[\/latex], so [latex]\\frac{n}{2} \\sin (\\frac{2\\pi }{n})\\le A[\/latex]. Similarly, the circle is contained inside [latex]n[\/latex] congruent triangles each of area [latex]\\frac{BH}{2}=\\frac{1}{2}( \\cos (\\frac{\\pi }{n})+ \\sin (\\frac{\\pi }{n}) \\tan (\\frac{\\pi }{n})) \\sin (\\frac{2\\pi }{n})[\/latex], so [latex]A\\le \\frac{n}{2} \\sin (\\frac{2\\pi }{n})( \\cos (\\frac{\\pi }{n}))+ \\sin (\\frac{\\pi }{n}) \\tan (\\frac{\\pi }{n})[\/latex]. As [latex]n\\to \\infty, \\, \\frac{n}{2} \\sin (\\frac{2\\pi }{n})=\\frac{\\pi \\sin \\left(\\frac{2\\pi }{n}\\right)}{(\\frac{2\\pi }{n})}\\to \\pi[\/latex], so we conclude [latex]\\pi \\le A[\/latex]. Also, as [latex]n\\to \\infty, \\, \\cos \\left(\\frac{\\pi }{n}\\right)+ \\sin \\left(\\frac{\\pi }{n}\\right) \\tan (\\frac{\\pi }{n})\\to 1[\/latex], so we also have [latex]A\\le \\pi[\/latex]. By the squeeze theorem for limits, we conclude that [latex]A=\\pi[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t

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