{"id":1155,"date":"2021-06-30T17:02:01","date_gmt":"2021-06-30T17:02:01","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/problem-set-integration-formulas-and-the-net-change-theorem\/"},"modified":"2021-11-17T01:49:20","modified_gmt":"2021-11-17T01:49:20","slug":"problem-set-integration-formulas-and-the-net-change-theorem","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/problem-set-integration-formulas-and-the-net-change-theorem\/","title":{"raw":"Problem Set: Integration Formulas and the Net Change Theorem","rendered":"Problem Set: Integration Formulas and the Net Change Theorem"},"content":{"raw":"

Use basic integration formulas to compute the following antiderivatives or definite integrals (1-6).<\/p>\r\n\r\n

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1.\u00a0<\/strong>[latex]\\displaystyle\\int (\\sqrt{x}-\\frac{1}{\\sqrt{x}})dx[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572558285\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572558285\"]\r\n

[latex]\\displaystyle\\int (\\sqrt{x}-\\frac{1}{\\sqrt{x}})dx=\\int {x}^{1\\text{\/}2}dx-\\int {x}^{-1\\text{\/}2}dx=\\frac{2}{3}{x}^{3\\text{\/}2}+{C}_{1}-2{x}^{1\\text{\/}2}+{C}_{2}=\\frac{2}{3}{x}^{3\\text{\/}2}-2{x}^{1\\text{\/}2}+C[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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2.\u00a0<\/strong>[latex]\\displaystyle\\int ({e}^{2x}-\\frac{1}{2}{e}^{x\\text{\/}2})dx[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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3.\u00a0<\/strong>[latex]\\displaystyle\\int \\frac{dx}{2x}[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572334239\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572334239\"]\r\n

[latex]\\displaystyle\\int \\frac{dx}{2x}=\\frac{1}{2}\\text{ln}|x|+C[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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4.\u00a0<\/strong>[latex]\\displaystyle\\int \\frac{x-1}{{x}^{2}}dx[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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5.\u00a0<\/strong>[latex]{\\int }_{0}^{\\pi }( \\sin x- \\cos x)dx[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572626594\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572626594\"]\r\n

[latex]{\\int }_{0}^{\\pi } \\sin xdx-{\\int }_{0}^{\\pi } \\cos xdx=\\text{\u2212} \\cos x{|}_{0}^{\\pi }-( \\sin x){|}_{0}^{\\pi }=(\\text{\u2212}(-1)+1)-(0-0)=2[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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6.\u00a0<\/strong>[latex]{\\int }_{0}^{\\pi \\text{\/}2}(x- \\sin x)dx[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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7.\u00a0<\/strong>Write an integral that expresses the increase in the perimeter [latex]P(s)[\/latex] of a square when its side length [latex]s[\/latex] increases from 2 units to 4 units and evaluate the integral.<\/p>\r\n[reveal-answer q=\"fs-id1170572293495\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572293495\"]\r\n

[latex]P(s)=4s,[\/latex] so [latex]\\frac{dP}{ds}=4[\/latex] and [latex]{\\int }_{2}^{4}4ds=8.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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\r\n\r\n8.\u00a0<\/strong>Write an integral that quantifies the change in the area [latex]A(s)={s}^{2}[\/latex] of a square when the side length doubles from S<\/em> units to 2S<\/em> units and evaluate the integral.\r\n\r\n<\/div>\r\n<\/div>\r\n
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\r\n\r\n9.\u00a0<\/strong>A regular N<\/em>-gon (an N<\/em>-sided polygon with sides that have equal length [latex]s[\/latex], such as a pentagon or hexagon) has perimeter Ns<\/em>. Write an integral that expresses the increase in perimeter of a regular N<\/em>-gon when the length of each side increases from 1 unit to 2 units and evaluate the integral.\r\n\r\n[reveal-answer q=\"fs-id1170572554425\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572554425\"]\r\n

[latex]{\\int }_{1}^{2}Nds=N[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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10. <\/strong>The area of a regular pentagon with side length [latex]a>0[\/latex] is pa<\/em>2<\/sup> with [latex]p=\\frac{1}{4}\\sqrt{5+\\sqrt{5+2\\sqrt{5}}}.[\/latex] The Pentagon in Washington, DC, has inner sides of length 360 ft and outer sides of length 920 ft. Write an integral to express the area of the roof of the Pentagon according to these dimensions and evaluate this area.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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\r\n\r\n11.\u00a0<\/strong>A dodecahedron is a Platonic solid with a surface that consists of 12 pentagons, each of equal area. By how much does the surface area of a dodecahedron increase as the side length of each pentagon doubles from 1 unit to 2 units?\r\n\r\n[reveal-answer q=\"fs-id1170572543702\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572543702\"]With [latex]p[\/latex] as in the previous exercise, each of the 12 pentagons increases in area from 2[latex]p[\/latex] to 4[latex]p[\/latex] units so the net increase in the area of the dodecahedron is 36[latex]p[\/latex] units.[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n
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12.\u00a0<\/strong>An icosahedron is a Platonic solid with a surface that consists of 20 equilateral triangles. By how much does the surface area of an icosahedron increase as the side length of each triangle doubles from [latex]a[\/latex] unit to 2[latex]a[\/latex] units?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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\r\n\r\n13.\u00a0<\/strong>Write an integral that quantifies the change in the area of the surface of a cube when its side length doubles from [latex]s[\/latex] unit to 2[latex]s[\/latex] units and evaluate the integral.\r\n\r\n[reveal-answer q=\"fs-id1170572184338\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572184338\"]\r\n

[latex]18{s}^{2}=6{\\int }_{s}^{2s}2xdx[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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14.\u00a0<\/strong>Write an integral that quantifies the increase in the volume of a cube when the side length doubles from [latex]s[\/latex] unit to 2[latex]s[\/latex] units and evaluate the integral.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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\r\n\r\n15.\u00a0<\/strong>Write an integral that quantifies the increase in the surface area of a sphere as its radius doubles from R<\/em> unit to 2R<\/em> units and evaluate the integral.\r\n\r\n[reveal-answer q=\"fs-id1170571653468\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571653468\"][latex]12\\pi {R}^{2}=8\\pi {\\int }_{R}^{2R}rdr[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n
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\r\n\r\n16.\u00a0<\/strong>Write an integral that quantifies the increase in the volume of a sphere as its radius doubles from R<\/em> unit to 2R<\/em> units and evaluate the integral.\r\n\r\n<\/div>\r\n<\/div>\r\n
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\r\n\r\n17.\u00a0<\/strong>Suppose that a particle moves along a straight line with velocity [latex]v(t)=4-2t,[\/latex] where [latex]0\\le t\\le 2[\/latex] (in meters per second). Find the displacement at time [latex]t[\/latex] and the total distance traveled up to [latex]t=2.[\/latex]\r\n\r\n[reveal-answer q=\"fs-id1170571788086\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571788086\"]\r\n

[latex]d(t)={\\int }_{0}^{t}v(s)ds=4t-{t}^{2}.[\/latex] The total distance is [latex]d(2)=4\\text{m}\\text{.}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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18.\u00a0<\/strong>Suppose that a particle moves along a straight line with velocity defined by [latex]v(t)={t}^{2}-3t-18,[\/latex] where [latex]0\\le t\\le 6[\/latex] (in meters per second). Find the displacement at time [latex]t[\/latex] and the total distance traveled up to [latex]t=6.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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\r\n\r\n19.\u00a0<\/strong>Suppose that a particle moves along a straight line with velocity defined by [latex]v(t)=|2t-6|,[\/latex] where [latex]0\\le t\\le 6[\/latex] (in meters per second). Find the displacement at time [latex]t[\/latex] and the total distance traveled up to [latex]t=6.[\/latex]\r\n\r\n[reveal-answer q=\"821823\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"821823\"]\r\n\r\n[latex]d(t)={\\displaystyle\\int _{0}^{t}}v(s)ds.[\/latex] For [latex]t<3,d(t)={\\displaystyle\\int _{0}^{t}}(6-2t)dt=6t-{t}^{2}.[\/latex] For [latex]t>3,d(t)=d(3)+{\\displaystyle\\int _{3}^{t}}(2t-6)dt=18+({t}^{2}-6t).[\/latex] The total distance is [latex]d(6)=18\\text{m}\\text{.}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n
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20.\u00a0<\/strong>Suppose that a particle moves along a straight line with acceleration defined by [latex]a(t)=t-3,[\/latex] where [latex]0\\le t\\le 6[\/latex] (in meters per second). Find the velocity and displacement at time [latex]t[\/latex] and the total distance traveled up to [latex]t=6[\/latex] if [latex]v(0)=3[\/latex] and [latex]d(0)=0.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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21.\u00a0<\/strong>A ball is thrown upward from a height of 1.5 m at an initial speed of 40 m\/sec. Acceleration resulting from gravity is \u22129.8 m\/sec2<\/sup>. Neglecting air resistance, solve for the velocity [latex]v(t)[\/latex] and the height [latex]h(t)[\/latex] of the ball [latex]t[\/latex] seconds after it is thrown and before it returns to the ground.<\/p>\r\n[reveal-answer q=\"fs-id1170572235316\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572235316\"]\r\n

[latex]v(t)=40-9.8t;h(t)=1.5+40t-4.9{t}^{2}[\/latex] m\/s<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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\r\n\r\n22.\u00a0<\/strong>A ball is thrown upward from a height of 3 m at an initial speed of 60 m\/sec. Acceleration resulting from gravity is \u22129.8 m\/sec2<\/sup>. Neglecting air resistance, solve for the velocity [latex]v(t)[\/latex] and the height [latex]h(t)[\/latex] of the ball [latex]t[\/latex] seconds after it is thrown and before it returns to the ground.\r\n\r\n<\/div>\r\n<\/div>\r\n
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23.\u00a0<\/strong>The area [latex]A(t)[\/latex] of a circular shape is growing at a constant rate. If the area increases from 4\u03c0<\/em> units to 9\u03c0<\/em> units between times [latex]t=2[\/latex] and [latex]t=3,[\/latex] find the net change in the radius during that time.<\/p>\r\n[reveal-answer q=\"fs-id1170572419265\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572419265\"]\r\n

The net increase is 1 unit.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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24.\u00a0<\/strong>A spherical balloon is being inflated at a constant rate. If the volume of the balloon changes from 36\u03c0<\/em> in.3<\/sup> to 288\u03c0<\/em> in.3<\/sup> between time [latex]t=30[\/latex] and [latex]t=60[\/latex] seconds, find the net change in the radius of the balloon during that time.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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25.\u00a0<\/strong>Water flows into a conical tank with cross-sectional area \u03c0x<\/em>2<\/sup> at height [latex]x[\/latex] and volume [latex]\\frac{\\pi {x}^{3}}{3}[\/latex] up to height [latex]x[\/latex]. If water flows into the tank at a rate of 1 m3<\/sup>\/min, find the height of water in the tank after 5 min. Find the change in height between 5 min and 10 min.<\/p>\r\n[reveal-answer q=\"fs-id1170571699492\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571699492\"]\r\n

At [latex]t=5,[\/latex] the height of water is [latex]x={(\\frac{15}{\\pi })}^{1\\text{\/}3}\\text{m}\\text{.}.[\/latex] The net change in height from [latex]t=5[\/latex] to [latex]t=10[\/latex] is [latex]{(\\frac{30}{\\pi })}^{1\\text{\/}3}-{(\\frac{15}{\\pi })}^{1\\text{\/}3}[\/latex] m.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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26.\u00a0<\/strong>A horizontal cylindrical tank has cross-sectional area [latex]A(x)=4(6x-{x}^{2}){m}^{2}[\/latex] at height [latex]x[\/latex] meters above the bottom when [latex]x\\le 3.[\/latex]<\/p>\r\n\r\n

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  1. The volume V<\/em> between heights [latex]a[\/latex] and [latex]b[\/latex] is [latex]{\\int }_{a}^{b}A(x)dx.[\/latex] Find the volume at heights between 2 m and 3 m.<\/li>\r\n \t
  2. Suppose that oil is being pumped into the tank at a rate of 50 L\/min. Using the chain rule, [latex]\\frac{dx}{dt}=\\frac{dx}{dV}\\frac{dV}{dt},[\/latex] at how many meters per minute is the height of oil in the tank changing, expressed in terms of [latex]x[\/latex], when the height is at [latex]x[\/latex] meters?<\/li>\r\n \t
  3. How long does it take to fill the tank to 3 m starting from a fill level of 2 m?<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n
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    \r\n\r\n27.\u00a0<\/strong>The following table lists the electrical power in gigawatts\u2014the rate at which energy is consumed\u2014used in a certain city for different hours of the day, in a typical 24-hour period, with hour 1 corresponding to midnight to 1 a.m.\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\r\n\r\n
    Hour<\/th>\r\nPower<\/th>\r\nHour<\/th>\r\nPower<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    1<\/td>\r\n28<\/td>\r\n13<\/td>\r\n48<\/td>\r\n<\/tr>\r\n
    2<\/td>\r\n25<\/td>\r\n14<\/td>\r\n49<\/td>\r\n<\/tr>\r\n
    3<\/td>\r\n24<\/td>\r\n15<\/td>\r\n49<\/td>\r\n<\/tr>\r\n
    4<\/td>\r\n23<\/td>\r\n16<\/td>\r\n50<\/td>\r\n<\/tr>\r\n
    5<\/td>\r\n24<\/td>\r\n17<\/td>\r\n50<\/td>\r\n<\/tr>\r\n
    6<\/td>\r\n27<\/td>\r\n18<\/td>\r\n50<\/td>\r\n<\/tr>\r\n
    7<\/td>\r\n29<\/td>\r\n19<\/td>\r\n46<\/td>\r\n<\/tr>\r\n
    8<\/td>\r\n32<\/td>\r\n20<\/td>\r\n43<\/td>\r\n<\/tr>\r\n
    9<\/td>\r\n34<\/td>\r\n21<\/td>\r\n42<\/td>\r\n<\/tr>\r\n
    10<\/td>\r\n39<\/td>\r\n22<\/td>\r\n40<\/td>\r\n<\/tr>\r\n
    11<\/td>\r\n42<\/td>\r\n23<\/td>\r\n37<\/td>\r\n<\/tr>\r\n
    12<\/td>\r\n46<\/td>\r\n24<\/td>\r\n34<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

    Find the total amount of power in gigawatt-hours (gW-h) consumed by the city in a typical 24-hour period.<\/p>\r\n[reveal-answer q=\"fs-id1170571543238\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571543238\"]\r\n

    The total daily power consumption is estimated as the sum of the hourly power rates, or 911 gW-h.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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    28.\u00a0<\/strong>The average residential electrical power use (in hundreds of watts) per hour is given in the following table.<\/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\r\n\r\n\r\n
    Hour<\/th>\r\nPower<\/th>\r\nHour<\/th>\r\nPower<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    1<\/td>\r\n8<\/td>\r\n13<\/td>\r\n12<\/td>\r\n<\/tr>\r\n
    2<\/td>\r\n6<\/td>\r\n14<\/td>\r\n13<\/td>\r\n<\/tr>\r\n
    3<\/td>\r\n5<\/td>\r\n15<\/td>\r\n14<\/td>\r\n<\/tr>\r\n
    4<\/td>\r\n4<\/td>\r\n16<\/td>\r\n15<\/td>\r\n<\/tr>\r\n
    5<\/td>\r\n5<\/td>\r\n17<\/td>\r\n17<\/td>\r\n<\/tr>\r\n
    6<\/td>\r\n6<\/td>\r\n18<\/td>\r\n19<\/td>\r\n<\/tr>\r\n
    7<\/td>\r\n7<\/td>\r\n19<\/td>\r\n18<\/td>\r\n<\/tr>\r\n
    8<\/td>\r\n8<\/td>\r\n20<\/td>\r\n17<\/td>\r\n<\/tr>\r\n
    9<\/td>\r\n9<\/td>\r\n21<\/td>\r\n16<\/td>\r\n<\/tr>\r\n
    10<\/td>\r\n10<\/td>\r\n22<\/td>\r\n16<\/td>\r\n<\/tr>\r\n
    11<\/td>\r\n10<\/td>\r\n23<\/td>\r\n13<\/td>\r\n<\/tr>\r\n
    12<\/td>\r\n11<\/td>\r\n24<\/td>\r\n11<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
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    1. Compute the average total energy used in a day in kilowatt-hours (kWh).<\/li>\r\n \t
    2. If a ton of coal generates 1842 kWh, how long does it take for an average residence to burn a ton of coal?<\/li>\r\n \t
    3. Explain why the data might fit a plot of the form [latex]p(t)=11.5-7.5 \\sin (\\frac{\\pi t}{12}).[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n
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      29.\u00a0<\/strong>The data in the following table are used to estimate the average power output produced by Peter Sagan for each of the last 18 sec of Stage 1 of the 2012 Tour de France<\/span>.<\/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
      Average Power OutputSource<\/em>: sportsexercisengineering.com<\/caption>\r\n
      Second<\/th>\r\nWatts<\/th>\r\nSecond<\/th>\r\nWatts<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
      1<\/td>\r\n600<\/td>\r\n10<\/td>\r\n1200<\/td>\r\n<\/tr>\r\n
      2<\/td>\r\n500<\/td>\r\n11<\/td>\r\n1170<\/td>\r\n<\/tr>\r\n
      3<\/td>\r\n575<\/td>\r\n12<\/td>\r\n1125<\/td>\r\n<\/tr>\r\n
      4<\/td>\r\n1050<\/td>\r\n13<\/td>\r\n1100<\/td>\r\n<\/tr>\r\n
      5<\/td>\r\n925<\/td>\r\n14<\/td>\r\n1075<\/td>\r\n<\/tr>\r\n
      6<\/td>\r\n950<\/td>\r\n15<\/td>\r\n1000<\/td>\r\n<\/tr>\r\n
      7<\/td>\r\n1050<\/td>\r\n16<\/td>\r\n950<\/td>\r\n<\/tr>\r\n
      8<\/td>\r\n950<\/td>\r\n17<\/td>\r\n900<\/td>\r\n<\/tr>\r\n
      9<\/td>\r\n1100<\/td>\r\n18<\/td>\r\n780<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

      Estimate the net energy used in kilojoules (kJ), noting that 1W = 1 J\/s, and the average power output by Sagan during this time interval.<\/p>\r\n[reveal-answer q=\"fs-id1170572399028\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572399028\"]\r\n

      17 kJ<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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      \r\n\r\n30. <\/strong>The data in the following table are used to estimate the average power output produced by Peter Sagan for each 15-min interval of Stage 1 of the 2012 Tour de France.\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
      Average Power OutputSource<\/em>: sportsexercisengineering.com<\/caption>\r\n
      Minutes<\/th>\r\nWatts<\/th>\r\nMinutes<\/th>\r\nWatts<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
      15<\/td>\r\n200<\/td>\r\n165<\/td>\r\n170<\/td>\r\n<\/tr>\r\n
      30<\/td>\r\n180<\/td>\r\n180<\/td>\r\n220<\/td>\r\n<\/tr>\r\n
      45<\/td>\r\n190<\/td>\r\n195<\/td>\r\n140<\/td>\r\n<\/tr>\r\n
      60<\/td>\r\n230<\/td>\r\n210<\/td>\r\n225<\/td>\r\n<\/tr>\r\n
      75<\/td>\r\n240<\/td>\r\n225<\/td>\r\n170<\/td>\r\n<\/tr>\r\n
      90<\/td>\r\n210<\/td>\r\n240<\/td>\r\n210<\/td>\r\n<\/tr>\r\n
      105<\/td>\r\n210<\/td>\r\n255<\/td>\r\n200<\/td>\r\n<\/tr>\r\n
      120<\/td>\r\n220<\/td>\r\n270<\/td>\r\n220<\/td>\r\n<\/tr>\r\n
      135<\/td>\r\n210<\/td>\r\n285<\/td>\r\n250<\/td>\r\n<\/tr>\r\n
      150<\/td>\r\n150<\/td>\r\n300<\/td>\r\n400<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

      Estimate the net energy used in kilojoules, noting that 1W = 1 J\/s.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

      \r\n
      \r\n

      31.\u00a0<\/strong>The distribution of incomes as of 2012 in the United States in $5000 increments is given in the following table. The [latex]k[\/latex]th row denotes the percentage of households with incomes between [latex]$5000xk[\/latex] and [latex]5000xk+4999.[\/latex] The row [latex]k=40[\/latex] contains all households with income between $200,000 and $250,000 and [latex]k=41[\/latex] accounts for all households with income exceeding $250,000.<\/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\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
      Income DistributionsSource<\/em>: http:\/\/www.census.gov\/prod\/2013pubs\/p60-245.pdf<\/caption>\r\n
      0<\/td>\r\n3.5<\/td>\r\n21<\/td>\r\n1.5<\/td>\r\n<\/tr>\r\n
      1<\/td>\r\n4.1<\/td>\r\n22<\/td>\r\n1.4<\/td>\r\n<\/tr>\r\n
      2<\/td>\r\n5.9<\/td>\r\n23<\/td>\r\n1.3<\/td>\r\n<\/tr>\r\n
      3<\/td>\r\n5.7<\/td>\r\n24<\/td>\r\n1.3<\/td>\r\n<\/tr>\r\n
      4<\/td>\r\n5.9<\/td>\r\n25<\/td>\r\n1.1<\/td>\r\n<\/tr>\r\n
      5<\/td>\r\n5.4<\/td>\r\n26<\/td>\r\n1.0<\/td>\r\n<\/tr>\r\n
      6<\/td>\r\n5.5<\/td>\r\n27<\/td>\r\n0.75<\/td>\r\n<\/tr>\r\n
      7<\/td>\r\n5.1<\/td>\r\n28<\/td>\r\n0.8<\/td>\r\n<\/tr>\r\n
      8<\/td>\r\n4.8<\/td>\r\n29<\/td>\r\n1.0<\/td>\r\n<\/tr>\r\n
      9<\/td>\r\n4.1<\/td>\r\n30<\/td>\r\n0.6<\/td>\r\n<\/tr>\r\n
      10<\/td>\r\n4.3<\/td>\r\n31<\/td>\r\n0.6<\/td>\r\n<\/tr>\r\n
      11<\/td>\r\n3.5<\/td>\r\n32<\/td>\r\n0.5<\/td>\r\n<\/tr>\r\n
      12<\/td>\r\n3.7<\/td>\r\n33<\/td>\r\n0.5<\/td>\r\n<\/tr>\r\n
      13<\/td>\r\n3.2<\/td>\r\n34<\/td>\r\n0.4<\/td>\r\n<\/tr>\r\n
      14<\/td>\r\n3.0<\/td>\r\n35<\/td>\r\n0.3<\/td>\r\n<\/tr>\r\n
      15<\/td>\r\n2.8<\/td>\r\n36<\/td>\r\n0.3<\/td>\r\n<\/tr>\r\n
      16<\/td>\r\n2.5<\/td>\r\n37<\/td>\r\n0.3<\/td>\r\n<\/tr>\r\n
      17<\/td>\r\n2.2<\/td>\r\n38<\/td>\r\n0.2<\/td>\r\n<\/tr>\r\n
      18<\/td>\r\n2.2<\/td>\r\n39<\/td>\r\n1.8<\/td>\r\n<\/tr>\r\n
      19<\/td>\r\n1.8<\/td>\r\n40<\/td>\r\n2.3<\/td>\r\n<\/tr>\r\n
      20<\/td>\r\n2.1<\/td>\r\n41<\/td>\r\n<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
        \r\n \t
      1. Estimate the percentage of U.S. households in 2012 with incomes less than $55,000.<\/li>\r\n \t
      2. What percentage of households had incomes exceeding $85,000?<\/li>\r\n \t
      3. Plot the data and try to fit its shape to that of a graph of the form [latex]a(x+c){e}^{\\text{\u2212}b(x+e)}[\/latex] for suitable [latex]a,b,c.[\/latex]<\/li>\r\n<\/ol>\r\n
        \r\n\r\n[reveal-answer q=\"fs-id1170572589227\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572589227\"]\r\n

        a. 54.3%; b. 27.00%; c. The curve in the following plot is [latex]2.35(t+3){e}^{-0.15(t+3)}.[\/latex]<\/p>\r\n\"A<\/span>[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

        \r\n
        \r\n

        32.\u00a0<\/strong>Newton\u2019s law of gravity states that the gravitational force exerted by an object of mass M<\/em> and one of mass [latex]m[\/latex] with centers that are separated by a distance [latex]r[\/latex] is [latex]F=G\\frac{mM}{{r}^{2}},[\/latex] with G<\/em> an empirical constant [latex]G=6.67x{10}^{-11}{m}^{3}\\text{\/}(kg\u00b7{s}^{2}).[\/latex] The work done by a variable force over an interval [latex]\\left[a,b\\right][\/latex] is defined as [latex]W={\\int }_{a}^{b}F(x)dx.[\/latex] If Earth has mass [latex]5.97219\u00d7{10}^{24}[\/latex] and radius 6371 km, compute the amount of work to elevate a polar weather satellite of mass 1400 kg to its orbiting altitude of 850 km above Earth.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

        \r\n
        \r\n

        33.\u00a0<\/strong>For a given motor vehicle, the maximum achievable deceleration<\/span> from braking is approximately 7 m\/sec2<\/sup> on dry concrete. On wet asphalt, it is approximately 2.5 m\/sec2<\/sup>. Given that 1 mph corresponds to 0.447 m\/sec, find the total distance that a car travels in meters on dry concrete after the brakes are applied until it comes to a complete stop if the initial velocity is 67 mph (30 m\/sec) or if the initial braking velocity is 56 mph (25 m\/sec). Find the corresponding distances if the surface is slippery wet asphalt.<\/p>\r\n[reveal-answer q=\"fs-id1170572379142\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572379142\"]In dry conditions, with initial velocity [latex]{v}_{0}=30[\/latex] m\/s, [latex]D=64.3[\/latex] and, if [latex]{v}_{0}=25,D=44.64.[\/latex] In wet conditions, if [latex]{v}_{0}=30,[\/latex] and [latex]D=180[\/latex] and if [latex]{v}_{0}=25,D=125.[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

        \r\n
        \r\n

        34.\u00a0<\/strong>John is a 25-year old man who weighs 160 lb. He burns [latex]500-50t[\/latex] calories\/hr while riding his bike for [latex]t[\/latex] hours. If an oatmeal cookie has 55 cal and John eats 4[latex]t[\/latex] cookies during the [latex]t[\/latex]th hour, how many net calories has he lost after 3 hours riding his bike?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

        \r\n
        \r\n

        35.\u00a0<\/strong>Sandra is a 25-year old woman who weighs 120 lb. She burns [latex]300-50t[\/latex] cal\/hr while walking on her treadmill. Her caloric intake from drinking Gatorade is 100[latex]t[\/latex] calories during the [latex]t[\/latex]th hour. What is her net decrease in calories after walking for 3 hours?<\/p>\r\n[reveal-answer q=\"fs-id1170571613635\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571613635\"]\r\n

        225 cal<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

        \r\n
        \r\n

        36.\u00a0<\/strong>A motor vehicle has a maximum efficiency of 33 mpg at a cruising speed of 40 mph. The efficiency drops at a rate of 0.1 mpg\/mph between 40 mph and 50 mph, and at a rate of 0.4 mpg\/mph between 50 mph and 80 mph. What is the efficiency in miles per gallon if the car is cruising at 50 mph? What is the efficiency in miles per gallon if the car is cruising at 80 mph? If gasoline costs $3.50\/gal, what is the cost of fuel to drive 50 mi at 40 mph, at 50 mph, and at 80 mph?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

        \r\n
        \r\n

        37.\u00a0<\/strong>Although some engines are more efficient at given a horsepower than others, on average, fuel efficiency decreases with horsepower at a rate of [latex]1\\text{\/}25[\/latex] mpg\/horsepower. If a typical 50-horsepower engine has an average fuel efficiency of 32 mpg, what is the average fuel efficiency of an engine with the following horsepower: 150, 300, 450?<\/p>\r\n[reveal-answer q=\"fs-id1170572434983\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572434983\"]\r\n

        [latex]E(150)=28,E(300)=22,E(450)=16[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

        \r\n
        \r\n

        38. [T]<\/strong> The following table lists the 2013 schedule of federal income tax<\/span> versus taxable income.<\/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
        Federal Income Tax Versus Taxable IncomeSource<\/em>: http:\/\/www.irs.gov\/pub\/irs-prior\/i1040tt--2013.pdf.<\/caption>\r\n
        Taxable Income Range<\/th>\r\nThe Tax Is \u2026<\/th>\r\n\u2026 Of the Amount Over<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
        $0\u2013$8925<\/td>\r\n10%<\/td>\r\n$0<\/td>\r\n<\/tr>\r\n
        $8925\u2013$36,250<\/td>\r\n$892.50 + 15%<\/td>\r\n$8925<\/td>\r\n<\/tr>\r\n
        $36,250\u2013$87,850<\/td>\r\n$4,991.25 + 25%<\/td>\r\n$36,250<\/td>\r\n<\/tr>\r\n
        $87,850\u2013$183,250<\/td>\r\n$17,891.25 + 28%<\/td>\r\n$87,850<\/td>\r\n<\/tr>\r\n
        $183,250\u2013$398,350<\/td>\r\n$44,603.25 + 33%<\/td>\r\n$183,250<\/td>\r\n<\/tr>\r\n
        $398,350\u2013$400,000<\/td>\r\n$115,586.25 + 35%<\/td>\r\n$398,350<\/td>\r\n<\/tr>\r\n
        > $400,000<\/td>\r\n$116,163.75 + 39.6%<\/td>\r\n$400,000<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

        Suppose that Steve just received a $10,000 raise. How much of this raise is left after federal taxes if Steve\u2019s salary before receiving the raise was $40,000? If it was $90,000? If it was $385,000?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

        \r\n
        \r\n

        39. [T]<\/strong> The following table provides hypothetical data regarding the level of service for a certain highway.<\/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
        Highway Speed Range (mph)<\/th>\r\nVehicles per Hour per Lane<\/th>\r\nDensity Range (vehicles\/mi)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
        > 60<\/td>\r\n< 600<\/td>\r\n< 10<\/td>\r\n<\/tr>\r\n
        60\u201357<\/td>\r\n600\u20131000<\/td>\r\n10\u201320<\/td>\r\n<\/tr>\r\n
        57\u201354<\/td>\r\n1000\u20131500<\/td>\r\n20\u201330<\/td>\r\n<\/tr>\r\n
        54\u201346<\/td>\r\n1500\u20131900<\/td>\r\n30\u201345<\/td>\r\n<\/tr>\r\n
        46\u201330<\/td>\r\n1900\u2013<\/strong>2100<\/td>\r\n45\u201370<\/td>\r\n<\/tr>\r\n
        <30<\/td>\r\nUnstable<\/td>\r\n70\u2013200<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
          \r\n \t
        1. Plot vehicles per hour per lane on the [latex]x[\/latex]-axis and highway speed on the [latex]y[\/latex]-axis.<\/li>\r\n \t
        2. Compute the average decrease in speed (in miles per hour) per unit increase in congestion (vehicles per hour per lane) as the latter increases from 600 to 1000, from 1000 to 1500, and from 1500 to 2100. Does the decrease in miles per hour depend linearly on the increase in vehicles per hour per lane?<\/li>\r\n \t
        3. Plot minutes per mile (60 times the reciprocal of miles per hour) as a function of vehicles per hour per lane. Is this function linear?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1170571712876\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571712876\"]\r\n

          a.<\/p>\r\n\"A<\/span>\r\nb. Between 600 and 1000 the average decrease in vehicles per hour per lane is \u22120.0075. Between 1000 and 1500 it is \u22120.006 per vehicles per hour per lane, and between 1500 and 2100 it is \u22120.04 vehicles per hour per lane. c.\r\n\"A<\/span>\r\nThe graph is nonlinear, with minutes per mile increasing dramatically as vehicles per hour per lane reach 2000.[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

          For the next two exercises use the data in the following table, which displays bald eagle<\/span> populations from 1963 to 2000 in the continental United States.<\/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
          Population of Breeding Bald Eagle PairsSource<\/em>: http:\/\/www.fws.gov\/Midwest\/eagle\/population\/chtofprs.html.<\/caption>\r\n
          Year<\/th>\r\nPopulation of Breeding Pairs of Bald Eagles<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
          1963<\/td>\r\n487<\/td>\r\n<\/tr>\r\n
          1974<\/td>\r\n791<\/td>\r\n<\/tr>\r\n
          1981<\/td>\r\n1188<\/td>\r\n<\/tr>\r\n
          1986<\/td>\r\n1875<\/td>\r\n<\/tr>\r\n
          1992<\/td>\r\n3749<\/td>\r\n<\/tr>\r\n
          1996<\/td>\r\n5094<\/td>\r\n<\/tr>\r\n
          2000<\/td>\r\n6471<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
          \r\n
          \r\n\r\n40. [T]<\/strong> The graph below plots the quadratic [latex]p(t)=6.48{t}^{2}-80.31t+585.69[\/latex] against the data in preceding table, normalized so that [latex]t=0[\/latex] corresponds to 1963. Estimate the average number of bald eagles per year present for the 37 years by computing the average value of [latex]p[\/latex] over [latex]\\left[0,37\\right].[\/latex]\r\n\r\n\"A<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n
          \r\n
          \r\n

          41. [T]<\/strong> The graph below plots the cubic [latex]p(t)=0.07{t}^{3}+2.42{t}^{2}-25.63t+521.23[\/latex] against the data in the preceding table, normalized so that [latex]t=0[\/latex] corresponds to 1963. Estimate the average number of bald eagles per year present for the 37 years by computing the average value of [latex]p[\/latex] over [latex]\\left[0,37\\right].[\/latex]<\/p>\r\n\"A<\/span>\r\n\r\n[reveal-answer q=\"fs-id1170572223553\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572223553\"]\r\n

          [latex]\\frac{1}{37}{\\int }_{0}^{37}p(t)dt=\\frac{0.07{(37)}^{3}}{4}+\\frac{2.42{(37)}^{2}}{3}-\\frac{25.63(37)}{2}+521.23\\approx 2037[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

          \r\n
          \r\n

          42. [T]<\/strong> Suppose you go on a road trip and record your speed at every half hour, as compiled in the following table. The best quadratic fit to the data is [latex]q(t)=5{x}^{2}-11x+49\\text{,}[\/latex] shown in the accompanying graph. Integrate [latex]q[\/latex] to estimate the total distance driven over the 3 hours.<\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
          Time (hr)<\/th>\r\nSpeed (mph)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
          0 (start)<\/td>\r\n50<\/td>\r\n<\/tr>\r\n
          1<\/td>\r\n40<\/td>\r\n<\/tr>\r\n
          2<\/td>\r\n50<\/td>\r\n<\/tr>\r\n
          3<\/td>\r\n60<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n\"A<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n

          As a car accelerates, it does not accelerate at a constant rate; rather, the acceleration is variable. For the following exercises, use the following table, which contains the acceleration measured at every second as a driver merges onto a freeway.<\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
          Time (sec)<\/th>\r\nAcceleration (mph\/sec)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
          1<\/td>\r\n11.2<\/td>\r\n<\/tr>\r\n
          2<\/td>\r\n10.6<\/td>\r\n<\/tr>\r\n
          3<\/td>\r\n8.1<\/td>\r\n<\/tr>\r\n
          4<\/td>\r\n5.4<\/td>\r\n<\/tr>\r\n
          5<\/td>\r\n0<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
          \r\n
          \r\n

          43. [T]<\/strong> The accompanying graph plots the best quadratic fit, [latex]a(t)=-0.70{t}^{2}+1.44t+10.44,[\/latex] to the data from the preceding table. Compute the average value of [latex]a(t)[\/latex] to estimate the average acceleration between [latex]t=0[\/latex] and [latex]t=5.[\/latex]<\/p>\r\n\"A\r\n\r\n[reveal-answer q=\"742509\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"742509\"]Average acceleration is [latex]A=\\frac{1}{5}{\\int }_{0}^{5}a(t)dt=-\\frac{0.7({5}^{2})}{3}+\\frac{1.44(5)}{2}+10.44\\approx 8.2[\/latex] mph\/s[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

          \r\n
          \r\n

          44. [T]<\/strong> Using your acceleration equation from the previous exercise, find the corresponding velocity equation. Assuming the final velocity is 0 mph, find the velocity at time [latex]t=0.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

          \r\n
          \r\n

          45. [T]<\/strong> Using your velocity equation from the previous exercise, find the corresponding distance equation, assuming your initial distance is 0 mi. How far did you travel while you accelerated your car? (Hint:<\/em> You will need to convert time units.)<\/p>\r\n[reveal-answer q=\"fs-id1170571610313\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571610313\"]\r\n

          [latex]d(t)={\\int }_{0}^{1}|v(t)|dt={\\int }_{0}^{t}(\\frac{7}{30}{t}^{3}-0.72{t}^{2}-10.44t+41.033)dt=\\frac{7}{120}{t}^{4}-0.24{t}^{3}-5.22{t}^{3}+41.033t.[\/latex] Then, [latex]d(5)\\approx 81.12[\/latex] mph [latex]\u00d7 \\sec \\approx 119[\/latex] feet.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

          \r\n
          \r\n

          46. [T]<\/strong> The number of hamburgers sold at a restaurant throughout the day is given in the following table, with the accompanying graph plotting the best cubic fit to the data, [latex]b(t)=0.12{t}^{3}-2.13{t}^{3}+12.13t+3.91,[\/latex] with [latex]t=0[\/latex] corresponding to 9 a.m. and [latex]t=12[\/latex] corresponding to 9 p.m. Compute the average value of [latex]b(t)[\/latex] to estimate the average number of hamburgers sold per hour.<\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
          Hours Past Midnight<\/th>\r\nNo. of Burgers Sold<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
          9<\/td>\r\n3<\/td>\r\n<\/tr>\r\n
          12<\/td>\r\n28<\/td>\r\n<\/tr>\r\n
          15<\/td>\r\n20<\/td>\r\n<\/tr>\r\n
          18<\/td>\r\n30<\/td>\r\n<\/tr>\r\n
          21<\/td>\r\n45<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n\"A<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n
          \r\n
          \r\n

          47. [T]\u00a0\u00a0<\/strong>An athlete runs by a motion detector, which records her speed, as displayed in the following table. The best linear fit to this data, [latex]\\ell (t)=-0.068t+5.14\\text{,}[\/latex] is shown in the accompanying graph. Use the average value of [latex]\\ell (t)[\/latex] between [latex]t=0[\/latex] and [latex]t=40[\/latex] to estimate the runner\u2019s average speed.<\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
          Minutes<\/th>\r\nSpeed (m\/sec)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
          0<\/td>\r\n5<\/td>\r\n<\/tr>\r\n
          10<\/td>\r\n4.8<\/td>\r\n<\/tr>\r\n
          20<\/td>\r\n3.6<\/td>\r\n<\/tr>\r\n
          30<\/td>\r\n3.0<\/td>\r\n<\/tr>\r\n
          40<\/td>\r\n2.5<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n\"A<\/span>\r\n\r\n[reveal-answer q=\"fs-id1170572480382\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572480382\"]\r\n

          [latex]\\frac{1}{40}{\\int }_{0}^{40}(-0.068t+5.14)dt=-\\frac{0.068(40)}{2}+5.14=3.78[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>","rendered":"

          Use basic integration formulas to compute the following antiderivatives or definite integrals (1-6).<\/p>\n

          \n
          \n

          1.\u00a0<\/strong>[latex]\\displaystyle\\int (\\sqrt{x}-\\frac{1}{\\sqrt{x}})dx[\/latex]<\/p>\n

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

          [latex]\\displaystyle\\int (\\sqrt{x}-\\frac{1}{\\sqrt{x}})dx=\\int {x}^{1\\text{\/}2}dx-\\int {x}^{-1\\text{\/}2}dx=\\frac{2}{3}{x}^{3\\text{\/}2}+{C}_{1}-2{x}^{1\\text{\/}2}+{C}_{2}=\\frac{2}{3}{x}^{3\\text{\/}2}-2{x}^{1\\text{\/}2}+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

          \n
          \n

          2.\u00a0<\/strong>[latex]\\displaystyle\\int ({e}^{2x}-\\frac{1}{2}{e}^{x\\text{\/}2})dx[\/latex]<\/p>\n<\/div>\n<\/div>\n

          \n
          \n

          3.\u00a0<\/strong>[latex]\\displaystyle\\int \\frac{dx}{2x}[\/latex]<\/p>\n

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

          [latex]\\displaystyle\\int \\frac{dx}{2x}=\\frac{1}{2}\\text{ln}|x|+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

          \n
          \n

          4.\u00a0<\/strong>[latex]\\displaystyle\\int \\frac{x-1}{{x}^{2}}dx[\/latex]<\/p>\n<\/div>\n<\/div>\n

          \n
          \n

          5.\u00a0<\/strong>[latex]{\\int }_{0}^{\\pi }( \\sin x- \\cos x)dx[\/latex]<\/p>\n

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

          [latex]{\\int }_{0}^{\\pi } \\sin xdx-{\\int }_{0}^{\\pi } \\cos xdx=\\text{\u2212} \\cos x{|}_{0}^{\\pi }-( \\sin x){|}_{0}^{\\pi }=(\\text{\u2212}(-1)+1)-(0-0)=2[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

          \n
          \n

          6.\u00a0<\/strong>[latex]{\\int }_{0}^{\\pi \\text{\/}2}(x- \\sin x)dx[\/latex]<\/p>\n<\/div>\n<\/div>\n

          \n
          \n

          7.\u00a0<\/strong>Write an integral that expresses the increase in the perimeter [latex]P(s)[\/latex] of a square when its side length [latex]s[\/latex] increases from 2 units to 4 units and evaluate the integral.<\/p>\n

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

          [latex]P(s)=4s,[\/latex] so [latex]\\frac{dP}{ds}=4[\/latex] and [latex]{\\int }_{2}^{4}4ds=8.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

          \n
          \n

          8.\u00a0<\/strong>Write an integral that quantifies the change in the area [latex]A(s)={s}^{2}[\/latex] of a square when the side length doubles from S<\/em> units to 2S<\/em> units and evaluate the integral.<\/p>\n<\/div>\n<\/div>\n

          \n
          \n

          9.\u00a0<\/strong>A regular N<\/em>-gon (an N<\/em>-sided polygon with sides that have equal length [latex]s[\/latex], such as a pentagon or hexagon) has perimeter Ns<\/em>. Write an integral that expresses the increase in perimeter of a regular N<\/em>-gon when the length of each side increases from 1 unit to 2 units and evaluate the integral.<\/p>\n

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

          [latex]{\\int }_{1}^{2}Nds=N[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

          \n
          \n

          10. <\/strong>The area of a regular pentagon with side length [latex]a>0[\/latex] is pa<\/em>2<\/sup> with [latex]p=\\frac{1}{4}\\sqrt{5+\\sqrt{5+2\\sqrt{5}}}.[\/latex] The Pentagon in Washington, DC, has inner sides of length 360 ft and outer sides of length 920 ft. Write an integral to express the area of the roof of the Pentagon according to these dimensions and evaluate this area.<\/p>\n<\/div>\n<\/div>\n

          \n
          \n

          11.\u00a0<\/strong>A dodecahedron is a Platonic solid with a surface that consists of 12 pentagons, each of equal area. By how much does the surface area of a dodecahedron increase as the side length of each pentagon doubles from 1 unit to 2 units?<\/p>\n

          Show Solution<\/span><\/p>\n
          With [latex]p[\/latex] as in the previous exercise, each of the 12 pentagons increases in area from 2[latex]p[\/latex] to 4[latex]p[\/latex] units so the net increase in the area of the dodecahedron is 36[latex]p[\/latex] units.<\/div>\n<\/div>\n<\/div>\n<\/div>\n
          \n
          \n

          12.\u00a0<\/strong>An icosahedron is a Platonic solid with a surface that consists of 20 equilateral triangles. By how much does the surface area of an icosahedron increase as the side length of each triangle doubles from [latex]a[\/latex] unit to 2[latex]a[\/latex] units?<\/p>\n<\/div>\n<\/div>\n

          \n
          \n

          13.\u00a0<\/strong>Write an integral that quantifies the change in the area of the surface of a cube when its side length doubles from [latex]s[\/latex] unit to 2[latex]s[\/latex] units and evaluate the integral.<\/p>\n

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

          [latex]18{s}^{2}=6{\\int }_{s}^{2s}2xdx[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

          \n
          \n

          14.\u00a0<\/strong>Write an integral that quantifies the increase in the volume of a cube when the side length doubles from [latex]s[\/latex] unit to 2[latex]s[\/latex] units and evaluate the integral.<\/p>\n<\/div>\n<\/div>\n

          \n
          \n

          15.\u00a0<\/strong>Write an integral that quantifies the increase in the surface area of a sphere as its radius doubles from R<\/em> unit to 2R<\/em> units and evaluate the integral.<\/p>\n

          Show Solution<\/span><\/p>\n
          [latex]12\\pi {R}^{2}=8\\pi {\\int }_{R}^{2R}rdr[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n
          \n
          \n

          16.\u00a0<\/strong>Write an integral that quantifies the increase in the volume of a sphere as its radius doubles from R<\/em> unit to 2R<\/em> units and evaluate the integral.<\/p>\n<\/div>\n<\/div>\n

          \n
          \n

          17.\u00a0<\/strong>Suppose that a particle moves along a straight line with velocity [latex]v(t)=4-2t,[\/latex] where [latex]0\\le t\\le 2[\/latex] (in meters per second). Find the displacement at time [latex]t[\/latex] and the total distance traveled up to [latex]t=2.[\/latex]<\/p>\n

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

          [latex]d(t)={\\int }_{0}^{t}v(s)ds=4t-{t}^{2}.[\/latex] The total distance is [latex]d(2)=4\\text{m}\\text{.}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

          \n
          \n

          18.\u00a0<\/strong>Suppose that a particle moves along a straight line with velocity defined by [latex]v(t)={t}^{2}-3t-18,[\/latex] where [latex]0\\le t\\le 6[\/latex] (in meters per second). Find the displacement at time [latex]t[\/latex] and the total distance traveled up to [latex]t=6.[\/latex]<\/p>\n<\/div>\n<\/div>\n

          \n
          \n

          19.\u00a0<\/strong>Suppose that a particle moves along a straight line with velocity defined by [latex]v(t)=|2t-6|,[\/latex] where [latex]0\\le t\\le 6[\/latex] (in meters per second). Find the displacement at time [latex]t[\/latex] and the total distance traveled up to [latex]t=6.[\/latex]<\/p>\n

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

          [latex]d(t)={\\displaystyle\\int _{0}^{t}}v(s)ds.[\/latex] For [latex]t<3,d(t)={\\displaystyle\\int _{0}^{t}}(6-2t)dt=6t-{t}^{2}.[\/latex] For [latex]t>3,d(t)=d(3)+{\\displaystyle\\int _{3}^{t}}(2t-6)dt=18+({t}^{2}-6t).[\/latex] The total distance is [latex]d(6)=18\\text{m}\\text{.}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

          \n
          \n

          20.\u00a0<\/strong>Suppose that a particle moves along a straight line with acceleration defined by [latex]a(t)=t-3,[\/latex] where [latex]0\\le t\\le 6[\/latex] (in meters per second). Find the velocity and displacement at time [latex]t[\/latex] and the total distance traveled up to [latex]t=6[\/latex] if [latex]v(0)=3[\/latex] and [latex]d(0)=0.[\/latex]<\/p>\n<\/div>\n<\/div>\n

          \n
          \n

          21.\u00a0<\/strong>A ball is thrown upward from a height of 1.5 m at an initial speed of 40 m\/sec. Acceleration resulting from gravity is \u22129.8 m\/sec2<\/sup>. Neglecting air resistance, solve for the velocity [latex]v(t)[\/latex] and the height [latex]h(t)[\/latex] of the ball [latex]t[\/latex] seconds after it is thrown and before it returns to the ground.<\/p>\n

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

          [latex]v(t)=40-9.8t;h(t)=1.5+40t-4.9{t}^{2}[\/latex] m\/s<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

          \n
          \n

          22.\u00a0<\/strong>A ball is thrown upward from a height of 3 m at an initial speed of 60 m\/sec. Acceleration resulting from gravity is \u22129.8 m\/sec2<\/sup>. Neglecting air resistance, solve for the velocity [latex]v(t)[\/latex] and the height [latex]h(t)[\/latex] of the ball [latex]t[\/latex] seconds after it is thrown and before it returns to the ground.<\/p>\n<\/div>\n<\/div>\n

          \n
          \n

          23.\u00a0<\/strong>The area [latex]A(t)[\/latex] of a circular shape is growing at a constant rate. If the area increases from 4\u03c0<\/em> units to 9\u03c0<\/em> units between times [latex]t=2[\/latex] and [latex]t=3,[\/latex] find the net change in the radius during that time.<\/p>\n

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

          The net increase is 1 unit.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

          \n
          \n

          24.\u00a0<\/strong>A spherical balloon is being inflated at a constant rate. If the volume of the balloon changes from 36\u03c0<\/em> in.3<\/sup> to 288\u03c0<\/em> in.3<\/sup> between time [latex]t=30[\/latex] and [latex]t=60[\/latex] seconds, find the net change in the radius of the balloon during that time.<\/p>\n<\/div>\n<\/div>\n

          \n
          \n

          25.\u00a0<\/strong>Water flows into a conical tank with cross-sectional area \u03c0x<\/em>2<\/sup> at height [latex]x[\/latex] and volume [latex]\\frac{\\pi {x}^{3}}{3}[\/latex] up to height [latex]x[\/latex]. If water flows into the tank at a rate of 1 m3<\/sup>\/min, find the height of water in the tank after 5 min. Find the change in height between 5 min and 10 min.<\/p>\n

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

          At [latex]t=5,[\/latex] the height of water is [latex]x={(\\frac{15}{\\pi })}^{1\\text{\/}3}\\text{m}\\text{.}.[\/latex] The net change in height from [latex]t=5[\/latex] to [latex]t=10[\/latex] is [latex]{(\\frac{30}{\\pi })}^{1\\text{\/}3}-{(\\frac{15}{\\pi })}^{1\\text{\/}3}[\/latex] m.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

          \n
          \n

          26.\u00a0<\/strong>A horizontal cylindrical tank has cross-sectional area [latex]A(x)=4(6x-{x}^{2}){m}^{2}[\/latex] at height [latex]x[\/latex] meters above the bottom when [latex]x\\le 3.[\/latex]<\/p>\n

            \n
          1. The volume V<\/em> between heights [latex]a[\/latex] and [latex]b[\/latex] is [latex]{\\int }_{a}^{b}A(x)dx.[\/latex] Find the volume at heights between 2 m and 3 m.<\/li>\n
          2. Suppose that oil is being pumped into the tank at a rate of 50 L\/min. Using the chain rule, [latex]\\frac{dx}{dt}=\\frac{dx}{dV}\\frac{dV}{dt},[\/latex] at how many meters per minute is the height of oil in the tank changing, expressed in terms of [latex]x[\/latex], when the height is at [latex]x[\/latex] meters?<\/li>\n
          3. How long does it take to fill the tank to 3 m starting from a fill level of 2 m?<\/li>\n<\/ol>\n<\/div>\n<\/div>\n
            \n
            \n

            27.\u00a0<\/strong>The following table lists the electrical power in gigawatts\u2014the rate at which energy is consumed\u2014used in a certain city for different hours of the day, in a typical 24-hour period, with hour 1 corresponding to midnight to 1 a.m.<\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
            Hour<\/th>\nPower<\/th>\nHour<\/th>\nPower<\/th>\n<\/tr>\n<\/thead>\n
            1<\/td>\n28<\/td>\n13<\/td>\n48<\/td>\n<\/tr>\n
            2<\/td>\n25<\/td>\n14<\/td>\n49<\/td>\n<\/tr>\n
            3<\/td>\n24<\/td>\n15<\/td>\n49<\/td>\n<\/tr>\n
            4<\/td>\n23<\/td>\n16<\/td>\n50<\/td>\n<\/tr>\n
            5<\/td>\n24<\/td>\n17<\/td>\n50<\/td>\n<\/tr>\n
            6<\/td>\n27<\/td>\n18<\/td>\n50<\/td>\n<\/tr>\n
            7<\/td>\n29<\/td>\n19<\/td>\n46<\/td>\n<\/tr>\n
            8<\/td>\n32<\/td>\n20<\/td>\n43<\/td>\n<\/tr>\n
            9<\/td>\n34<\/td>\n21<\/td>\n42<\/td>\n<\/tr>\n
            10<\/td>\n39<\/td>\n22<\/td>\n40<\/td>\n<\/tr>\n
            11<\/td>\n42<\/td>\n23<\/td>\n37<\/td>\n<\/tr>\n
            12<\/td>\n46<\/td>\n24<\/td>\n34<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

            Find the total amount of power in gigawatt-hours (gW-h) consumed by the city in a typical 24-hour period.<\/p>\n

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

            The total daily power consumption is estimated as the sum of the hourly power rates, or 911 gW-h.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

            \n
            \n

            28.\u00a0<\/strong>The average residential electrical power use (in hundreds of watts) per hour is given in the following table.<\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
            Hour<\/th>\nPower<\/th>\nHour<\/th>\nPower<\/th>\n<\/tr>\n<\/thead>\n
            1<\/td>\n8<\/td>\n13<\/td>\n12<\/td>\n<\/tr>\n
            2<\/td>\n6<\/td>\n14<\/td>\n13<\/td>\n<\/tr>\n
            3<\/td>\n5<\/td>\n15<\/td>\n14<\/td>\n<\/tr>\n
            4<\/td>\n4<\/td>\n16<\/td>\n15<\/td>\n<\/tr>\n
            5<\/td>\n5<\/td>\n17<\/td>\n17<\/td>\n<\/tr>\n
            6<\/td>\n6<\/td>\n18<\/td>\n19<\/td>\n<\/tr>\n
            7<\/td>\n7<\/td>\n19<\/td>\n18<\/td>\n<\/tr>\n
            8<\/td>\n8<\/td>\n20<\/td>\n17<\/td>\n<\/tr>\n
            9<\/td>\n9<\/td>\n21<\/td>\n16<\/td>\n<\/tr>\n
            10<\/td>\n10<\/td>\n22<\/td>\n16<\/td>\n<\/tr>\n
            11<\/td>\n10<\/td>\n23<\/td>\n13<\/td>\n<\/tr>\n
            12<\/td>\n11<\/td>\n24<\/td>\n11<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
              \n
            1. Compute the average total energy used in a day in kilowatt-hours (kWh).<\/li>\n
            2. If a ton of coal generates 1842 kWh, how long does it take for an average residence to burn a ton of coal?<\/li>\n
            3. Explain why the data might fit a plot of the form [latex]p(t)=11.5-7.5 \\sin (\\frac{\\pi t}{12}).[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n
              \n
              \n

              29.\u00a0<\/strong>The data in the following table are used to estimate the average power output produced by Peter Sagan for each of the last 18 sec of Stage 1 of the 2012 Tour de France<\/span>.<\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n\n
              Average Power OutputSource<\/em>: sportsexercisengineering.com<\/caption>\n
              Second<\/th>\nWatts<\/th>\nSecond<\/th>\nWatts<\/th>\n<\/tr>\n<\/thead>\n
              1<\/td>\n600<\/td>\n10<\/td>\n1200<\/td>\n<\/tr>\n
              2<\/td>\n500<\/td>\n11<\/td>\n1170<\/td>\n<\/tr>\n
              3<\/td>\n575<\/td>\n12<\/td>\n1125<\/td>\n<\/tr>\n
              4<\/td>\n1050<\/td>\n13<\/td>\n1100<\/td>\n<\/tr>\n
              5<\/td>\n925<\/td>\n14<\/td>\n1075<\/td>\n<\/tr>\n
              6<\/td>\n950<\/td>\n15<\/td>\n1000<\/td>\n<\/tr>\n
              7<\/td>\n1050<\/td>\n16<\/td>\n950<\/td>\n<\/tr>\n
              8<\/td>\n950<\/td>\n17<\/td>\n900<\/td>\n<\/tr>\n
              9<\/td>\n1100<\/td>\n18<\/td>\n780<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

              Estimate the net energy used in kilojoules (kJ), noting that 1W = 1 J\/s, and the average power output by Sagan during this time interval.<\/p>\n

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

              17 kJ<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

              \n
              \n

              30. <\/strong>The data in the following table are used to estimate the average power output produced by Peter Sagan for each 15-min interval of Stage 1 of the 2012 Tour de France.<\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
              Average Power OutputSource<\/em>: sportsexercisengineering.com<\/caption>\n
              Minutes<\/th>\nWatts<\/th>\nMinutes<\/th>\nWatts<\/th>\n<\/tr>\n<\/thead>\n
              15<\/td>\n200<\/td>\n165<\/td>\n170<\/td>\n<\/tr>\n
              30<\/td>\n180<\/td>\n180<\/td>\n220<\/td>\n<\/tr>\n
              45<\/td>\n190<\/td>\n195<\/td>\n140<\/td>\n<\/tr>\n
              60<\/td>\n230<\/td>\n210<\/td>\n225<\/td>\n<\/tr>\n
              75<\/td>\n240<\/td>\n225<\/td>\n170<\/td>\n<\/tr>\n
              90<\/td>\n210<\/td>\n240<\/td>\n210<\/td>\n<\/tr>\n
              105<\/td>\n210<\/td>\n255<\/td>\n200<\/td>\n<\/tr>\n
              120<\/td>\n220<\/td>\n270<\/td>\n220<\/td>\n<\/tr>\n
              135<\/td>\n210<\/td>\n285<\/td>\n250<\/td>\n<\/tr>\n
              150<\/td>\n150<\/td>\n300<\/td>\n400<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

              Estimate the net energy used in kilojoules, noting that 1W = 1 J\/s.<\/p>\n<\/div>\n<\/div>\n

              \n
              \n

              31.\u00a0<\/strong>The distribution of incomes as of 2012 in the United States in $5000 increments is given in the following table. The [latex]k[\/latex]th row denotes the percentage of households with incomes between [latex]$5000xk[\/latex] and [latex]5000xk+4999.[\/latex] The row [latex]k=40[\/latex] contains all households with income between $200,000 and $250,000 and [latex]k=41[\/latex] accounts for all households with income exceeding $250,000.<\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
              Income DistributionsSource<\/em>: http:\/\/www.census.gov\/prod\/2013pubs\/p60-245.pdf<\/caption>\n
              0<\/td>\n3.5<\/td>\n21<\/td>\n1.5<\/td>\n<\/tr>\n
              1<\/td>\n4.1<\/td>\n22<\/td>\n1.4<\/td>\n<\/tr>\n
              2<\/td>\n5.9<\/td>\n23<\/td>\n1.3<\/td>\n<\/tr>\n
              3<\/td>\n5.7<\/td>\n24<\/td>\n1.3<\/td>\n<\/tr>\n
              4<\/td>\n5.9<\/td>\n25<\/td>\n1.1<\/td>\n<\/tr>\n
              5<\/td>\n5.4<\/td>\n26<\/td>\n1.0<\/td>\n<\/tr>\n
              6<\/td>\n5.5<\/td>\n27<\/td>\n0.75<\/td>\n<\/tr>\n
              7<\/td>\n5.1<\/td>\n28<\/td>\n0.8<\/td>\n<\/tr>\n
              8<\/td>\n4.8<\/td>\n29<\/td>\n1.0<\/td>\n<\/tr>\n
              9<\/td>\n4.1<\/td>\n30<\/td>\n0.6<\/td>\n<\/tr>\n
              10<\/td>\n4.3<\/td>\n31<\/td>\n0.6<\/td>\n<\/tr>\n
              11<\/td>\n3.5<\/td>\n32<\/td>\n0.5<\/td>\n<\/tr>\n
              12<\/td>\n3.7<\/td>\n33<\/td>\n0.5<\/td>\n<\/tr>\n
              13<\/td>\n3.2<\/td>\n34<\/td>\n0.4<\/td>\n<\/tr>\n
              14<\/td>\n3.0<\/td>\n35<\/td>\n0.3<\/td>\n<\/tr>\n
              15<\/td>\n2.8<\/td>\n36<\/td>\n0.3<\/td>\n<\/tr>\n
              16<\/td>\n2.5<\/td>\n37<\/td>\n0.3<\/td>\n<\/tr>\n
              17<\/td>\n2.2<\/td>\n38<\/td>\n0.2<\/td>\n<\/tr>\n
              18<\/td>\n2.2<\/td>\n39<\/td>\n1.8<\/td>\n<\/tr>\n
              19<\/td>\n1.8<\/td>\n40<\/td>\n2.3<\/td>\n<\/tr>\n
              20<\/td>\n2.1<\/td>\n41<\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
                \n
              1. Estimate the percentage of U.S. households in 2012 with incomes less than $55,000.<\/li>\n
              2. What percentage of households had incomes exceeding $85,000?<\/li>\n
              3. Plot the data and try to fit its shape to that of a graph of the form [latex]a(x+c){e}^{\\text{\u2212}b(x+e)}[\/latex] for suitable [latex]a,b,c.[\/latex]<\/li>\n<\/ol>\n
                \n
                Show Solution<\/span><\/p>\n
                \n

                a. 54.3%; b. 27.00%; c. The curve in the following plot is [latex]2.35(t+3){e}^{-0.15(t+3)}.[\/latex]<\/p>\n

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

                \n
                \n

                32.\u00a0<\/strong>Newton\u2019s law of gravity states that the gravitational force exerted by an object of mass M<\/em> and one of mass [latex]m[\/latex] with centers that are separated by a distance [latex]r[\/latex] is [latex]F=G\\frac{mM}{{r}^{2}},[\/latex] with G<\/em> an empirical constant [latex]G=6.67x{10}^{-11}{m}^{3}\\text{\/}(kg\u00b7{s}^{2}).[\/latex] The work done by a variable force over an interval [latex]\\left[a,b\\right][\/latex] is defined as [latex]W={\\int }_{a}^{b}F(x)dx.[\/latex] If Earth has mass [latex]5.97219\u00d7{10}^{24}[\/latex] and radius 6371 km, compute the amount of work to elevate a polar weather satellite of mass 1400 kg to its orbiting altitude of 850 km above Earth.<\/p>\n<\/div>\n<\/div>\n

                \n
                \n

                33.\u00a0<\/strong>For a given motor vehicle, the maximum achievable deceleration<\/span> from braking is approximately 7 m\/sec2<\/sup> on dry concrete. On wet asphalt, it is approximately 2.5 m\/sec2<\/sup>. Given that 1 mph corresponds to 0.447 m\/sec, find the total distance that a car travels in meters on dry concrete after the brakes are applied until it comes to a complete stop if the initial velocity is 67 mph (30 m\/sec) or if the initial braking velocity is 56 mph (25 m\/sec). Find the corresponding distances if the surface is slippery wet asphalt.<\/p>\n

                Show Solution<\/span><\/p>\n
                In dry conditions, with initial velocity [latex]{v}_{0}=30[\/latex] m\/s, [latex]D=64.3[\/latex] and, if [latex]{v}_{0}=25,D=44.64.[\/latex] In wet conditions, if [latex]{v}_{0}=30,[\/latex] and [latex]D=180[\/latex] and if [latex]{v}_{0}=25,D=125.[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n
                \n
                \n

                34.\u00a0<\/strong>John is a 25-year old man who weighs 160 lb. He burns [latex]500-50t[\/latex] calories\/hr while riding his bike for [latex]t[\/latex] hours. If an oatmeal cookie has 55 cal and John eats 4[latex]t[\/latex] cookies during the [latex]t[\/latex]th hour, how many net calories has he lost after 3 hours riding his bike?<\/p>\n<\/div>\n<\/div>\n

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                35.\u00a0<\/strong>Sandra is a 25-year old woman who weighs 120 lb. She burns [latex]300-50t[\/latex] cal\/hr while walking on her treadmill. Her caloric intake from drinking Gatorade is 100[latex]t[\/latex] calories during the [latex]t[\/latex]th hour. What is her net decrease in calories after walking for 3 hours?<\/p>\n

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

                225 cal<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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                \n

                36.\u00a0<\/strong>A motor vehicle has a maximum efficiency of 33 mpg at a cruising speed of 40 mph. The efficiency drops at a rate of 0.1 mpg\/mph between 40 mph and 50 mph, and at a rate of 0.4 mpg\/mph between 50 mph and 80 mph. What is the efficiency in miles per gallon if the car is cruising at 50 mph? What is the efficiency in miles per gallon if the car is cruising at 80 mph? If gasoline costs $3.50\/gal, what is the cost of fuel to drive 50 mi at 40 mph, at 50 mph, and at 80 mph?<\/p>\n<\/div>\n<\/div>\n

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                \n

                37.\u00a0<\/strong>Although some engines are more efficient at given a horsepower than others, on average, fuel efficiency decreases with horsepower at a rate of [latex]1\\text{\/}25[\/latex] mpg\/horsepower. If a typical 50-horsepower engine has an average fuel efficiency of 32 mpg, what is the average fuel efficiency of an engine with the following horsepower: 150, 300, 450?<\/p>\n

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

                [latex]E(150)=28,E(300)=22,E(450)=16[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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                38. [T]<\/strong> The following table lists the 2013 schedule of federal income tax<\/span> versus taxable income.<\/p>\n\n\n\n\n\n\n\n\n\n\n\n
                Federal Income Tax Versus Taxable IncomeSource<\/em>: http:\/\/www.irs.gov\/pub\/irs-prior\/i1040tt–2013.pdf.<\/caption>\n
                Taxable Income Range<\/th>\nThe Tax Is \u2026<\/th>\n\u2026 Of the Amount Over<\/th>\n<\/tr>\n<\/thead>\n
                $0\u2013$8925<\/td>\n10%<\/td>\n$0<\/td>\n<\/tr>\n
                $8925\u2013$36,250<\/td>\n$892.50 + 15%<\/td>\n$8925<\/td>\n<\/tr>\n
                $36,250\u2013$87,850<\/td>\n$4,991.25 + 25%<\/td>\n$36,250<\/td>\n<\/tr>\n
                $87,850\u2013$183,250<\/td>\n$17,891.25 + 28%<\/td>\n$87,850<\/td>\n<\/tr>\n
                $183,250\u2013$398,350<\/td>\n$44,603.25 + 33%<\/td>\n$183,250<\/td>\n<\/tr>\n
                $398,350\u2013$400,000<\/td>\n$115,586.25 + 35%<\/td>\n$398,350<\/td>\n<\/tr>\n
                > $400,000<\/td>\n$116,163.75 + 39.6%<\/td>\n$400,000<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                Suppose that Steve just received a $10,000 raise. How much of this raise is left after federal taxes if Steve\u2019s salary before receiving the raise was $40,000? If it was $90,000? If it was $385,000?<\/p>\n<\/div>\n<\/div>\n

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                39. [T]<\/strong> The following table provides hypothetical data regarding the level of service for a certain highway.<\/p>\n\n\n\n\n\n\n\n\n\n\n
                Highway Speed Range (mph)<\/th>\nVehicles per Hour per Lane<\/th>\nDensity Range (vehicles\/mi)<\/th>\n<\/tr>\n<\/thead>\n
                > 60<\/td>\n< 600<\/td>\n< 10<\/td>\n<\/tr>\n
                60\u201357<\/td>\n600\u20131000<\/td>\n10\u201320<\/td>\n<\/tr>\n
                57\u201354<\/td>\n1000\u20131500<\/td>\n20\u201330<\/td>\n<\/tr>\n
                54\u201346<\/td>\n1500\u20131900<\/td>\n30\u201345<\/td>\n<\/tr>\n
                46\u201330<\/td>\n1900\u2013<\/strong>2100<\/td>\n45\u201370<\/td>\n<\/tr>\n
                <30<\/td>\nUnstable<\/td>\n70\u2013200<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
                  \n
                1. Plot vehicles per hour per lane on the [latex]x[\/latex]-axis and highway speed on the [latex]y[\/latex]-axis.<\/li>\n
                2. Compute the average decrease in speed (in miles per hour) per unit increase in congestion (vehicles per hour per lane) as the latter increases from 600 to 1000, from 1000 to 1500, and from 1500 to 2100. Does the decrease in miles per hour depend linearly on the increase in vehicles per hour per lane?<\/li>\n
                3. Plot minutes per mile (60 times the reciprocal of miles per hour) as a function of vehicles per hour per lane. Is this function linear?<\/li>\n<\/ol>\n
                  Show Solution<\/span><\/p>\n
                  \n

                  a.<\/p>\n

                  \"A<\/span>
                  \nb. Between 600 and 1000 the average decrease in vehicles per hour per lane is \u22120.0075. Between 1000 and 1500 it is \u22120.006 per vehicles per hour per lane, and between 1500 and 2100 it is \u22120.04 vehicles per hour per lane. c.
                  \n\"A<\/span>
                  \nThe graph is nonlinear, with minutes per mile increasing dramatically as vehicles per hour per lane reach 2000.<\/div>\n<\/div>\n<\/div>\n<\/div>\n

                  For the next two exercises use the data in the following table, which displays bald eagle<\/span> populations from 1963 to 2000 in the continental United States.<\/p>\n\n\n\n\n\n\n\n\n\n\n\n
                  Population of Breeding Bald Eagle PairsSource<\/em>: http:\/\/www.fws.gov\/Midwest\/eagle\/population\/chtofprs.html.<\/caption>\n
                  Year<\/th>\nPopulation of Breeding Pairs of Bald Eagles<\/th>\n<\/tr>\n<\/thead>\n
                  1963<\/td>\n487<\/td>\n<\/tr>\n
                  1974<\/td>\n791<\/td>\n<\/tr>\n
                  1981<\/td>\n1188<\/td>\n<\/tr>\n
                  1986<\/td>\n1875<\/td>\n<\/tr>\n
                  1992<\/td>\n3749<\/td>\n<\/tr>\n
                  1996<\/td>\n5094<\/td>\n<\/tr>\n
                  2000<\/td>\n6471<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
                  \n
                  \n

                  40. [T]<\/strong> The graph below plots the quadratic [latex]p(t)=6.48{t}^{2}-80.31t+585.69[\/latex] against the data in preceding table, normalized so that [latex]t=0[\/latex] corresponds to 1963. Estimate the average number of bald eagles per year present for the 37 years by computing the average value of [latex]p[\/latex] over [latex]\\left[0,37\\right].[\/latex]<\/p>\n

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

                  \n
                  \n

                  41. [T]<\/strong> The graph below plots the cubic [latex]p(t)=0.07{t}^{3}+2.42{t}^{2}-25.63t+521.23[\/latex] against the data in the preceding table, normalized so that [latex]t=0[\/latex] corresponds to 1963. Estimate the average number of bald eagles per year present for the 37 years by computing the average value of [latex]p[\/latex] over [latex]\\left[0,37\\right].[\/latex]<\/p>\n

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

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

                  [latex]\\frac{1}{37}{\\int }_{0}^{37}p(t)dt=\\frac{0.07{(37)}^{3}}{4}+\\frac{2.42{(37)}^{2}}{3}-\\frac{25.63(37)}{2}+521.23\\approx 2037[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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                  42. [T]<\/strong> Suppose you go on a road trip and record your speed at every half hour, as compiled in the following table. The best quadratic fit to the data is [latex]q(t)=5{x}^{2}-11x+49\\text{,}[\/latex] shown in the accompanying graph. Integrate [latex]q[\/latex] to estimate the total distance driven over the 3 hours.<\/p>\n\n\n\n\n\n\n\n\n
                  Time (hr)<\/th>\nSpeed (mph)<\/th>\n<\/tr>\n<\/thead>\n
                  0 (start)<\/td>\n50<\/td>\n<\/tr>\n
                  1<\/td>\n40<\/td>\n<\/tr>\n
                  2<\/td>\n50<\/td>\n<\/tr>\n
                  3<\/td>\n60<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

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

                  As a car accelerates, it does not accelerate at a constant rate; rather, the acceleration is variable. For the following exercises, use the following table, which contains the acceleration measured at every second as a driver merges onto a freeway.<\/p>\n\n\n\n\n\n\n\n\n\n
                  Time (sec)<\/th>\nAcceleration (mph\/sec)<\/th>\n<\/tr>\n<\/thead>\n
                  1<\/td>\n11.2<\/td>\n<\/tr>\n
                  2<\/td>\n10.6<\/td>\n<\/tr>\n
                  3<\/td>\n8.1<\/td>\n<\/tr>\n
                  4<\/td>\n5.4<\/td>\n<\/tr>\n
                  5<\/td>\n0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
                  \n
                  \n

                  43. [T]<\/strong> The accompanying graph plots the best quadratic fit, [latex]a(t)=-0.70{t}^{2}+1.44t+10.44,[\/latex] to the data from the preceding table. Compute the average value of [latex]a(t)[\/latex] to estimate the average acceleration between [latex]t=0[\/latex] and [latex]t=5.[\/latex]<\/p>\n

                  \"A<\/p>\n

                  Show Solution<\/span><\/p>\n
                  Average acceleration is [latex]A=\\frac{1}{5}{\\int }_{0}^{5}a(t)dt=-\\frac{0.7({5}^{2})}{3}+\\frac{1.44(5)}{2}+10.44\\approx 8.2[\/latex] mph\/s<\/div>\n<\/div>\n<\/div>\n<\/div>\n
                  \n
                  \n

                  44. [T]<\/strong> Using your acceleration equation from the previous exercise, find the corresponding velocity equation. Assuming the final velocity is 0 mph, find the velocity at time [latex]t=0.[\/latex]<\/p>\n<\/div>\n<\/div>\n

                  \n
                  \n

                  45. [T]<\/strong> Using your velocity equation from the previous exercise, find the corresponding distance equation, assuming your initial distance is 0 mi. How far did you travel while you accelerated your car? (Hint:<\/em> You will need to convert time units.)<\/p>\n

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

                  [latex]d(t)={\\int }_{0}^{1}|v(t)|dt={\\int }_{0}^{t}(\\frac{7}{30}{t}^{3}-0.72{t}^{2}-10.44t+41.033)dt=\\frac{7}{120}{t}^{4}-0.24{t}^{3}-5.22{t}^{3}+41.033t.[\/latex] Then, [latex]d(5)\\approx 81.12[\/latex] mph [latex]\u00d7 \\sec \\approx 119[\/latex] feet.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

                  \n
                  \n

                  46. [T]<\/strong> The number of hamburgers sold at a restaurant throughout the day is given in the following table, with the accompanying graph plotting the best cubic fit to the data, [latex]b(t)=0.12{t}^{3}-2.13{t}^{3}+12.13t+3.91,[\/latex] with [latex]t=0[\/latex] corresponding to 9 a.m. and [latex]t=12[\/latex] corresponding to 9 p.m. Compute the average value of [latex]b(t)[\/latex] to estimate the average number of hamburgers sold per hour.<\/p>\n\n\n\n\n\n\n\n\n\n
                  Hours Past Midnight<\/th>\nNo. of Burgers Sold<\/th>\n<\/tr>\n<\/thead>\n
                  9<\/td>\n3<\/td>\n<\/tr>\n
                  12<\/td>\n28<\/td>\n<\/tr>\n
                  15<\/td>\n20<\/td>\n<\/tr>\n
                  18<\/td>\n30<\/td>\n<\/tr>\n
                  21<\/td>\n45<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

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

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                  47. [T]\u00a0\u00a0<\/strong>An athlete runs by a motion detector, which records her speed, as displayed in the following table. The best linear fit to this data, [latex]\\ell (t)=-0.068t+5.14\\text{,}[\/latex] is shown in the accompanying graph. Use the average value of [latex]\\ell (t)[\/latex] between [latex]t=0[\/latex] and [latex]t=40[\/latex] to estimate the runner\u2019s average speed.<\/p>\n\n\n\n\n\n\n\n\n\n
                  Minutes<\/th>\nSpeed (m\/sec)<\/th>\n<\/tr>\n<\/thead>\n
                  0<\/td>\n5<\/td>\n<\/tr>\n
                  10<\/td>\n4.8<\/td>\n<\/tr>\n
                  20<\/td>\n3.6<\/td>\n<\/tr>\n
                  30<\/td>\n3.0<\/td>\n<\/tr>\n
                  40<\/td>\n2.5<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

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

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

                  [latex]\\frac{1}{40}{\\int }_{0}^{40}(-0.068t+5.14)dt=-\\frac{0.068(40)}{2}+5.14=3.78[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t

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