MA123, Chapter 8: Idea of the Integral (pp. 155-187, Go ot ma n)Chapter Goals:• Understand the relationship between the area under a curve and thedefinite integral.• Understand the relationship between velocity (speed), distance and thedefinite integral.• Estimate the value of a definite integral.• Understand the summation, or Σ, notation.• Understand the formal d efinition of the definite integral.Assignments:Assignment 18 Assignment 19◮ The basic idea: The first two problems are easy to solve as certain “problem ingredients” are constant.Example 1 (Easy area problem):Find the area of the region in the x y-plane bound ed above by the graph of the function f (x) = 2, below by the x-axis,on the left by the line x = 1, and on th e right by the line x = 5.xy0 1 52Example 2 (Easy distance traveled problem):Suppose a car is traveling due east at a constantvelocity of 55 miles per hour. How far does th e car travel between noon an d 2:00 pm?General philosophy:By means of the integral, p roblems similar to the previous on es can be solved whenthe ingredients of the problem are variable. In this Chapter, we learn how to estimate a solution to these morecomplex problems. The key idea is to notice that the value of the function does not vary very much over asmall interval, and so it is approximately constant over a small interval. By the end of Chapter 9 we will beable to solve these problems exactly, an d by the end of Chapter 10 we will be able to solve them both exactlyand easily.Example 3:Estimate the area under the graph of y = x2+12x for x between0 and 2 in two different ways:(a) Subdivide the interval [0, 2] into four equal sub intervals and use the leftendpoint of each subinterval as “sample point”.(b) Subdivide the interval [0, 2] into four equal subintervals and use the rightendpoint of each subinterval as “sample point”.Find the difference b etween the two estimates (right en dpoint estimate m inusleft endpoint estimate).xy0 1 20.51.535xy01 20.51.53585Example 4: Estimate the area under th e graph of y = 3xfor x between0 and 2. Use a partition that consists of four equal subintervals of [0, 2] an duse the left endpoint of each subinterval as a s ample point.xy0 1 211.73235.1969Note: In the previous two examples we systematically ch ose the value of the function at one of the endpointsof each subinterval. However, since the guiding idea is that we are assuming that the values of the functionover a small s ubinterval do not change by very much, then we could take the value of the function at anypointof the subinterval as a good sample or representative value of the function. We could also have chosen smallsubintervals of different lengths. However, we are trying to establish a systematic procedure that works well ingeneral.Getting bette r estimates:We can only expect the previous answers to be approximations of the correct answers. This is because thevalues of the function do change on each subinterval, even though they do not change by much.If, however, we replace the subintervals we used by “smaller” subintervals we can reasonably expect the valuesof the function to vary by much less on each thinner subinterval. Thus, we can reasonably expect that the areaof each thinner vertical strip under the grap h of the function to be more accurately approximated by the areaof these thinner rectangles. Then if we add up the areas of all these th inner rectangles, we should get a muchmore accurate estimate for the area of the original region.Here is Example 3(b), revisited:xy01 2y = x2+12x on [0, 2]n = 4 equal sub intervalsArea ≈ 5xy01 2y = x2+12x on [0, 2]n = 8 equal sub intervalsArea ≈ 4.3125We will see later that the exact value of the area under consideration in Example 3 is113≈ 3.66.86Example 5: Estimate the area of the ellipse given by the equation4x2+ y2= 49as follows : The area of the ellipse is 4 times the area of the part of theellipse in the first quadrant (x and y positive). Estimate the area of theellipse in the first quadrant by solving for y in terms of x. Estimate thearea under the graph of y by dividin g the interval [0, 3.5] into four equalsubintervals and using the left endpoint of each subinterval.xy0The area of the ellipse (using the above method) is approximatelyTrapezoids versus rectangles:We could use trapezoids instead of rectangles to obtain better estimates,even though the calculations get a little bit more complicated. This willoccur in some of the latter examples. We recall that the area of a trapezoidisArea of a tr apezoid =(h1+ h2) ·b2.h1h2bh2h1bExample 6: A train travels in a straight westward direction along atrack. The velocity of th e train varies, but it is measured at regular timeintervals of 1/10 hour. The measurements for the first half hour aretime0 0.1 0.2 0.3 0.4 0.5velocity 0 10 15 18 20 25We will see later that th e total distance traveled by the train is equal tothe area underneath the graph of the velocity function and lying abovethe t-axis. Compute the total distance traveled by the train during thefirst half hour by assuming the velocity is a lin ear function of t on thesubintervals. (The velocity in the table is given in miles per hour.)tv010151820250.1 0.2 0.3 0.4 0.587Example 7: Estimate the area under the graph of y =1xfor x between 1 and 31 in two different ways:(a) Subdivide the interval [1, 31] into 30 equal subintervals and use the left endpoint of each su binterval assample point.(b) Subdivide the interval [1, 31] into 30 equ al subintervals and use the right endpoint of each subinterval assample point.Find the difference between the two estimates (left endpoint estimate minus right endpoint estimate).◮ Sigma (Σ) nota tion:In appr oximating areas we have encountered sums with many terms. A convenientway of wr iting such sums uses the Greek letter Σ (which corresponds to our capital S) and is called sigmanotation. More precisely, if a1, a2, . . . , anare real numbers we denote the suma1+ a2+ ··· + anby using th e notationnXk=1ak.The integer k is called an index or counter and takes on (in this case) the values 1, 2, . . . , n.For example,6Xk=1k2= 12+ 22+ 32+ 42+ 52+ 62= 1 + 4 + 9 + 16 + 25 + 36 = 91whereas6Xk=3k2= 32+ 42+ 52+ 62= 9 + 16 + 25 + 36 = 86.Example 8:Evaluate the sum5Xk=1(2k − 1).88Example 9: Evaluate the sum6Xk=2(6k3+ 3).Example 10:Evaluate the sum5Xk=1(3k2+ k).Example 11:Evaluate the sum112Xk=175.Example 12:Evaluate the sum273Xk=1523.89The idea we have us ed so far is to break