ࡱ>  xy `bjbj {{M `"v"KKKKKK86LJN<KrP p[p[([^-\E`w222222$#VK<-V""[^aRkX"[K^wwYj@ ^`+s?Aqc0y[Ѿ[[KVV[ &!: Definite Integrals As an area under the curve In the previous chapter we found how to take an antiderivative and investigated the indefinite integral. In this chapter the connection between antiderivatives and definite integrals is established as we try to solve one of the most famous problems in mathematics, finding the area under a given curve. 9.1 Approximating Area Under a Curve When it comes to finding the area of basic geometric shapes such as circles, squares, rectangles, triangles, and trapezoids, we can rely on geometric formulas to calculate the area. ( Example 9.1 Find the area of the shaded regions. a. b.  ( Solution a. The shaded region is half of a circle with a radius of 3. Thus, we will use the formula for the area of a circle ( EMBED Equation.DSMT4 ), multiply it by 0.5 and use EMBED Equation.DSMT4 .  EMBED Equation.DSMT4  b. The shaded region is made up of a rectangle  EMBED Equation.DSMT4  and a triangle  EMBED Equation.DSMT4 . If we add the areas of the two geometric shapes, we will have found the area of the shaded region. ( Figure 9.1 Graph for Example 9.1a  EMBED Equation.DSMT4  The area of the trapezoid is therefore 7.5 square units. ( Finding the area between the x-axis and a curve  EMBED Equation.DSMT4  on a given interval is a bit more challenging if the region formed is not a basic geometric shape. For example, the area under the curve  EMBED Equation.DSMT4  on [0 ,4] forms the shape shown in Figure 9.2. ( Figure 9.2 Graph of  EMBED Equation.DSMT4 .  We can see from the figure that the area between the x-axis and  EMBED Equation.DSMT4  is not a shape that has a familiar formula for finding the area. When this occurs, we use rectangles to approximate the area of the region. If we draw four rectangles, as seen in Figure 9.3, we can sum up the area of the rectangles (R1 + R2 + R3 + R4) and obtain an approximation of the area under the curve. ( Figure 9.3 Graph of  EMBED Equation.DSMT4  with 4 right endpointsH.  We will, however, find an overestimate of the area because the rectangles extend above the curve. Nonetheless, we will have some idea of the area under the curve. To find the area of each rectangle in Figure 9.3 we need to find the base and height of each rectangle. The base of each rectangle, EMBED Equation.DSMT4 , is found by taking the length of the given interval,  EMBED Equation.DSMT4  and dividing it by the number of rectangles constructed,  EMBED Equation.DSMT4 . This leads to the following calculation.  EMBED Equation.DSMT4 . The height of each rectangle is the value of the function from the right end of each interval. Figure 9.4 below shows the dimensions of each rectangle. ( Figure 9.4 Dimensions of R1, R2, R3, and R4.  The sum of the rectangles are found in the table below. The area, A, is the base times the height. Rectangle #Base,  EMBED Equation.DSMT4 Right Endpt,  EMBED Equation.DSMT4 Height,  EMBED Equation.DSMT4  EMBED Equation.DSMT4 R111 EMBED Equation.DSMT4 (1)(2) = 2R212 EMBED Equation.DSMT4 (1)(2.41) = 2.41R313 EMBED Equation.DSMT4 (1)(2.73) = 2.73R414 EMBED Equation.DSMT4 (1)(3) = 3Total Area (Right)10.14 Since this is an overestimate, the area under the curve is less then 10.14 units. We can also approximate the area under the curve using left endpoint rectangles as shown in figure 9.4. This approximation will give us an underestimate because the rectangles do not fill the entire area under the curve. ( Figure 9.5 Graph of  EMBED Equation.DSMT4  with 4 left endpoint rectangles. (rename rectangles!!)  The base of each rectangle is still 1 unit but the height of each rectangle is the value of the function from the left end of each interval. The sum of the rectangles is found in the table below. Rectangle #Base EMBED Equation.DSMT4 Left Endpt, EMBED Equation.DSMT4 Height,  EMBED Equation.DSMT4  EMBED Equation.DSMT4 R110 EMBED Equation.DSMT4 (1)(1) = 1R211 EMBED Equation.DSMT4 (1)(2.41) = 2.41R312 EMBED Equation.DSMT4 (1)(2.73) = 2.73R413 EMBED Equation.DSMT4 (1)(3) = 3Total Area (Left)9.14 Thus, the area must be greater than 9.14 units. In general, we can find an approximation for the area under a continuous curve EMBED Equation.DSMT4  on  EMBED Equation.DSMT4 by drawing n equally spaced right (or left) endpoint rectangles under the curve and then finding the sum of the area of the rectangles. If  EMBED Equation.DSMT4 is the width of each rectangle and  EMBED Equation.DSMT4 an endpoint where  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4 , then the sum of the area of n rectangles is  EMBED Equation.DSMT4  For right endpoint rectangles the sum of the area rectangles can be denoted as H Total Area (Right) =  EMBED Equation.DSMT4  For left endpoint rectangles, the sum of the area of the rectangles can be denoted as Total Area (Left) =  EMBED Equation.DSMT4  ( Example 9.2 Approximate the area under the curve  EMBED Equation.DSMT4  on  EMBED Equation.DSMT4  using a. 4 right endpoint rectangles b. 8 left endpoint rectangles. State if the estimate is an overestimate or an underestimate. ( Solution a. The graph of EMBED Equation.DSMT4  on [0, 2] with the 4 right endpoint rectangles is shown in Figure 9.6. ( Figure 9.6 Graph of 4 right endpoint rectangles  The base of each rectangle is  EMBED Equation.DSMT4  and the height is  EMBED Equation.DSMT4  where  EMBED Equation.DSMT4  is the right endpoint of each interval. The calculation below shows the sum of the areas of the four rectangles.  EMBED Equation.DSMT4  Thus, the area under the curve is less than 7.75 square units. b. The graph of EMBED Equation.DSMT4  on [0, 2] with the 8 left endpoint rectangles is shown in Figure 9.7. ( Figure 9.7 Graph of 8 left endpoint rectangles. (rename rectangles!!)  The base of each rectangle is  EMBED Equation.DSMT4  and the height is  EMBED Equation.DSMT4  where  EMBED Equation.DSMT4  is the left endpoint of each interval. The table below shows the sum of the areas of the eight rectangles.  EMBED Equation.DSMT4  Since these rectangles all lie below the curve, the estimate for the area under the curve is an underestimate. ( There are numerous methods of using rectangles to approximate the area under a curve. A few of the other methods are shown in Figure 9.8 below. ( Figure 9.8 Other methods to approximate  EMBED Equation.DSMT4 . (a) Lower Sum Method all rectangles lie below the curve.(b) Midpoint Method the midpoint of all rectangles are touching the curve.(c) Upper Sum Method all rectangles lie above the curve.9.2 Definite Integrals and the Fundamental Theorem of Calculus The methods used in the previous section allow us to obtain a good approximation of the area under a curve, but can we make this approximation better? If we take thinner and thinner rectangles, we can make the approximation of the area under the curve more accurate. In Example 9.2 we found an approximation of 6.1875 square units for the area under the curve using 8 left endpoint rectangles. If we would have used 4 left endpoint rectangles our approximation would have been 5.75 square units. The approximation with 8 rectangles was more accurate simply because more rectangles were used. Compare the rectangles in Figure 9.9. ( Figure 9.8 Graphs of  EMBED Equation.DSMT4  with 8 and 16 right endpoint rectangles.  It appears that the amount of excess area made by the 16 rectangles is considerably less than the excess area made by the 8 rectangles. One can imagine that the approximation would be even better if we could fit 100 rectangles or even 1000 rectangles under the curve. What if we had an infinite number of rectangles drawn under the curve? As one might hypothesize, the sum of an infinite number of rectangles does accurately find the area under a curve, and we represent the area under a curve using the definite integral. The Definite Integral For a continuous function f on the interval  EMBED Equation.DSMT4 H let  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4 be the right endpoint of the n intervals. Then the definite integral of f is  EMBED Equation.DSMT4  Some useful properties of definite integrals are listed in Table 9.1. ( Table 9.1 Properties of Definite Integrals  EMBED Equation.DSMT4  Using the definition of the definite integral the area in Figure 9.9 is represented as  EMBED Equation.DSMT4  ( Example 9.3 Represent the area of the shaded regions from Example 9.1 as definite integrals. ( Solution a.  EMBED Equation.DSMT4  b.  EMBED Equation.DSMT4  ( From Example 9.1 (b) we found the area to be exactly  EMBED Equation.DSMT4  and from Example 9.3 (b) we found that the area can be represented as a definite integral. We can put the two of these together and conclude  EMBED Equation.DSMT4 . Next we can connect the notion of an antiderivative and a definite integral. Take the antiderivative of  EMBED Equation.DSMT4 ,  EMBED Equation.DSMT4  Note that  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4  is  EMBED Equation.DSMT4  now find  EMBED Equation.DSMT4 ,  EMBED Equation.DSMT4 . We can see that finding the antiderivative  EMBED Equation.DSMT4  of a function and then evaluating  EMBED Equation.DSMT4  gives the exact area under the curve. This process is an important theorem in calculus known as the Fundamental Theorem of Calculus. The Fundamental Theorem of Calculus If f is a continuous function defined on a closed interval  EMBED Equation.DSMT4  and F is an antiderivative of f, then  EMBED Equation.DSMT4  ( Example 9.4 Draw a geometric representation of each definite integral and then evaluate the definite integral using the Fundamental Theorem of Calculus. a.  EMBED Equation.DSMT4  b.  EMBED Equation.DSMT4  ( Solution a. The shaded region in the graph below shows the geometric representation. Use the Fundamental Theorem of Calculus to find the value of the definite integral.  EMBED Equation.DSMT4  b. The shaded region in the graph below shows the geometric representation. Use the Fundamental Theorem of Calculus to find the value of the definite integral.  EMBED Equation.DSMT4  ( Example 9.5 Graph EMBED Equation.DSMT4  and use the graph to find  EMBED Equation.DSMT4 . ( Solution The graph of  EMBED Equation.DSMT4  is shown below. To find  EMBED Equation.DSMT4 , we need to write an integral that represents R1 and another to represent R2. This is necessary because R1 and R2 are bounded by different functions.  EMBED Equation.DSMT4  ( All of the integrals we have considered thus far have been positive. That is the graphs of the functions lied strictly above the x axis. The next example demonstrates what happens when a shaded region lies strictly below the x axis. ( Example 9.6 Draw a geometric representation of  EMBED Equation.DSMT4 then evaluate the definite integral using the Fundamental Theorem of Calculus. ( Solution The shaded region in the graph below shows the geometric representation.  EMBED Equation.DSMT4  This definite integral is negative because the shaded area lies below the x-axis. ( When a definite integral represents a portion of the graph that lies above as well as below the x-axis we can calculate two types of areas, gross area and net area. The gross area is the total amount of area that lies between the curve and the x-axis while the net area calculates how much more area lies above or below x-axis. Figure 9.7 shows the different values of the net area. ( Figure 9.?? Net Area. Net area is positive because more area lies above the x-axis.Net area is negative because more area lies below the x-axisNet area is zero because the area above and below the x-axis is the same. ( Example 9.7 Draw a geometric representation of  EMBED Equation.DSMT4  and then calculate the net and gross areas. ( Solution The shaded region in the graph below shows the geometric representation.  To find the gross area we need to evaluate the integral that represents each shaded region.  EMBED Equation.DSMT4  In Example 9.6, we found the area of region 2 to be  EMBED Equation.DSMT4 . Therefore the gross area is  EMBED Equation.DSMT4  The net area is just the sum of the two integrals,  EMBED Equation.DSMT4 . Since the net area is zero, we know there is the same amount of area above the x-axis as there is below the x-axis. Notice that calculating the function over the entire interval is another method of obtaining net area. 9.3 Area Between Two Curves Suppose we are to find the area of the shaded region shown in Figure 9.??. Figure 9.?? The area between f(x) and g(x).  The area under  EMBED Equation.DSMT4  on  EMBED Equation.DSMT4  is shown in Figure 9.?? (a) and the area under  EMBED Equation.DSMT4  on  EMBED Equation.DSMT4  is shown in Figure 9.?? (b). If the area under  EMBED Equation.DSMT4  is taken away from the area under  EMBED Equation.DSMT4  we obtain the area in Figure 9.?? (c) which is the area we were trying to find in Figure 9.??. ( Figure 9.?? Area Between Two Curves area under  EMBED Equation.DSMT4  on  EMBED Equation.DSMT4 area under  EMBED Equation.DSMT4  on  EMBED Equation.DSMT4 area under  EMBED Equation.DSMT4  with area under  EMBED Equation.DSMT4  taken away Thus, we can find the area between two curves if we find the area under the top curve and subtract off the area under the bottom curve. Area Between Two Curves On the closed interval  EMBED Equation.DSMT4 , the area between two continuous functions f(x) and g(x), where  EMBED Equation.DSMT4 , is given by  EMBED Equation.DSMT4  The area between two curves can be remembered as  EMBED Equation.DSMT4  ( Example 9.8 Find the area between  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4  on [0, 2]. ( Solution First lets graph both functions over [0 2].  Since  EMBED Equation.DSMT4  is the top function and  EMBED Equation.DSMT4  is the bottom function, the definite integral, and thus the area between the two curves is  EMBED Equation.DSMT4  Sometimes the two given curves will intersect at one or more points, thus forming an area bounded by the curves as shown in Figure 9.??. Figure 9.??  To find the area bounded by two curves we need to find the limits of integration. We do this by locating the points where the curves intersect. The definite integral for Figure 9.10 is represented by  EMBED Equation.DSMT4  ( Example 9.9 Find the area of the region bounded by  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4 . ( Solution First, we need to graph the two functions on the same coordinate plane.  From the graph we notice that  EMBED Equation.DSMT4 is the top function and  EMBED Equation.DSMT4 is the bottom function. In addition, the points of intersection show that the lower limit of integration is x = 1 and the upper limit of integration is x = 5. Thus, the definite integral is  EMBED Equation.DSMT4  ( Example 9.10 Find the area of the region bounded by  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4 . ( Solution First, we need to graph the two functions on the same coordinate plane.  There are two bounded regions (R1 and R2) produced by these curves. Notice that the top function of R1 is g(x) and the top function of R2 is f(x). Consequently we will need to set up an integral to find the area of R1, another integral to find the area of R2, and then add the results.  EMBED Equation.DSMT4   EMBED Equation.DSMT4  Now adding the two results together we get  EMBED Equation.DSMT4  9.4 Applications of Definite Integrals Consumers and Producers Surplus Suppose you worked all summer and put away $800 to buy a new stereo system for your dorm room. When you went shopping to buy the stereo system you found exactly what you wanted for only $650. Thus, we could say that you saved $150. If we could find all the consumers who were willing to pay over $650 for this stereo system and calculate the total savings of all consumers, we will have found the consumers surplus. Figure 9.11(a) shows the graph of a supply curve,  EMBED Equation.DSMT4 , and a demand XE "Demand"  curve,  EMBED Equation.DSMT4 . The dotted lines represents the equilibrium price,  EMBED Equation.DSMT4 , and the equilibrium quantity  EMBED Equation.DSMT4 ,. The area above the dotted line, but below  EMBED Equation.DSMT4 , would represent the consumers surplus. Figure 9.??  Now lets say you are the producer of the stereo systems and are willing to supply the stereos for $500. If, however, you end up selling the stereos for $650, you have gained $150. The total amount gained over all possible prices is the producers surplus. Figure 9.11 also shows the graph of the producers surplus. If  EMBED Equation.DSMT4 is the demand XE "Demand"  equation,  EMBED Equation.DSMT4  the supply equation, and  EMBED Equation.DSMT4 is the equilibrium point then the consumers surplus is given by  EMBED Equation.DSMT4  the producers surplus is given by  ( Example 9.11 A company has determined that its supply and demand XE "Demand"  equations can be modeled by  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4 where x represents the number of units supplied each week and p is the selling price (in hundreds of dollars) for each unit. Find the consumers and producers surplus. ( Solution First we need to graph the supply and demand XE "Demand"  functions and find the equilibrium point. The equilibrium point is found by setting  EMBED Equation.DSMT4    EMBED Equation.DSMT4  The consumers surplus is  EMBED Equation.DSMT4  So the consumers saved approximately $266.67 per week when the selling price was $500. The producers surplus is   EMBED Equation.DSMT4  So the producers saved approximately $533.33 per week when the selling price was $500. ( Sample Quiz Question 9.1 Find an approximation for the area under  EMBED Equation.DSMT4  on  EMBED Equation.DSMT4  using 4 left endpoint rectangles and 4 right endpoint rectangles. Which is an overestimate and which is an underestimate? Question 9.2 Write a definite integral that represents the shaded area. Question 9.3 Evaluate  EMBED Equation.DSMT4 . Question 9.4 Draw a graph of  EMBED Equation.DSMT4  and then find  EMBED Equation.DSMT4 . Question 9.5 Evaluate  EMBED Equation.DSMT4 . Question 9.6 Calculate the net and gross areas of  EMBED Equation.DSMT4 . Question 9.7 Find the area between  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4  on  EMBED Equation.DSMT4 . Question 9.8 Find area bounded by  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4  Question 9.9 Find the area bounded by  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4 . Question 9.10 A company has determined its demand XE "Demand"  equation can be modeled by  EMBED Equation.DSMT4 and its supply equation can be modeled by  EMBED Equation.DSMT4  where x is the number of units sold per day and p is the selling price in hundreds of dollars. Find the consumers and producers surplus. H The rectangles have been constructed such that the right endpoint,  EMBED Equation.DSMT4 , of the interval touches the curve. Because of this, we call these rectangles right endpoint rectangles H The symbol  EMBED Equation.DSMT4 is the notation for the sum of H Note: a is referred to as the lower limit and b as the upper limit. Together, a and b are known as the limits of integration.     x 3 3 3 x 3 3 x 3 3 2 1 3 x 3 5 1 1 3 1 1 x 3 5 1 1 3 1 1 R1 R2 R3 R4 R1 R2 R3 R4  EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4  EMBED Equation.DSMT4   EMBED Equation.DSMT4  EMBED Equation.DSMT4   EMBED Equation.DSMT4  EMBED Equation.DSMT4   EMBED Equation.DSMT4  EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4  x 3 5 1 1 3 1 1 R0 R1 R2 R3 x 2 1 6 2 1 R1 R2 R3 R4 x 2 1 6 2 1 R0 R1 R2 R3 R4 R5 R6 R7 x 3 5 1 1 3 1 1 x 3 5 1 1 3 1 1 x 3 2 1 1 6 2 1 10 x e 1 1 1 1 3 1 x 2 1 1 3 5 y  EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4  R1 R2 x 3 2 1 1 3 1 1 x 3 2 1 1 3 1 1 x 3 2 1 1 1 -1 3 x 3 2 1 1 3 1 1 x 3 2 1 1 3 1 1 Region 1 Region 2 x b a  EMBED Equation.3   EMBED Equation.3  x b a  EMBED Equation.3  x b a  EMBED Equation.3  x b a  EMBED Equation.3  x 2 1 1 3 1 1 5  EMBED Equation.3   EMBED Equation.3   EMBED Equation.3  x  EMBED Equation.3  a b x 3 2 1 -2 2  EMBED Equation.3  6 -6 5  EMBED Equation.3  (5, 7) (1, 5) x 4 2 2 -2 2 2 -6  EMBED Equation.3  -4  EMBED Equation.3  (3, 3) (3, 3) R1 R2  EMBED Equation.3  x  EMBED Equation.3  p Consumers Surplus x0 p0  EMBED Equation.3  x  EMBED Equation.3  p Producers Surplus x0 p0 x 3 1 1 6 2 1 10  EMBED Equation.3   EMBED Equation.3  5 (2, 5) p x 3 1 1 6 2 1 10  EMBED Equation.3   EMBED Equation.3  5 (2, 5) p x 3 1 1 6 2 1 10  EMBED Equation.3   EMBED Equation.3  5 (2, 5) p 6 2 10 x 1 3 -1 3 p .= > ? @ K M N s t u z { }     ! > ? 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Equation MathType EFEquation.DSMT49q_1103611647F m?A m?AOle tCompObjuiObjInfowEquation Native x]_1103611652F m?A m?AOle CompObji!A|<TDSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  pDxf(x i )==Dxf(x 0 )++Dxf(x 1 )++Dxf(x 2 )++...++Dxf(x n"-1 ) i==0n"-1 " FMathType 5.0 Equation MathType EFEquation.DSMT49q|LTDSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!ED/APObjInfoEquation Native _1103611654F m?A m?AOle G_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  f(x)==x 2 ++2 FMathType 5.0 Equation MathType EFEquation.DSMT49q| TDSMT5WinAllBasicCodePagesCompObjiObjInfoEquation Native _1103611657F m?A m?ATimes New RomanSymbolCourier NewMT Extra!ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  [0,2] FMathType 5.0 Equation MathType EFEquation.DSMT49q|,TDSMT5WinAllBasicCodePagesOle CompObjiObjInfoEquation Native Times New RomanSymbolCourier NewMT Extra!ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  f(x)==x 2 ++2 FMathType 5.0 Equation MathType EFEquation.DSMT49q_1103611659F m?A m?AOle CompObjiObjInfoEquation Native j_1103611661F m?A m?AOle CompObjiN|DTDSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  Dx== b"-an== 2"-04== 12==0.5 FMathType 5.0 Equation MathType EFEquation.DSMT49q|4TDSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  f(x i )ObjInfoEquation Native _1103611663F m?A m?AOle CompObjiObjInfoEquation Native _1106411469F m?A0m?A FMathType 5.0 Equation MathType EFEquation.DSMT49q|<TDSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  x iOle CompObjiObjInfoEquation Native  FMathType 5.0 Equation MathType EFEquation.DSMT49qy8 DSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  Total Area (Right)=pDxf(x i ) i==14 " ()==0.5"fx 1 ()++0.5"fx 2 ()++0.5"fx 3 ()++0.5"fx 4 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EFEquation.DSMT49q|,TDSMT5WinAllBasicCodePagesCompObj$iObjInfo&Equation Native '_1103611740IF@m?A@m?ATimes New RomanSymbolCourier NewMT Extra!ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  f(x)== x  ++1 FMathType 5.0 Equation MathType EFEquation.DSMT49qOle ,CompObj-iObjInfo/Equation Native 0|DTDSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  [a,b] FMathType 5.0 Equation MathType EFEquation.DSMT49q_1103611742F@m?A@m?AOle 4CompObj5iObjInfo 7Equation Native 8_11036117450#F@m?A@m?AOle =CompObj"$>i|4TDSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  Dx==  (b"-a)n FMathType 5.0 Equation MathType EFEquation.DSMT49qObjInfo%@Equation Native A_1106413757(F@m?A@m?AOle E|<TDSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  x i FMathType 5.0 Equation MathType EFEquation.DSMT49qCompObj')FiObjInfo*HEquation Native I_1106414113E*-F@m?A@m?A8 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EFEquation.DSMT49q|LTDSMT5WinAllBasicCodePagesCompObjOQiObjInfoREquation Native _1103813681]UFP"n?AP"n?ATimes New RomanSymbolCourier NewMT Extra!ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  F(x)== 13 12x 2 ()++2x++C== 16x 2 ++2x++C FMathType 5.0 Equation MathTyOle CompObjTViObjInfoWEquation Native pe EFEquation.DSMT49q8 DSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  F(0)==C FMathType 5.0 Equation MathTy_1103611770NbZFP"n?AP"n?AOle CompObjY[iObjInfo\pe EFEquation.DSMT49q|,TDSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  F(3) FMathType 5.0 Equation MathTyEquation Native _1103813711_FP"n?AP"n?AOle CompObj^`ipe EFEquation.DSMT49qf8 DSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  F(3)== 16(3 2 )++2(3)++C== 152++CObjInfoaEquation Native _1103611777dFP"n?A`In?AOle  FMathType 5.0 Equation MathType 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S  S"`?;-   S  S"`?3   S  S"`? p  `B  c $D KL>f  C 3  \ `B  c $D B C`B  c $D     S  S"`?56c   S  S"`?y6 `B  c $D KL`B  c $D 6   0e0e    BC DE(F 5% 8c8c     ?1 d0u0@Ty2 NP'p<'pA)BCD|E|| qX  c "a @   c"$`;*0  C 0e0e    B}CPDEtFVA 5% 8c8c     ?1 d0u0@Ty2 NP'p<'pA)BCD|E|| }kg!M*B;M->68:,B,B(N"MT_ZcE)G=PG2;<B(,@     `Large confettic"$`.0<j u [1,9 C  AA%T "/6H7r8A6SA6SH6SC!6S3N3K:AHr8%H44!HAc"$?   $0e0e    B.CvDEFA$ 5% 8c8c     ?1 d0u0@Ty2 NP'p<'pA)BCD|E|| l.vl @`Dashed horizontalc"$`%6)b9p   0e0e    B.CvDEFA 5% 8c8c     ?1 d0u0@Ty2 NP'p<'pA)BCD|E|| l.vl @`50%c"$`\!4%6B  HD1 ?#[1@#9B  HD1  6+6ZB  S D %6%6ZB  S D W!6X!6  S  S"`?:+C6,7 ZB  S D '6'6ZB  S D )6)6  S  S"`?')6*7   S  S"`?,'6(7   S  S"`?u 6"8   S  S"`?$6d&7   S  S"`?E"2#3 `B  c $D >#&8#'8>f  C  3  D#73#5`B   c $D B C`B   c $D      S   S"`?"5#>6    S   S"`?"7#8 `B  c $D E#&8#'8`B  c $D E#p4#q4Z =19 FC  II"/<5H7A6SI6SX6S6SXKFBF-AI=CIr8J6C3(0"/V(4 % t% =XIc"$? 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ZB X S D dX e  Y S Y S"`?$ @ SummaryInformation(DocumentSummaryInformation8CompObjrNew Chapter 3 ACOW GroupNormalShahzad Muhammad Tanvir2Microsoft Office Word@F#@(W@8A@8A  B՜.+,D՜.+,L hp  Texas A&M University'M New Chapter 3 TitleH 6> MTWinEqns   F Microsoft Word 97-2003 Document MSWordDocWord.Document.89q Z S Z S"`? o  R [ 3 B C DEdF, 8"rrO,h V9  "3   _  / 3  f6 h 4ax7 T @        "`tR~B \ BD Z+ ] 0] S"`?Y \2 ^ 3  "`R  \2 _ 3  "`} ` S ` S"`?MEO  a S a S"`?j"  . R >! b3 H6 &(#i(% 15'^)-I.4B d;Q|?AC"H#H"GP#S$Sb&S/OK2H3A6I.J-J(7'8-!?zFY&F:;b8$ & & $i#ic"$? c  t0e0e    BC0DEFR 5%      !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRST 8c8c     ?1 d0u0@Ty2 NP'p<'pA)BCD|E|| ""Z0xo5E#  &((-'3+7/;78:<@E>QDZJcSxSx;5pZ&(@        `c"$`LW| d  t0e0e    BC0DEFR 5% 8c8c     ?1 d0u0@Ty2 NP'p<'pA)BCD|E|| ""Z0xo5E#  &((-'3+7/;78:<@E>QDZJcSxSx;5pZ&(@        `c"$` B e HD1 v:? f S f S"`?! 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[ @VerdanaA$BCambria Math#ph3232$rF B' B'$24MM ;QXR _2!xx  New Chapter 3 ACOW GroupShahzad Muhammad Tanvir