Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Limit of a function wikipedia , lookup
Function of several real variables wikipedia , lookup
History of calculus wikipedia , lookup
Series (mathematics) wikipedia , lookup
Divergent series wikipedia , lookup
Path integral formulation wikipedia , lookup
Itô calculus wikipedia , lookup
Multiple integral wikipedia , lookup
Chapter 4 Integrals 4.1 Sigma Notation This notation is very important. You will use it in this chapter as well as later on, when you study sequences and series. 4.1.1 Theory De…nition 225 (sigma notation) Let am ; am+1 ; :::; an be real numbers and let m and n be integers such that m n. 1. We represent am + am+1 + ::: + an 1 + an by n X ai and we say ”The sum i=m as i ranges from m to n of a sub i”. 2. i is called the index of summation. 3. n X ai is the sum written in ”sigma” notation. am + am+1 + ::: + an 1 + an i=m is the sum written in ”expanded” form. Remark 226 Let us make the following remarks. 1. When writing n X ai , the index of summation takes on all integer values i=m between m and n: 2. The name of the index of summation is not important as it is being replaced by integers when expanding the sum. In other words n X ai = i=m n X aj j=m Usually, we use the letters i, j or k for index of summation. 155 156 CHAPTER 4. INTEGRALS Example 227 10 X i2 = 12 + 22 + 32 + ::: + 102 i=1 Example 228 10 X j 2 = 12 + 22 + 32 + ::: + 102 j=1 Example 229 10 X 2i = 2 + 22 + 23 + ::: + 210 i=1 Example 230 10 X i ( 1) 2i = 2 + 22 23 + ::: + 210 i=1 Remark 231 If you compare the last two examples, you will see that the difi ference between them is the presence of the term ( 1) . The e¤ ect of this term is the minus sign which appears every other term. You should remember this. i Whenever you need to generate a minus sign every other term, use ( 1) . Example 232 5 X i ( 1) x2i = x2 + x4 x6 + x8 x10 i=1 Remark 233 To generate only even powers, use 2i instead of i. Example 234 n X xi i=0 4.1.2 i! =1+x+ x2 x3 xn + + ::: + 2! 3! n! Some Results Worth Remembering Proposition 235 If C is any constant, then n X Cai = C n X ai i=1 i=1 Proof. This is simply factoring C. n X Cai = Ca1 + Ca2 + ::: + Can i=1 = C (a1 + a2 + ::: + an ) n X = C ai i=1 4.1. SIGMA NOTATION Proposition 236 n X 157 ai + i=1 n X bi = i=1 n X (ai + bi ) i=1 Proof. See problems at the end of the section. n X n (n + 1) 2 i=1 Proof. Let us …rst remark that the above formula simply says that 1+2+:::+n = n (n + 1) . To prove it, we write 2 Proposition 237 (Sum of the …rst n integers) S = 1 + 2 + ::: + (n i= 1) + n which we can also write as S = n + (n 1) + ::: + 2 + 1 If we add the two equalities, we obtain S + S = (1 + n) + (2 + (n 1)) + ::: + ((n 1) + 2) + n + 1 2S = (n + 1) + (n + 1) + ::: + (n + 1) + (n + 1) We are adding n terms, which are all equal to n + 1, therefore 2S = n (n + 1) S= n (n + 1) 2 Proposition 238 There are similar results for the sum of the …rst n squares and n cubes which we give without proof. 1. Sum of the …rst n squares: n X i2 = i=1 n (n + 1) (2n + 1) 6 2. Sum of the …rst n cubes: n X i=1 4.1.3 i3 = n (n + 1) 2 2 Things to know: Given a sum in sigma notation, know how to write it in expanded form. Given a sum in expanded form, know how to write it in sigma notation. Be able to do problems like the ones below. 158 4.1.4 CHAPTER 4. INTEGRALS Problems 1. Write the sum in expanded form (a) 5 X p i i=1 (b) 5 X 3i i=4 (c) 4 X 2k k=0 (d) n X1 1 2k + 1 ( 1) j j=0 2. Write the sum in sigma notation 1 2 3 4 19 + + + + ::: + 2 3 4 5 20 (b) 2 + 4 + 6 + 8 + ::: + 2n (a) (c) 1 + 3 + 5 + ::: + (2n 1) (d) x + x2 + x3 + ::: + xn (e) 1 x + x2 3. Prove that n X i=1 4.1.5 n x3 + ::: + ( 1) xn ai + n X bi = i=1 n X (ai + bi ). i=1 Answers 1. Write the sum in expanded form (a) 5 X p i i=1 5 X p i= p 1+ p 2+ p 3+ i=1 (b) 5 X 3i i=4 5 X i=4 3i = 34 + 35 p 4+ p 5 4.1. SIGMA NOTATION (c) 4 X 2k k=0 159 1 2k + 1 4 X 2k k=0 (d) n X1 ( 1) 1 1 3 5 7 1 = + + + + 2k + 1 1 3 5 7 9 j j=0 n X1 j 0 1 2 ( 1) = ( 1) + ( 1) + ( 1) + ::: + ( 1) n 1 j=0 2. Write the sum in sigma notation (a) 19 1 2 3 4 + + + + ::: + 2 3 4 5 20 19 19 X i 1 2 3 4 + + + + ::: + = 2 3 4 5 20 i+1 i=1 (b) 2 + 4 + 6 + 8 + ::: + 2n 2 + 4 + 6 + 8 + ::: + 2n = n X 2i i=1 (c) 1 + 3 + 5 + ::: + (2n 1) 1 + 3 + 5 + ::: + (2n 1) = n X (2i 1) i=1 (d) x + x2 + x3 + ::: + xn x + x2 + x3 + ::: + xn = n X xi i=1 (e) 1 n x + x2 x3 + ::: + ( 1) xn 1 x + x2 n x3 + ::: + ( 1) xn = n X i=0 3. No answer to write. i ( 1) xi 160 CHAPTER 4. INTEGRALS Figure 4.1: Partition with 6 subintervals 4.2 Area and Riemann Sums 4.2.1 Preliminary De…nitions De…nition 239 (partition) In this section, unless stated otherwise, we assume that a and b are …nite real numbers. 1. Given an interval [a; b] ; a partition P is a subdivision of the interval into n subintervals. If we denote the subintervals [x0 ; x1 ], [x1 ; x2 ], :::, [xn 1 ; xn ] then P = fx0 ; x1 ; :::; xn g. With this notation, we always have a = x0 and b = xn , and n is the number of subintervals. Figure 4.1 shows an interval [a; b] and a partition P = fx0 ; x1 ; x2 ; x3 ; x4 ; x5 ; x6 g of the interval with n = 6. You will note that the number of points is always one more than the number of subintervals that is if there are n intervals, there will be n + 1 points . You will also note that the …rst subinterval is always [x0 ; x1 ]. The second subinterval is [x1 ; x2 ]. In general, the ith subinterval will be [xi 1 ; xi ]. The last subinterval will be the nth subinterval, that is [xn 1 ; xn ]. It is important to understand this notation in order to be able to understand what follows. 2. If all the subintervals have the same length, the partition is called a regular partition. The length of each subinterval, denoted x is then x= b a n Figure 4.2 shows an interval and a regular partition P = fx0 ; x1 ; x2 ; x3 ; x4 ; x5 ; x6 g of the interval with n = 6. Example 240 A possible partition of the interval [0; 4] with n = 4 is P = f0; 0:5; 1; 3:5; 4g. The subintervals obtained are [0; 0:5] ; [0:5; 1] ; [1; 3:5] ; [3:5; 4]. This is not a regular partition since not all the subintervals have the same length. There are many other ways of partitioning [0; 4]. 4.2. AREA AND RIEMANN SUMS 161 Figure 4.2: Regular partition with 6 subintervals Figure 4.3: Area below a graph Example 241 A regular partition of [0; 4] with n = 4 is P = f0; 1; 2; 3; 4g. The subintervals obtained are [0; 1] ; [1; 2] ; [2; 3] ; [3; 4]. This is a regular partition since all the intervals have the same length. 4.2.2 Area Below a Graph Let f be a positive function (its graph is above the x-axis), assume that f is continuous on the interval [a; b]. The goal is to …nd the area of the region bounded by the graph of y = f (x), the x-axis, the vertical lines x = a and x = b. In other words, we want to …nd the area of the shaded region shown on Figure 4.3. We often say that we want to …nd the area below the graph of f between a and b.In the case that the graph of y = f (x) is a straight line, the corresponding region may be a rectangle, or a polygon. In this case, we know how to compute its area. In general though, the region will not correspond to a region for which we have a formula for its area. A purpose of this section is to devise a way to compute its area. We do it in two steps. First, we will describe a procedure to approximate the area. Then, we see how to …nd the exact value 162 CHAPTER 4. INTEGRALS Figure 4.4: Approximating an area with rectangles of the area. 4.2.3 Approximation of the Area below a Graph To better understand the procedure we are about to describe, you can refer to Figure 4.4. It illustrates the procedure in the case n = 4. Problem: Given a positive function f and an interval [a; b], approximate the area (denoted A) below the graph of f between a and b. We will approximate the region with rectangles and the area of the region by the sum of the areas of the rectangles. Since a rectangle is de…ned by its base and its height, we describe how to derive the base and the height of each rectangle used. We proceed as follows: 1. Deriving the base of each rectangle. Divide the interval [a; b] into n subintervals (we will see later which number to choose for n). Though the intervals do not have to have equal length, to simplify the procedure, we will only consider subintervals of equal length. Thus, we obtain a regular partition P = fa = x0 ; x1 ; :::; xn = bg. Also, the length of each subinterval b a will be x = . The subintervals obtained are [x0 ; x1 ], [x1 ; x2 ], :::, n [xn 1 ; xn ]. In other words, the subintervals are of the form [xi 1 ; xi ] for i = 1; 2; :::; n. The points xi can be generated by the formula a n Though these formulae are not needed to carry out the procedure, they are useful if one were to write a program to simulate the procedure. These subintervals [xi 1 ; xi ] will form the base of our rectangles. xi = a + i b 4.2. AREA AND RIEMANN SUMS 163 2. Deriving the height of each rectangle. In each interval [xi 1 ; xi ], select a point denoted xi . In theory, xi can be selected anywhere in [xi 1 ; xi ]. In practice, it will usually be one of the end points, or the midpoint, or the point at which f is either maximum or minimum. Thus, in the …rst interval, the interval [x0 ; x1 ] we select a point that we call x1 . In the second interval, the interval [x1 ; x2 ], we select a point we call x2 , and so on. The height of our rectangles will be f (xi ). If xi is the right endpoint of [xi If xi is the left endpoint of [xi If xi is the midpoint of [xi 1 ; xi ] 1 ; xi ] then xi = xi then xi = xi 1 ; xi ] then xi = xi +xi 2 1 1 3. Forming the rectangles. In each subinterval [xi 1 ; xi ], consider the rectangle Ri with base the width of the interval, and height f (xi ). Let Ai denote its area. We can compute Ai easily. Ai = width height = (xi xi 1) f (xi ) = xf (xi ) b a f (xi ) = n 4. Approximating the area. We approximate the area below the graph (denoted A) by adding the area of all the rectangles. Thus, A A1 + A2 + ::: + An n X Ai i=1 n X f (xi ) x i=1 x n X f (xi ) i=1 b n n aX f (xi ) i=1 Example 242 Approximate the area below the graph of y = x2 between x = 0 and x = 4, using 4 subintervals and by selecting xi to be the right end point of each subinterval. Repeat the procedure by selecting xi to be the left end point. In this case, a = 0, b = 4 and n = 4. If we select xi to be the right end point of each subinterval, then x1 = 1, x2 = 2, x3 = 3 and x4 = 4. The function f is f (x) = x2 . We can now 164 CHAPTER 4. INTEGRALS Figure 4.5: Rectangles using right end points apply the formula. A n b n 4 aX f (xi ) i=1 0 (f (x1 ) + f (x2 ) + f (x3 ) + f (x4 )) 4 f (1) + f (2) + f (3) + f (4) 1 + 4 + 9 + 16 30 Figure 4.5 illustrates this procedure. You will note that the exact value of the area is less than the approximation. Thus, we know that A < 30. If we select xi to be the left end point of each subinterval, then x1 = 0, x2 = 1, x3 = 2 and x4 = 3. The function f is f (x) = x2 . We can now apply the formula. A n b n 4 aX f (xi ) i=1 0 (f (x1 ) + f (x2 ) + f (x3 ) + f (x4 )) 4 f (0) + f (1) + f (2) + f (3) 0+1+4+9 14 Figure 4.6 illustrates this procedure. You will note that the exact value of the area is more than the approximation. Thus, we know that A > 14. 4.2. AREA AND RIEMANN SUMS 165 Figure 4.6: Rectangles using left end points Remark 243 From the two methods above used to approximate the area, we see that 14 < A < 30. Therefore, we have an upper bound as well as a lower bound on the area. They are not very good, since they are far apart. The next example will show that we can do better. Example 244 Same as the previous example, using 8 subintervals. In this case, a = 0, b = 4, n = 8. If we select xi to be the right end point of each subinterval, then x1 = 0:5, x2 = 1, x3 = 1:5, x4 = 2, x5 = 2:5, x6 = 3, x7 = 3:5 and x8 = 4. The function f is f (x) = x2 . We can now apply the formula. A n b n 4 aX 0 8 f (xi ) i=1 (f (x1 ) + f (x2 ) + f (x3 ) + f (x4 ) + f (x5 ) + f (x6 ) + f (x7 ) + f (x8 )) 1 (f (0:5) + f (1) + f (1:5) + f (2) + f (2:5) + f (3) + f (3:5) + f (4)) 2 1 1 9 25 49 +1+ +4+ +9+ + 16 2 4 4 4 4 25:5 Figure 4.7 illustrates this procedure. You will note that the exact value of the area is less than the approximation. Thus, we know that A < 25:5. If we select xi to be the left end point of each subinterval, then x1 = 0, x2 = 0:5, x3 = 1, x4 = 1:5, x5 = 2, x6 = 2:5, x7 = 3, x8 = 3:5. The 166 CHAPTER 4. INTEGRALS Figure 4.7: Rectangles using right end points function f is f (x) = x2 . We can now apply the formula. A n b n 4 aX 0 8 f (xi ) i=1 (f (x1 ) + f (x2 ) + f (x3 ) + f (x4 ) + f (x5 ) + f (x6 ) + f (x7 ) + f (x8 )) 1 (f (0) + f (0:5) + f (1) + f (1:5) + f (2) + f (2:5) + f (3) + f (3:5)) 2 1 1 9 25 49 +1+ +4+ +9+ 2 4 4 4 4 17:5 Figure 4.8 illustrates this procedure. You will note that the exact value of the area is more than the approximation. Thus, we know that A > 17:5. Remark 245 This time, we see that 17:5 < A < 25:5, which is slightly better than what we had before. Therefore, increasing the number of subintervals seems to give a better approximation. In fact, it can be proven (this is done in a more advanced class such as Real Analysis) that as n gets larger, our approximation becomes closer and closer to the exact value of the area. 4.2.4 Exact Value of the Area From the remark made above, we de…ne the area below a graph as follows: De…nition 246 (area below a graph) Let f be a positive function and a and b two …nite real numbers such that a < b. 4.2. AREA AND RIEMANN SUMS 167 Figure 4.8: Rectangles using left end points 1. The area A below the graph of y = f (x) between x = a and x = b is A = lim n!1 n X f (xi ) x i=1 providing the limit exists. In the above formula, xi is a point chosen in [xi 1 ; xi ]. 2. The sum in the above formula is called a Riemann sum. If xi corresponds to a maximum of f on [xi 1 ; xi ] then the sum is called an upper Riemann sum. If xi corresponds to a minimum of f on [xi 1 ; xi ] then the sum is called an lower Riemann sum. In practice, …nding A using the de…nition is very di¢ cult. First, the expression resulting from the sum is usually very complicated. Then, we have to take its limit as n approaches in…nity. We will not spend any more time on it, but you should remember the formula. 4.2.5 Use of Technology Carrying out the computations involved in the procedure described above is very tedious. I have designed a Java applet to assist students in this procedure. It can be found at http://science.kennesaw.edu/~plaval/tools/index.html under integration. 4.2.6 Things to know Know how to approximate an area using Riemann sums.. If you are given a partition, use it, otherwise make up your own partition. 168 4.2.7 CHAPTER 4. INTEGRALS Problems Following the procedure described in this section, approximate the area below the graph of the given function f (x) in the given interval [a; b] using the given number of subdivisions n and the given rule to determine the height of the rectangle for each situation below. 1. f (x) = 16 x2 for x in [0; 4] using 4 subdivisions and selecting xi to be the left endpoint of each subinterval. 2. f (x) = 16 x2 for x in [0; 4] using 4 subdivisions and selecting xi to be the right endpoint of each subinterval. 3. f (x) = 16 x2 for x in [0; 4] using 4 subdivisions and selecting xi to be the midpoint of each subinterval. 4. f (x) = 16 x2 for x in [0; 4] using 8 subdivisions and selecting xi to be the left endpoint of each subinterval. 5. f (x) = 16 x2 for x in [0; 4] using 8 subdivisions and selecting xi to be the right endpoint of each subinterval. 6. f (x) = 16 x2 for x in [0; 4] using 8 subdivisions and selecting xi to be the midpoint of each subinterval. 7. Using the applet mentioned in the notes, approximate the area below the graph of f (x) = sin x between x = 0 and x = by using upper and lower Riemann sums and various values for the number of subintervals. 8. Using the applet mentioned in the notes, approximate the area below the graph of f (x) = ex between x = 0 and x = 1 by using upper and lower Riemann sums and various values for the number of subintervals. 4.2.8 Answers Following the procedure described in this section, approximate the area below the graph of the given function f (x) in the given interval [a; b] using the given number of subdivisions n and the given rule to determine the height of the rectangle for each situation below. 1. f (x) = 16 x2 for x in [0; 4] using 4 subdivisions and selecting xi to be the left endpoint of each subinterval. A = 50 2. f (x) = 16 x2 for x in [0; 4] using 4 subdivisions and selecting xi to be the right endpoint of each subinterval. A 34 4.3. THE DEFINITE INTEGRAL 169 3. f (x) = 16 x2 for x in [0; 4] using 4 subdivisions and selecting xi to be the midpoint of each subinterval. A 43 4. f (x) = 16 x2 for x in [0; 4] using 8 subdivisions and selecting xi to be the left endpoint of each subinterval. A 46:5 5. f (x) = 16 x2 for x in [0; 4] using 8 subdivisions and selecting xi to be the right endpoint of each subinterval. A 38:5 6. f (x) = 16 x2 for x in [0; 4] using 8 subdivisions and selecting xi to be the midpoint of each subinterval. A 42:75 7. Using the applet mentioned in the notes, approximate the area below the graph of f (x) = sin x between x = 0 and x = by using upper and lower Riemann sums and various values for the number of subintervals. The area is approximately 2. 8. Using the applet mentioned in the notes, approximate the area below the graph of f (x) = ex between x = 0 and x = 1 by using upper and lower Riemann sums and various values for the number of subintervals. The area is approximately 1:7. 4.3 4.3.1 The De…nite Integral Theory De…nition 247 (de…nite integral) The de…nite integral of f from a to b is de…ned by Zb n X f (x) dx = lim f (xi ) x n!1 a i=1 if this limit exists. When the limit exists, f is said to be integrable. The number a is called the lower limit of integration, b is the upper limit of integration. f is called the integrand. Remark 248 You may ask what functions are integrable. It turns out that the answer is not simple. You will study this question in a more advanced class such as Real Analysis. For now, we will only say that if f is continuous then f is integrable. Keep in mind that this does not mean if f is not continuous it is not integrable. 170 CHAPTER 4. INTEGRALS Regarding areas, as we discussed in the previous section, we see that if f happens to be a positive function, then the integral is the area below the graph. If the function is entirely negative, then the above sum will be negative since the term f (xi ) will be negative while x > 0. However, an area is always positive. Therefore, the integral will be the negative of the area. The proposition below summarizes the relation between integral and area. Proposition 249 The relation between an integral and the area of the region bounded by y = f (x), the x-axis, the lines x = a and x = b is as follows: 1. If f (x) 0 then Rb f (x) dx is the area of the region between the graph of a f and the x-axis and between x = a and x = b. 2. If f (x) 0 then Rb f (x) dx is the negative of the area of the region between a the graph of f and the x-axis and between x = a and x = b. 3. In general, Rb f (x) dx = A1 A2 where A1 is the area of the region above a the x-axis, below the graph of f and A2 is the area of the region below the x-axis, above the graph of f . If the relationship between integral and area is still not entirely clear, Figure 4.9 along with the explanations below should clarify everything. 1. Assuming A1 , A2 , and A3 are known, then we could use this knowledge to compute the following integrals: (a) R1 f (x) dx = A1 0 (b) R3 f (x) dx = A2 (remember, the numbers Ai represent areas, they 1 are positive). R5 (c) f (x) dx = A3 3 (d) R3 f (x) dx = A1 A2 0 (e) R5 f (x) dx = A2 + A3 1 (f) R5 f (x) dx = A1 A2 + A3 0 2. If, on the other hand we could compute Rb f (x) dx for any a and b, then a we could use this knowledge to compute areas as follows: 4.3. THE DEFINITE INTEGRAL 171 Figure 4.9: Relationship between integral and area (a) A1 = R1 f (x) dx 0 (b) A2 = R3 f (x) dx 1 (c) A3 = R5 f (x) dx 3 (d) area of the shaded region = R1 0 f (x) dx R3 1 f (x) dx + R5 f (x) dx 3 Thus, we see that our knowledge of areas can be used to compute integrals. Our ability to compute integrals can also be used to …nd the area of certain regions. Once we know how to compute integrals e¢ ciently, we will use integrals to compute areas. Integrals have the following properties: Proposition 250 Assuming all the integrals below exist, and a 1. Rb f (x) dx = a 2. Ra a Ra b f (x) dx = 0 f (x) dx b we have: 172 3. CHAPTER 4. INTEGRALS Rb cdx = c (b a) a 4. Rb [f (x) g (x)] dx = a 5. Rb a 6. Rb Rb Rb f (x) dx a g (x) dx a Rb cf (x) dx = c f (x) dx a f (x) dx = a 7. If f (x) Rc f (x) dx + Rb f (x) dx c a 0 for x in [a; b] then Rb f (x) dx 0 a 8. If f (x) g (x) for x in [a; b] then Rb f (x) dx a Rb g (x) dx a 9. If f has a maximum value M on [a; b] and a minimum value m on [a; b] Rb then m (b a) f (x) dx M (b a) a Proof. Though we will not prove these properties, they are not very hard. Most of them follow from the de…nition. For the moment, we have two ways of computing integrals. We can either approximate them using Riemann sums, or we can computing them using the area interpretation of integrals. Both methods are very limited, we illustrate them with a few examples. In the next section (4.4), we will see an easier way of computing integrals. Example 251 Approximate R4 x2 dx using Riemann sums. You will use 4 subin- 0 tervals, and select xi to be the right end point of each subinterval. This is precisely the work we did for one of the examples on area above. The answer we found was 30. Example 252 Approximate R cos xdx using Riemann sums. You will use 4 0 subintervals, and select xi to be the right end point of each subinterval. Figure 4.10 illustrates this. The general formula we use is Z 0 cos xdx 4 X f (xi ) x i=1 4 4 0X i=1 cos xi 4.3. THE DEFINITE INTEGRAL Figure 4.10: Approximating R 173 cos xdx using 4 rectangles and right end points 0 We need to …gure out what the xi are. For this, we need to …gure out what the xi are. Since we divide the interval [0; ] into four subintervals, we will have 3 , …ve points: x0 ; x1 ; :::; x4 . These points are: x0 = 0, x1 = , x2 = , x3 = 4 2 4 3 x4 = . Since we pick right end points, we have x1 = , x2 = , x3 = and 4 2 4 x4 = . It follows that Z 0 cos xdx 4 cos 4 + cos 2 + cos 3 + cos 4 0 = Example 253 Compute 4 R2 xdx using the area interpretation of the integral. 0 For this, we begin by sketching the graph of f (x) = x (see Figure 4.11). Between 0 and 2, we see that the integrand is positive, therefore the integral is exactly the area of the region below the graph between 0 and 2. The region is a triangle, 174 CHAPTER 4. INTEGRALS Figure 4.11: Integral and area its base is 2, its height is 2, therefore Z2 xdx = area of shaded region 0 1 2 2 =2 2 = Example 254 Compute R2 xdx using the area interpretation of the integral. 1 Again, we sketch the graph of the function, see Figure 4.12. According to the area interpretation of the integral, we have: Z2 xdx = A2 A1 1 1 2 2 3 = 2 = Example 255 Compute R2 p 4 2 1 1 2 1 x2 dx using the area interpretation of the inte- 0 gral. p The function f (x) = 4 x2 is the upper half of the circle of radius 2, centered R2 p at the origin. Therefore, the integral 4 x2 dx corresponds to the area of the 0 4.3. THE DEFINITE INTEGRAL 175 Figure 4.12: Integral and area shaded region shown on Figure 4.13, that is one fourth of the area of a circle of radius 2. It follows that Z2 p 4 x2 dx = 1 2 (2) 4 0 = 1 4 4 = Example 256 Compute R2 x+ p 4 x2 dx using the area interpretation of the 0 integral. Using the properties of the integral, we have: Z2 0 p x+ 4 x2 dx = Z2 0 xdx + Z2 p 4 x2 dx 0 We computed each integral in the examples above, so Z2 x+ 0 4.3.2 p 4 x2 dx = 2 + Things to Know Given a partition and a function, be able to evaluate Rb a Riemann sums. f (x) dx with 176 CHAPTER 4. INTEGRALS Figure 4.13: Integral and area Be able to select your own partition to evaluate f (x) dx with Riemann a sums. Be able to evaluate Rb Rb f (x) dx by interpreting it in terms of areas. a Know and be able to use the properties of the de…nite integral. 4.3.3 Problems R2 R4 R4 1. Assuming that 1 f (x) dx = 1, 1 f (x) dx = 3 and 1 g (x) dx = 2, compute the de…nite integrals below using the properties in integrals. State which properties are being used. R1 (a) 1 f (x) dx R1 (b) 4 f (x) dx R1 (c) 4 5g (x) dx R4 (d) 1 (f (x) g (x)) dx R4 (e) 1 (2f (x) + 3g (x)) dx R4 (f) 2 f (x) dx 2. Use the area interpretation of the integral to evaluate the integrals below. R3 (a) 0 xdx R3 (b) xdx 1 4.3. THE DEFINITE INTEGRAL (c) 177 R3 jxj dx 2 R3 p (d) 9 x2 dx 3 R3p (e) 0 9 x2 dx p R3 (f) 0 x + 9 x2 dx 3. Write the area of each region below as a de…nite integral or a sum of de…nite integrals. You do not need to evaluate the integrals. (a) The region between the x-axis and the graph of sin x between x = 0 and x = . 2 (b) The region between the x-axis and the graph of sin x between x = and x = 2 . (c) The region between the x-axis and the graph of sin x between x = 0 and x = 2 . (d) The upper half of a circle of radius 16. 4.3.4 Answers R2 R4 R4 1. Assuming that 1 f (x) dx = 1, 1 f (x) dx = 3 and 1 g (x) dx = 2, compute the de…nite integrals below using the properties in integrals. State which properties are being used. (a) (b) (c) (d) (e) (f) R1 1 f (x) dx = 0 4 f (x) dx = 4 5g (x) dx = 10 1 (f (x) 1 (2f (x) + 3g (x)) dx = 0 2 f (x) dx = 2 R1 R1 R4 R4 R4 3 g (x)) dx = 5 2. Use the area interpretation of the integral to evaluate the integrals below. (a) R3 0 xdx Z 3 xdx = 0 (b) R3 1 xdx Z 9 2 3 xdx = 4 1 178 CHAPTER 4. INTEGRALS (c) (d) (e) R3 2 jxj dx R3 p 3 R3p 0 Z x2 dx 9 Z x2 dx 9 3 3 0 (f) R3 0 x+ p 9 jxj dx = 2 p 13 2 9 x2 dx = 9 2 p 9 x2 dx = 9 4 p x2 dx = 3 Z 3 x2 dx Z 0 3 x+ 9 9 9 + 2 4 3. Write the area of each region below as a de…nite integral or a sum of de…nite integrals. You do not need to evaluate the integrals. (a) The region between the x-axis and the graph of sin x between x = 0 and x = . 2 Z 2 Area = sin xdx 0 (b) The region between the x-axis and the graph of sin x between x = and x = 2 . Z 2 Area = sin xdx (c) The region between the x-axis and the graph of sin x between x = 0 and x = 2 . Z Z 2 Area = sin xdx sin xdx 0 (d) The upper half of a circle of radius 16. Area = 4.4 4.4.1 Z 4 4 p 16 x2 dx The Fundamental Theorem of Calculus Theory De…nition 257 An antiderivative of a function f is a function F such that F 0 (x) = f (x). 4.4. THE FUNDAMENTAL THEOREM OF CALCULUS 179 1 Example 258 Since the derivative of ln x is , it means that an antiderivative x 1 of is ln x x 0 Example 259 Since (sin x) = cos x, it means that an antiderivative of cos x is sin x. Remark 260 Antiderivatives are not unique. If F is an antiderivative of f (that is if F 0 = f ), then F + C where C is any constant is also an antiderivative 0 of f . To verify this, we need to show that (F + C) = f . 0 0 (F + C) = F 0 + (C) = F0 + 0 = F0 =f Remark 261 Because of the previous remark, antiderivatives are always given with a constant. For example, we say that an antiderivative for cos x is sin x + C. In the remaining part of this document, C will always be used to denote a constant. Antiderivatives play an important in evaluating integrals as the next theorem will show. Theorem 262 (Fundamental Theorem of Calculus) Let f be a continuous function on [a; b]. 1. The function g (x) = Rx f (t) dt for x in [a; b] is continuous and di¤ eren- a 0 tiable. Furthermore, g (x) = f (x) or an antiderivative of f . 2. If F is any antiderivative of f then Rb a d dx Rx f (t) dt = f (x) that is g is a b f (x) dx = F (x)ja = F (b) F (a). Remark 263 The following follows from the theorem: 1. Part 1 of the Fundamental Theorem of Calculus says that the derivative of the integral of a function is the function itself. 2. Part 2 of the Fundamental Theorem of Calculus provides a way to compute integrals. We simply have to …nd an antiderivative of the integrand, and plug in the limits of integration. 180 CHAPTER 4. INTEGRALS R De…nition 264 (inde…nite integral) f (x) dx is used to represent an antiderivative of f . That is Z f (x) dx = F (x) , F 0 (x) = f (x) It is a function, not a number. It is called the inde…nite integral of f . Remark 265 In the above de…nition, if we replace f (x) by F 0 (x), we obtain R 0 F (x) dx = F (x). In other words, the integral of the derivative of a function is the function itself. If we combine this with part 1 of the Fundamental Theorem of Calculus which says that the derivative of the integral of a function is the function itself, we see that integration and di¤ erentiation are inverse processes. One undoes what the other one does. At this point, you should know the following antiderivatives: 1. R un+1 + C if n 6= n+1 un du = 1 R 1 du = ln juj + C u R 3. eu du = eu + C 2. 4. 5. 6. 7. 8. 9. 10. 11. 12. R R R R R R R R R au du = au +C ln a sin udu = cos u + C cos udu = sin u + C sec2 udu = tan u + C csc2 udu = cot u + C sec u tan udu = sec u + C csc u cot udu = csc u + C 1 du = tan u2 + 1 1 p 1 1 u2 du = sin u+C 1 u+C In addition, the following properties of the integral are also often used: R R 13. Cf (x) dx = C f (x) dx R R R 14. [f (x) g (x)] dx = f (x) dx g (x) dx 4.4. THE FUNDAMENTAL THEOREM OF CALCULUS 4.4.2 181 Part 2 of the Fundamental Theorem of Calculus Let us …rst illustrate part 2 of the fundamental theorem of calculus. Recall it says that if F is any antiderivative of f then Zb b f (x) dx = F (x)ja = F (b) F (a) a Thus, the problem of computing a de…nite integral is reduced to the problem of …nding antiderivatives. Let us look at some examples. Example 266 Find R3 x2 dx 1 Since an antiderivative of x2 is Z3 x3 (formula 1 above with n = 2), it follows that 3 x2 dx = x3 3 1 3 1 3 = = Example 267 Find Z 13 3 3 3 26 3 sin xdx 0 Z sin xdx = ( cos x)j0 0 = cos = ( 1) + 1 =2 ( cos 0) 182 CHAPTER 4. INTEGRALS Example 268 Find Z2 x + x2 dx 0 Z2 x + x2 dx = x3 x2 + 2 3 0 2 2 0 3 02 03 + 2 3 2 2 + 2 3 8 =2+ 3 14 = 3 = 4.4.3 Part 1 of the Fundamental Theorem of Calculus We now illustrate part 1 of the fundamental theorem of calculus. Recall, it says Rx Rx 0 d that if g (x) = f (t) dt then g (x) = f (x) or dx f (t) dt = f (x) . Note that a a in order to be able to apply it, the lower limit of integration must be a constant, we must be di¤erentiating with respect to the upper limit of integration. If we combine this with the chain rule, then we have d dx Zu f (t) dt = f (u) du dx (4.1) a where u is a function of x. 1 0 x Z d @ Example 269 Find sin tdtA dx 0 x 1 Z d @ By part 1 of the Fundamental Theorem of Calculus, sin tdtA = sin x. dx 0 0 0 0 1 Z d @ sin tdtA Example 270 Find dx x Before we can apply the Fundamental Theorem of Calculus, we must switch the 4.4. THE FUNDAMENTAL THEOREM OF CALCULUS 183 limits of integration, which we can do by one of the properties of the integral. 1 1 0 0 0 x Z Z d d @ @ sin tdtA = sin tdtA (property of the integral) dx dx x 0 11 0 0 x Z d = @ @ sin tdtAA (property of the derivative) dx 0 = sin x (by the previous example) 1 0 2 Zx d B C Example 271 Find @ sec tdtA dx 1 The upper limit of integration is not x but a function of x, so we must use formula 4.1. Therefore 0 2 1 Zx 2 d B C 2 dx @ sec tdtA = sec x dx dx 1 = Example 272 Find 0 d B @ dx Zx x 2x sec x2 1 2 C sin tdtA Here, neither limit of integration is a constant. Using a property of integrals, we can break the interval of integration by introducing a constant. We have 0 2 1 0 2 1 0 0 1 Zx Zx Z d B d @ d B C C sin tdtA + @ sin tdtA = @ sin tdtA dx dx dx x x = 0 d @ dx = 0 Zx 0 1 sin tdtA + d B @ dx = 4.4.4 Zx 0 1 2 C sin tdtA 0 2 1 0 x 11 Z Zx C @ d @ sin tdtAA + d B @ sin tdtA dx dx 0 0 = 0 0 dx2 sin x + sin x2 dx sin x + 2x sin x2 Things to know Know part 1 of the Fundamental Theorem, be able to apply it. 184 CHAPTER 4. INTEGRALS Know part 2 of the Fundamental Theorem, be able to apply it to compute integrals. 4.4.5 Problems Evaluate the integrals below (problems 1-14) R4 1. 0 x3 dx R 2. 02 cos xdx R1 3. 0 ex dx R 4. 0 sin xdx R4p 5. 1 xdx R2 6. 0 x3 + 3x2 1 dx R0 7. (2x ex ) dx 1 R4 p 8. 0 x ( x + 1) dx R 9. 04 3 sec2 tdt R9 1 dx 1 2x R2 2 11. (3x + 1) dx 2 10. R e x2 + x + 1 dx 1 x R 1 + cos2 t dt 13. 04 cos2 t R2 14. 0 4x dx 12. 15. Find F 0 (x) for F (x) = 16. Find F 0 (x) for F (x) = 17. Find G0 (x) for G (x) = 18. Find G0 (x) for G (x) = Rxp 1 R2 x t2 sin tdt R x3 1 R x2 x 1 + 2tdt tan tdt ln tdt R x2 19. Find F 0 (x) for F (x) = 0 cos tdt two di¤erent ways. The …rst method, you evaluate the integral …rst, then compute the derivative of the result. The second method, you will use the Fundamental Theorem of Calculus. 20. Find the area of the region between the x-axis and the graph of y = x2 between x = 1 and x = 4. 4.4. THE FUNDAMENTAL THEOREM OF CALCULUS 4.4.6 Answers Evaluate the integrals below (problems 1-14) 1. 2. 3. 4. 5. 6. 7. 8. 9. R4 x3 dx = 64 0 R cos xdx = 1 2 0 R1 ex dx = e 0 R sin xdx = 2 0 R4p 1 R2 1 R4 0 R xdx = 14 3 x3 + 3x2 0 R0 1 4 0 (2x 1 dx = 10 ex ) dx = 1 e 2 p 104 x ( x + 1) dx = 5 3 sec2 tdt = 3 R9 1 dx = ln 3 1 2x R2 2 11. (3x + 1) dx = 52 2 10. 12. 13. 14. R e x2 + x + 1 1 dx = e + e2 1 x 2 R 4 0 R2 0 1 2 1 + cos2 t 1 dt = +1 cos2 t 4 4x dx = 15 2 ln 2 15. Find F 0 (x) for F (x) = Rxp 1 1 + 2tdt F 0 (x) = 16. Find F 0 (x) for F (x) = 17. Find G0 (x) for G (x) = R2 x 1 + 2x t2 sin tdt F 0 (x) = R x3 1 p x2 sin x tan tdt G0 (x) = 3x2 tan x3 185 186 CHAPTER 4. INTEGRALS 18. Find G0 (x) for G (x) = R x2 x ln tdt G0 (x) = ln x (4x 1) R x2 19. Find F 0 (x) for F (x) = 0 cos tdt two di¤erent ways. The …rst method, you evaluate the integral …rst, then compute the derivative of the result. The second method, you will use the Fundamental Theorem of Calculus. (a) Method 1: F 0 (x) = 2x cos x2 (b) Method 2: F 0 (x) = 2x cos x2 20. Find the area of the region between the x-axis and the graph of y = x2 between x = 1 and x = 4. Area = 21