# Lecture Notes on Integral Calculus

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Lecture Notes on Integral Calculus UBC Math 103 Lecture Notes by Yue-Xian Li (Spring, 2004) 1 Introduction and highlights Di erential calculus you learned in the past term was about di erentiation. You may feel embarrassed to nd out that you have already forgotten a number of things that you learned di erential calculus. However, if you still remember that di erential calculus was about the rate of change, the slope of a graph, and the tangent of a curve, you are probably OK.  The essence of di erentiation is nding the ratio between the di erence in the value of f(x) and the increment in x. Remember, the derivative or the slope of a function is given by df f(x + x) f(x) 0 f (x) = = lim : (1) x0 dx x Integral calculus that we are beginning to learn now is called integral calculus. It will be mostly about adding an incremental process to arrive at a \total". It will cover three major aspects of integral calculus: 1. The meaning of integration.  We'll learn that integration and di erentiation are inverse operations of each other. They are simply two sides of the same coin (Fundamental Theorem of Caclulus). 2. The techniques for calculating integrals. 3. The applications. 2 Sigma Sum 2.1 Addition re-learned: adding a sequence of numbers In essence, integration is an advanced form of addition. We all started learning how to add two numbers since as young as we could remember. You might say \Are you kidding? Are you telling me that I have to start my university life by learning addition?". 1The answer is positive. You will nd out that doing addition is often much harder than calculating an integral. Some may even nd sigma sum is the most dicult thing to learn in integral calculus. Although this diculty is by-passed by using the Fundamental Theorem of Caclulus, you should NEVER forget that you are actually doing a sigma sum when you are calculating an integral. This is one secret for correctly formulating the integral in many applied problems with ease Now, I use a couple of examples to show that your skills in doing addition still need improve- ment. Example 1a: Find the total number of logs in a triangular pile of four layers (see gure). Solution 1a: Let the total number be S , where S' stands for Sum' and the subscript reminds 4 us that we are calculating the sum for a pile of 4 layers. S = 1 + 2 + 3 + 4 = 10: 4 z z z z in layer 1 in layer 2 in layer 3 in layer 4 A piece of cake Example 1b: Now, nd the total number of logs in a triangular pile of 50 layers, i.e. nd S (Give me the answer in a few seconds without using a calculator). 50 Solution 1b: Let's start by formulating the problem correctly. S = 1 + 2 + + 49 + 50 =? 50 z z z z in layer 1 in layer 2 in layer 49 in layer 50 where `' had to be used to represent the numbers between 3 and 48 inclusive. This is because there isn't enough space for writing all of them down. Even if there is enough space, it is tedious and unnecessary to write all of them down since the regularity of this sequence makes it very clear what are the numbers that are not written down. Still a piece of cake? Not really if you had not learned Gauss's formula. We'll have to leave it unanswered at the moment. Example 2: Finally, nd the total number of logs in a triangular pile of k layers, i.e. nd S k (k is any positive integer, e.g. k = 8; 888; 888 is one possible choice) Solution 2: This is equivalent to calculating the sum of the rst k positive integers. S = 1 + 2 + + (k 1) + k: k The only thing we can say now is that the answer must be a function of k which is the total number of integers we need to add. Again, we have to leave it unanswered at the moment. 22.2 Regular vs irregular sequences A sequence is a list of numbers written in a de nite order. A sequence is regular if each term of the sequence is uniquely determined, following a well-de ned rule, by its position/order in the sequence (often denoted by an integer i). Very often, each term can be generated by an explicit formula that is expressed as a function of the position i, e.g. f(i). We can call this formula the sequence generator or the general term. For example, the ith term in the sequence of integers is identical to its location in the sequence, thus its sequence generator is f(i) = i. Thus, the 9th term is 9 while the 109th term is equal to 109. Example 3: The sum of the rst ten odd numbers is O = 1 + 3 + 5 + + 19: 10 Find the sequence generator. Solution 3: Note that the ith odd number is equal to the ith even number minus 1. The ith even number is simply 2i. Thus, the ith odd number is 2i1, namely f(i) = 2i1. To verify, 5 is the 3rd odd number, i.e. i = 3. Thus, 2i 1 = 2 3 1 = 5 which is exactly the number we expect. Knowing the sequence generator, we can write down the sum of the rst k odd numbers for any positive integer k. O = 1 + 3 + 5 + + (2k 1): k Example 4: Find the sequence generator of the following sum of 100 products of subsequent pairs of integers. P = 1 2 + 2 3 + 3 4 + + 100 101 100 z z z z 1st term 2nd term 3rd term 100th term Solution 4: Since the ith term is equal to the number i multiplied by the subsequent integer which is equal to i + 1. Thus, f(i) = i(i + 1). Knowing the sequence generator, we can write down the sum of k such terms for any positive integer k. P = 1 2 + 2 3 + + k(k + 1): k Example 4: The sequence of the rst 8 digits of the irrational number  = 3:1415926 : : : is  = 3 + 1 + 4 + 1 + 5 + 9 + 2 + 6: 8 We cannot nd the sequence generator since the sequence is irregular, we cannot express the sum of the rst k digits of  (for arbitrary k). 32.3 The sigma notation In order to short-hand the mathematical exression of the sum of a regular sequence, a conve- nient notation is introduced. De nition ( sum): The sum of the rst k terms of a sequence generated by the sequence generator f(i) can be denoted by k X S = f(1) + f(2) + + f(k) f(i) k i=1 where the symbol  (the Greek equivalent of S reads \sigma") means \take the sum of", the general expression for the terms to be added or the sequence generator f(i) is called the summand, i is called the summation index, 1 and k are, respectively, the starting and the ending indices of the sum. Thus, k X f(i) i=1 means calculate the sum of all the terms generated by the sequence generator f(i) for all integers starting from i = 1 and ending at i = k. Note that the value of the sum is independent of the summation index i, hence i is called a \dummy" variable serving for the sole purpose of running the summation from the starting index to the ending index. Therefore, the sum only depends on the summand and both the starting and the ending indices. k P 2 Example 5: Express the sum S = i in an expanded form. k i=3 2 Solution 5: The sequence generator is f(i) = i . Note that the starting index is not 1 but 3. 2 Thus, the 1st term is f(3) = 3 . The subsequent terms can be determined accordngly. Thus, k X 2 2 2 2 2 S = i = f(3) + f(4) + f(5) + + f(k) = 3 + 4 + 5 + + k : k i=3 An easy check for a mistake that often occurs. If you still nd the \dummy" variable i in an expanded form or in the nal evaluation of the sum, your answer must be WRONG. 2.4 Gauss's formula and other formulas for simple sums Let us return to Examples 1 and 2 about the total number of logs in a triangular pile. Let's start with a pile of 4 layers. Imaging that you could (in a \thought-experiement") put an 4identical pile with up side down adjacent to the original pile, you obtain a pile that contains twice the number of logs that you want to calculate (see gure). The advantage of doing this is that, in this double-sized pile, each layer contains an equal number of logs. This number is equal to number on the 1st (top) layer plus the number on the 4th (bottom) layer. In the mean time, the height of the pile remains unchanged (4). Thus, the number in this double-sized pile is 4 (4 + 1) = 20. The sum S is just half of this number 4 which is 10. Let's apply this idea to nding the formula in the case of k layers. Note that (Original) S = 1 + 2 + + k 1 + k: (2) k (Inversed) S = k + k 1 + + 2 + 1: (3) k (Adding the two) 2S = (k + 1) + (k + 1) + + (k + 1) + (k + 1) = k(k + 1): (4) k z k terms in total Dividing both sides by 2, we obtain Gauss's formula for the sum of the rst k positive integers. k X 1 S = i = 1 + 2 + + k = k(k + 1): (5) k 2 i=1 This actaully answered the problem in Example 2. The answer to Example 1b is even simpler. 50 X 1 S = i =  50 51 = 1275: 50 2 i=1 The following are two important simple sums that we shall use later. One is the sum of the rst k integers squared. k X 1 2 2 2 2 S = i = 1 + 2 + + k = k(k + 1)(2k + 1): (6) k 6 i=1 The other is the sum of the rst k integers cubed.   k 2 X 1 3 3 3 3 S = i = 1 + 2 + + k = k(k + 1) : (7) k 2 i=1 We shall not illustrate how to derive these formulas. You can nd it in numerous calculus text books. To prove that these formulas work for arbitrarily large integers k, we can use a method called mathematical induction. To save time, we'll just outline the basic ideas here. 5The only way we can prove things concerning arbitrarily large numbers is to guarantee that this formula must be correct for k = N + 1 if it is correct for k = N. This is like trying to arrange a string/sequence of standing dominos. To guarantee that all the dominos in the string/sequence fall one after the other, we need to guarantee that the falling of each domino will necessarily cause the falling of the subsequent one. This is the essence of proving that if the formula is right for k = N, it must be right for k = N + 1. The last thing you need to do is to knock down the rst one and keep your ngers crossed. Knocking down the rst domino is equivalent to proving that the formula is correct for k = 1 which is very easy to check in all cases (see Keshet's notes for a prove of the sum of the rst k integers squared). 2.5 Important rules of sigma sums Rule 1: Summation involving constant summand. k X c = c + c + + c = kc: z i=1 k terms in total Note that: the total of terms=ending index - starting index +1. Rule 2: Constant multiplication: multiplying a sum by a constant is equal to multiplying each term of the sum by the same constant. k k X X c f(i) = cf(i) i=1 i=1 Rule 3: Adding two sums with identical starting and ending indices is equal to the sum of sums of the corresponding terms. k k k X X X f(i) + g(i) = f(i) + g(i): i=1 i=1 i=1 Rule 4: Break one sum into more than one pieces. k n k X X X f(i) = f(i) + f(i); (1 n k): i=1 i=1 i=n+1 Example 7: Let's go back to solve the sum of the rst k odd numbers. Solution 7: k k k k X X X X O = 1 + 3 + + (2k 1) = (2i 1) = (2i) 1 = 2 i k; k 1 1 1 1 6where Rules 1, 2, and 3 are used in the last two steps. Using the known formulas, we obtain k X 2 O = (2i 1) = k(k + 1) k = k : k 1 Example 8: Let us now go back to solve Example 4. It is the sum of the rst k products of pairs of subsequent integers. Solution 8: k k k k X X X X 2 2 P = 1 2 + 2 3 + + k (k + 1) = i(i + 1) = i + i = i + i; k 1 1 1 1 where Rule 3 was used in the last step. Applying the formulas, we learned k X 1 1 1 P = i(i + 1) = k(k + 1)(2k + 1) + k(k + 1) = k(k + 1)(k + 2): k 6 2 3 1 This is another simple sum that we can easily remember. k P 3 Example 9: Calculate the sum S = (i + 2) . 1 Solution 9: This problem can be solved in two di erent ways. The rst is to expand the 3 summand f(i) = (i + 2) which yield k k k k k X X X X X 3 3 2 3 2 S = (i + 2) = (i + 6i + 12i + 8) = i + 6 i + 12 i + 8k: 1 1 1 1 1 We can solve the resulting three sums separately using the known formulas. But there is a better way to solve this. This involves substituting the summation index. We nd it easier to see how substitution works by expanding the sum. k X 3 3 3 3 3 3 S = (i + 2) = 3 + 4 + + k + (k + 1) + (k + 2) : i=1 3 We see that this is simply a sum of integers cubed. But the sum does not start at 1 and end 3 at k like in the formula for the sum of the rst k integers cubed. Thus we re-write the sum with a sigma notation with an new index called l which starts at l = 3 and ends at l = k + 2 (there is no need to change the symbol for the index, you can keep calling it i if you do not feel any confusion). Thus, k k+2 X X 3 3 3 3 3 3 3 S = (i + 2) = 3 + 4 + + k + (k + 1) + (k + 2) = l : i=1 l=3 7We just nished doing a substitution of the summation index. It is equivalent to replacing i by l = i + 2. This relation also implies that i = 1 ) l = 3 and i = k ) l = k + 2. This is actually how you can determine the starting and the ending values of the new index. Now we can solve this sum using Rule 4 and the known formula.     k k+2 k+2 2 2 2 X X X X 1 1 3 3 3 3 2 3 S = (i+2) = l = l l = (k + 2)(k + 3) 1 2 = (k + 2)(k + 3) 9: 2 2 i=1 l=3 l=1 l=1 2.6 Applications of sigma sum The area under a curve We know that the area of a rectangle with length l and width w is A = w l. rect Starting from this formula we can calculate the area of a triangle and a trapezoid. This is because a triangle and a trapezoid can be transformed into a rectangle (see Figure). Thus, for a triangle of height h and base length b 1 A = hb: trig 2 Similarly, for a trapezoid with base length b, top length t, and height h 1 A = h(t + b): trap 2 Following a very similar idea, the sum of a trapezoid-shaped pile of logs with t logs on top layer, b logs on the bottom layer, and a height of h = b t + 1 layers (see gure) is b X 1 1 i = t + (t + 1) + + (b 1) + b = h(t + b) = (b t + 1)(t + b): (8) 2 2 i=t Now returning to the problem of calculating the area. Another important formula is for the area of a circle of radius r. 2 A = r : circ Now, once we learned sigma and/or integration, we can calculate the area under the curve of any function that is integrable. 2 Example 10: Calculate the area under the curve y = x between 0 and 2 (see gure). Solution 10: Remember always try to reduce a problem that you do not know how to solve into a problem that you know how to solve. 8Let the area be A. Let's divide A into 3 rectangles of equal width w = 2=3. Thus, (   2 2 1 1 2 2 2 2 2 wx + x + x = 0 + + ; left end approx:; 0 1 2 3 3 3 A wh +wh +wh =    1 2 3 2 2 2 2 2 2 1 1 2 3 wx + x + x = + + ; right end approx:: 1 2 3 3 3 3 3 By using these rectangles, we introduced large errors in our estimates using the heights based on both the left and the right endpoints of the subintervals. To increase accurary, we need to increase the number of rectangles by making each one thinner. Let us now divide A into n rectangles of equal width w = 2=n. Thus, using the height based on the right endpoint of each subinterval, we obtain n n n X X X 2 2 A S = A + A + + A = w h = w f(x ) = x : n 1 2 n i i i n i=1 i=1 i=1 It is very important to keep a clear account of the height of each rectangle. In this case, 2 h = f(x ) = x . Thus, the key is nding the x-coordinate of the right-end of each rectangle. i i i For rectangles of of equal width, x = x + iw = x + ix=n, where x is the length of the i 0 0 interval x is the left endpoint of the interval. x = 0 and x = 2 for this example. Therefore, 0 0 x = i(2=n). Substitute into the above equation i   n n 2 n X X X 2 2 2i 8 8 1 4 (n + 1)(2n + 1) 2 2 A S = x : = = i = n(n + 1)(2n + 1) = : n i 3 3 2 n n n n n 6 3 n i=1 i=1 i=1 Note that if we divide the area into in nitely many retangles with a width that is in nitely small, the approximate becomes accurate. 4 (n + 1)(2n + 1) 8 A = lim S = lim = : n 2 n1 n1 3 n 3 The volume of a solid revolution of a curve We know the volume of a rectangular block of height h, width w, and length l is V = w l h: rect The volume of a cylinder of thichness h and radius r 2 V = A  h = r h: cyl cross Example 11: Calculate the volume of the bowl-shaped solid obtained by rotating the curve 2 y = x on 0; 2 about the y-axis. 92.7 The sum of a geometric sequence 3 The De nite Integrals and the Fundamental Theorem 3.1 Riemann sums De nition 1: Suppose f(x) is nite-valued and piecewise continuous on a; b. Let P = fx = a; x ; x ; : : :; x = bg be a partition of a; b into n subintervals I = x ; x of width 0 1 2 n i i1 i x = x x , i = 1; 2; : : :; n. (Note in a special case, we can partition it into subintervals i i i1  of equal width: x = x x = w = (b a)=n for all i). Let x be a point in I such that i i i1 i i    x  x  x . (Here are some special ways to choose x : (i) left endpoint rule x = x , i1 i i1 i i i  and (ii) the right endpoint rule x = x ). i i The Riemann sums of f(x) on the interval a; b are de ned by: n n X X  R = h x = f(x )x n i i i i i=1 i=1 which approximate the area between f(x) and the x-axis by the sum of the areas of n thin rectangles (see gure). x Example 1: Approximate the area under the curve of y = e on 0; 1 by 10 rectangles of equal width using the left endpoint rule. Solution 1: We are actually calculating the Riemann sum R . The width of each rectangle 10  is x = w = 1=10 = 0:1 (for all i). The left endpoint of each subinterval is x = x 1 = i i i (i 1)=10. Thus, 10 10 9 X X X 1 e  (i1)=10 j=10 R = f(x )x = e (0:1) = 0:1 e = 0:1  1:6338: 10 i i 1=10 1 e i=1 i=1 j=0 3.2 The de nite integral De nition 1: Suppose f(x) is nite-valued and piecewise continuous on a; b. Let P =fx = 0 a; x ; x ; : : :; x = bg be a partition of a; b with a length de ned by jpj = Max fxg 1 2 n 1in i (i.e. the longest of all subintervals). The de nite integral of f(x) on a; b is b Z n X  f(x)dx = lim R = lim f(x )x n i i jpj0; n1 jpj0; n1 i=1 a R where the symbol means \to integrate", the function f(x) to be integrated is called the integrand, x is called the integration variable which is a \dummy" variable, a and b are, 10respectively, the lower (or the starting) limit and the upper (or the ending) limit of the integral. Thus, b Z f(x)dx a means integrate the function f(x) starting from x = a and ending at x = b. 2 Example 2: Calculate the de nite integral of f(x) = x on 0:2. (This is Example 10 of Lecture 1 reformulated in the form of a de nite integral). Solution 2: 2       Z n n 2 X X 2i 2 8 4 (n + 1)(2n + 1) 8 2 2 I = x dx = lim = lim i = lim = : 3 2 n1 n1 n1 n n n 3 n 3 i=1 i=1 z z 0  x f(x ) i i Important remarks on the relation between an area and a de nite integral:  An area, de ned as the physical measure of the size of a 2D domain, is always non- negative.  The value of a de nite integral, sometimes also referred to as an \area", can be both positive and negative.  This is because: a de nite integral = the limit of Riemann sums. But Riemann sums are de ned as n n X X th  R = (area of i rectangle) = f(x ) x : n i i z z i=1 i=1 width height   Note that both the hieght f(x ) and the width x can be negative implying that R i n i can have either signs. 3.3 The fundamental theorem of calculus 3.4 Areas between two curves 114 Applications of De nite Integrals: I 4.1 Displacement, velocity, and acceleration 4.2 Density and total mass 4.3 Rates of change and total change 4.4 The average value of a function 125 Di erentials De nition: The di erential, dF, of any di erentiable function F is an in nitely small incre- ment or change in the value of F. Remark: dF is measured in the same units as F itself. Example: If x is the position of a moving body measured in units of m (meters), then its di erential, dx, is also in units of m. dx is an in nitely small increment/change in the position x. Example: If t is time measured in units of s (seconds), then its di erential, dt, is also in units of s. dt is an in nitely small increment/change in time t. 2 Example: If A is area measured in units of m (square meters), then its di erential, dA, also 2 is in units of m . dA is an in nitely small increment/change in area A. 3 Example: If V is volume measured in units of m (cubic meters), then its di erential, dV , 3 also is in units of m . dV is an in nitely small increment/change in volume V . Example: If v is velocity measured in units of m=s (meters per second), then its di erential, dv, also is in units of m=s. dv is an in nitely small increment/change in velocity v. Example: If C is the concentration of a biomolecule in our body uid measured in units of 23 M (1 M = 1 molar = 1 mole=litre, where 1 mole is about 6:023 10 molecules), then its di erential, dC, also is in units of M. dC is an in nitely small increment/change in the concentration C. Example: If m is the mass of a rocket measured in units of kg (kilograms), then its di eren- tial, dm, also is in units of kg. dm is an in nitely small increment/change in the mass m. Example: If F(h) is the culmulative probability of nding a man in Canada whose height is smaller than h (meters), then dF is an in nitely small increment in the probability. De nition: The derivative of a function F with respect to another function x is de ned as the quotient between their di erentials: 13dF an infinitely small rise in F = : dx an infinitely small run in x Example: Velocity as the rate of change in position x with respect to time t can be expressed as infinitely small change in position dx v = = : infinitely small time interval dt Remark: Many physical laws are correct only when expressed in terms of di erentials. Example: The formula (distance) = (velocity) (time interval) is true either when the velocity is a constant or in terms of di erentials, i.e., (an infinitely small distance) = (velocity) (an infinitely short time interval). This is because in an in nitely short time interval, the velocity can be considered a constant. Thus, dx dx = dt = v(t)dt; dt which is nothing new but v(t) = dx=dt. Example: The formula (work) = (force) (distance) is true either when the force is a constant or in terms of di erentials, i.e., (an infinitely small work) = (force) (an infinitely small distance). This is because in an in nitely small distance, the force can be considered a constant. Thus, dx dW = fdx = f dt = f(t)v(t)dt; dt which simply implies: (1) f = dW=dx, i.e., force f is nothing but the rate of change in work W with respect to distance x; (2) dW=dt = f(t)v(t), i.e., the rate of change in work W with respect to time t is equal to the product between f(t) and v(t). Example: The formula (mass) = (density) (volume) is true either when the density is constant or in terms of di erentials, i.e., (an infinitely small mass) = (density) 14(volume of an infinitely small volume). This is because in an in nitely small piece of volume, the density can be considered a constant. Thus, dm = dV; which implies that  = dm=dV , i.e., density is nothing but the rate of change in mass with respect to volume. 6 The Chain Rule in Terms of Di erentials When we di erentiate a composite function, we need to use the Chain Rule. For example, 2 x f(x) = e is a composite function. This is because f is not an exponential function of x but 2 it is an exponential function of u = x which is itself a power function of x. Thus, 2 x u u df de de du de du 2 u 2 0 x = = = ( )( ) = e (x ) = 2xe ; dx dx dx du du dx 2 where a substitution u = x was used to change f(x) into a true exponential function f(u). Therefore, the Chain Rule can be simply interpreted as the quotient between df and dx is equal to the quotient between df and du multiplied by the quotient between du and dx. Or simply, divide df=dx by du and then multiply it by du. 2 2 However, if we simply regard x as a function di erent from x, the actual substitution u = x becomes unnecessary. Instead, the above derivative can be expressed as 2 2 2 x x 2 x 2 de de dx de dx 2 2 x 2 0 x = = ( )( ) = e (x ) = 2xe : 2 2 dx dx dx dx dx Generally, if f = f(g(x)) is a di erentiable function of g and g is a di erentiable function of x, then df df dg = ; dx dg dx where df, dg, and dx are, respectively, the di erentials of the functions f, g, and x. 2 x Example: Calculate df=dx for f(x) = sin(ln(x + e )). Solution: f is a composite function of another composite function 152 x 2 x 2 x df d sin(ln(x + e )) d ln(x + e ) d(x + e ) 1 2 x x = = cos(ln(x + e )) (2x + e ): 2 x 2 x 2 x dx d ln(x + e ) d(x + e ) dx (x + e ) 7 The Product Rule in Terms of Di erentials The Product Rule says if both u = u(x) and v = v(x) are di erentiable functions of x, then d(uv) du dv = v + u : dx dx dx Multiply both sides by the di erential dx, we obtain d(uv) = vdu + udv; which is the Product Rule in terms of di erentials. 2 2 sin(x ) Example: Let u = x and v = e , then 2 2 sin(x ) 2 2 sin(x ) d(uv) = vdu + udv = e d(x ) + x d(e ) 2 2 2 sin(x ) 2 sin(x ) 2 sin(x ) 2 2 = 2xe dx + x e cos(x )(2x)dx = 2xe 1 + x cos(x )dx: 8 Other Properties of Di erentials 0 0 1. For any di erentiable function F(x), dF = F (x)dx (Recall that F (x) = dF=dx). 2. For any constant C, dC = 0. 0 3. for any constant C and di erentiable function F(x), d(CF) = CdF = CF (x)dx. 4. For any di erentiable functions u and v, d(u v) = du dv. 169 The Fundamental Theorem in Terms of Di erentials Fundamental Theorem of Calculus: If F(x) is one antiderivative of the function f(x), i.e., 0 F (x) = f(x), then Z Z Z 0 f(x)dx = F (x)dx = dF = F(x) + C: Thus, the integral of the di erential of a function F is equal to the function itself plus an arbitrary constant. This is simply saying that di erential and integral are inverse math operations of each other. If we rst di erentiate a function F(x) and then integrate the 0 derivative F (x) = f(x), we obtain F(x) itself plus an arbitrary constant. The opposite also is true. If we rst integrate a function f(x) and then di erentiate the resulting integral F(x)+C, 0 we obtain F (x) = f(x) itself. Example: Z Z 6 6 x x 5 x dx = d( ) = + C: 6 6 Example: Z Z x x x e dx = d(e ) =e + C: Example: Z Z sin(3x) sin(3x) cos(3x)dx = d( ) = + C: 3 3 Example: Z Z 2 sec xdx = dtanx = tanx + C: Example: Z Z sinh(3x) sinh(3x) cosh(3x)dx = d( ) = + C: 3 3 17Example: Z Z 1 1 1 dx = dtan x = tan + C: 2 1 + x Example: Z Z Z 1 1 2 dx = dx = cosh xdx 2 2 1 tanh x sech x Z Z 1 + cosh(2x) 1 sinh(2x) 1 sinh(2x) = dx = d(x + ) = x + + C: 2 2 2 2 2 For more inde nite integrals involving elementary functions, look at the rst page of the table of integrals provided at the end of the notes. 10 Integration by Subsitution Substitution is a necessity when integrating a composite function since we cannot write down the antiderivative of a composite function in a straightforward manner. Many students nd it dicult to gure out the substitution since for di erent functions the subsitutions are also di erent. However, there is a general rule in substitution, namely, to change the composite function into a simple, elementary function. R p p Example: (sin( x)= x)dx. p Solution: Note that sin( x) is not an elementary sine function but a composite function. The p rst goal in solving this integral is to change sin( x) into an elementary sine function through p 0 substitution. Once you realize this, u = x is an obvious subsitution. Thus, du = u dx = p 1 p dx, or dx = 2 xdu = 2udu. Substitute into the integral, we obtain 2 x Z p Z Z Z p sin( x) sin(u) p dx = 2udu = 2 sin(u)du =2 dcos(u) =2cos(u)+C =2cos( x)+C: x u Once you become more experienced with subsitutions and di erentials, you do not need to do p 2 the actual substitution but only symbolically. Note that x = ( x) , 18p p p Z Z Z Z p p p p p p sin( x) sin( x) sin( x) 2 p dx = p d( x) = p 2 xd x = 2 sin( x)d x =2cos( x)+C: x x x p Thus, as soon as you realize that x is the substitution, your goal is to change the di erential p p in the integral dx into the di erential of x which is d x. If you feel that you cannot do it without the actual substitution, that is ne. You can always do the actual substitution. I here simply want to teach you a way that actual subsitution is not a necessity R 4 5 Example: x cos(x )dx. 5 Solution: Note that cos(x ) is a composite function that becomes a simple cosine function only 5 0 4 4 1 if the subsitution u = x is made. Since du = u dx = 5x dx, x dx = du. Thus, 5 Z Z 5 1 sin(u) sin(x ) 4 5 x cos(x )dx = cos(u)du = + C = + C: 5 5 5 Or alternatively, Z Z 1 1 4 5 5 5 5 x cos(x )dx = cos(x )dx = sin(x ) + C: 5 5 R p 2 Example: x x + 1dx. p 2 1=2 2 2 Solution: Note that x + 1 = (x + 1) is a composite function. We realize that u = x + 1 1 0 is a substitution. du = u dx = 2xdx implies xdx = du. Thus, 2 Z Z 2 3=2 p p 1 1 2 (x + 1) 3=2 2 x x + 1dx = udu = u + C = + C: 2 2 3 3 Or alternatively, Z Z 2 3=2 p p 2 (x + 1) 1 1 2 2 3=2 2 2 x x + 1dx = x + 1d(x + 1) = (x + 1) + C = + C: 2 2 3 3 19In many cases, substitution is required even no obvious composite function is involved. R Example: (ln x=x)dx (x 0). Solution: The integrand ln x=x is not a composite function. Nevertheless, its antiderivative is not obvious to calculate. We need to gure out that (1=x)dx = d lnx, thus by introducing the substitution u = lnx, we obtained a di erential of the function ln x which also appears in the integrand. Therefore, Z Z 2 u 1 2 (ln x=x)dx = udu = + C = (lnx) + C: 2 2 It is more natural to consider this substitution is an attempt to change the di erential dx into something that is identical to a function that appears in the integrand, namely d lnx. Thus, Z Z 1 2 (ln x=x)dx = ln xd ln x = (ln x) + C: 2 R Example: tanxdx. Solution: tanx is not a composite function. Nevertheless, it is not obvious to gure out tan x is the derivative of what function. However, if we write tanx = sin x= cos x, we can regard 1= cosx as a composite function. We see that u = cosx is a candidate for substitution and 0 du = u dx =sin(x)dx. Thus, Z Z Z sinx du tanxdx = dx = = lnjuj + C = lnjcosxj + C: cosx u Or alternatively, Z Z Z sinx dcosx tanxdx = dx = = lnjcosxj + C: cosx cosx Some substitutions are standard in solving speci c types of integrals. 2 2 1=2 Example: Integrands of the type (a x ) . 2 In this case both x = asinu and x = acosu will be good. x = a tanhu also works (1tanh u = 2 sech u). Let's pick x = asinu in this example. If you ask how can we nd out that x = asinu 20