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1 Indefinite Integrals November 16, 2019 MAT186 { Week 7 Justin Ko 1 Indefinite Integrals Definition 1. The antiderivative of f(x) on [a; b] is a function F (x) such that F 0(x) = f(x) for all x 2 [a; b]. The antiderivative is not unique since for any antiderivative F (x) of f(x), F (x) + C is also an antiderivative for all integration constants C 2 R. We will use the notation Z f(x) dx = F (x) + C called the indefinite integral of f to denote the family of all antiderivatives of f. 1.1 Table of Integrals (omitting the integration constant) Elementary Functions Z cot(x) dx = ln j sin(x)j (10) Z 1 xa dx = · xa+1; a 6= −1 (1) a + 1 Z Z 1 dx = ln jxj (2) sec(x) dx = ln j tan(x) + sec(x)j (11) x Exponential and Logarithms Z Z csc(x) dx = − ln j cot(x) + csc(x)j (12) ex dx = ex (3) Rational Functions Z ln(x) dx = x ln(x) − x (4) Z 1 dx = tan−1(x)(13) 1 + x2 Z ax ax dx = (5) Z 1 ln(a) p dx = sin−1(x)(14) 1 − x2 Z x ln(x) − x log (x) dx = (6) Z 1 a ln(a) p dx = sec−1(jxj)(15) x x2 − 1 Trigonometric Functions Hyperbolic Functions Z Z sin(x) dx = − cos(x)(7) sinh(x) dx = cosh(x)(16) Z Z cos(x) dx = sin(x)(8) cosh(x) dx = sinh(x)(17) Z Z tan(x) dx = − ln j cos(x)j (9) tanh(x) dx = ln j cosh(x)j (18) 1.2 Basic Property For constants a; b 2 R and integrable functions f and g, we have Linearity: Z Z Z (af(x) + bg(x)) dx = a f(x) dx + b g(x) dx: We will introduce more basic properties in the coming weeks. Page 1 of 10 November 16, 2019 MAT186 { Week 7 Justin Ko 1.3 Example Problems 1.3.1 Finding Indefinite Integrals Problem 1.1. (?) Find the indefinite integral Z e−7x dx: 1 −7x −7x Solution 1.1. It is easy to check that − 7 · e is an antiderivative of e . Therefore, Z 1 e−7x dx = − · e−7x + C: 7 Problem 1.2. (??) Find the indefinite integral Z x3 + 8 dx: x Solution 1.2. Using the linearity property, we have Z x3 + 8 Z Z 1 1 dx = x2 dx + 8 dx = x3 + 8 ln jxj + C: x x 3 1.3.2 Checking Antiderivatives Strategy: To check if a function F (x) is an antiderivative of f, it suffices to just differentiate F and check if F 0(x) = f(x). Problem 1.3. (??) Check that both − sin−1(x) and cos−1(x) are antiderivatives of −1 p on (−1; 1): 1 − x2 Show that π cos−1(x) + sin−1(x) = : 2 Solution 1.3. From the table of derivatives, we have d −1 d −1 (− sin−1(x)) = p and cos−1(x) = p : dx 1 − x2 dx 1 − x2 so both − sin−1(x) and cos−1(x) are antiderivatives of p −1 on its domain. We know that all an- 1−x2 tiderivatives on an interval differ by an integration constant, that is cos−1(x) = − sin−1(x) + C for x 2 (−1; 1). To find this integration constant, we can evaluate our functions at 0, π π cos−1(0) = and − sin−1(0) = 0 =) cos−1(0) = − sin−1(0) + C =) C = : 2 2 Since the difference of any two antiderivatives is a constant, we conclude that the two antiderivatives π must differ by 2 for all x in the domain, π π cos−1(x) = − sin−1(x) + =) cos−1(x) + sin−1(x) = : 2 2 Page 2 of 10 November 16, 2019 MAT186 { Week 7 Justin Ko 2 Differential Equations 2.1 Separable Differential Equations Suppose that we want to find the antiderivative of some special implicit functions. In particular, suppose that we want to find a function y such that it satisfies the differential equation dy = f(x)g(y) dx for some functions f(x) and g(y). We can find the solution by separating variables and integrating both sides, dy dy Z 1 Z = f(x)g(y) =) = f(x)dx =) dy = f(x) dx =) G(y) = F (x) + C dx g(y) g(y) 1 where G(y) is the antiderivative of g(y) and F (x) is the antiderivative of f(x). This procedure gives an implicit formula formula for a function y that satisfies the differential equation. If we are given an initial condition, then we can solve for the integration constant C. 2.2 Example Problems 2.2.1 Initial Value Problems Strategy: We first separate variables and integrate to recover a general form of the solution in terms of some yet to be determined integration constants. Next, we plug in the initial conditions to solve for the integration constants. Problem 2.1. (?) Suppose the velocity of a particle is given by v(t) = sin(t) − cos(t): Find the position function s(t) of the particle given that s(0) = 0. Solution 2.1. The rate of change of position is velocity, so we have the differential equation ds v(t) = = sin(t) − cos(t): dt Separating variables and integrating, we have ds Z Z = sin(t) − cos(t) ) ds = sin(t) − cos(t) dt ) s = − cos(t) − sin(t) + C: dt To solve for the integrating constant, since s(0) = 0, we have 0 = s(0) = − cos(0) − sin(0) + C = −1 + C ) C = 1: Therefore, the position is given by s(t) = − cos(t) − sin(t) + 1: Problem 2.2. (??) Suppose the acceleration of a particle is given by a(t) = t + 1: Find the position function s(t) of the particle given that s(0) = 0 and s(1) = 2. Page 3 of 10 November 16, 2019 MAT186 { Week 7 Justin Ko Solution 2.2. The rate of change of velocity is acceleration, so we have the differential equation dv = t + 1: dt Separating variables and integrating, we have dv Z Z t2 = t + 1 ) dv = t + 1 dt ) v = + t + C: dt 2 The rate of change of position is velocity, so using the fact above, we have the differential equation ds t2 = + t + C: dt 2 Separating variables and integrating, we have ds t2 Z Z t2 t3 t2 = + t + C ) ds = + t + C dt ) s = + + Ct + D: dt 2 2 6 2 To solve for the integrating constants C and D, since s(0) = 0, we have 0 = s(0) = D ) D = 0: And since s(1) = 2, we have 1 1 1 1 4 2 = s(1) = + + C ) C = 2 − − = 6 2 6 2 3 Therefore, the position is given by t3 t2 4t s(t) = + + : 6 2 3 2.2.2 Growth and Decay Problems Problem 2.3. (???) The rate of growth of a population P is modeled by dP = kP dt where k 6= 0. Suppose that the initial population P (0) = P0 for some constant P0 > 0. Find the population function P (t). How long will it take for the population to double? Solution 2.3. Separating variables and integrating, we have dP Z dP Z = kP ) = kdt ) ln jP j = kt + C dt P Since P (t) must be positive in a reasonable model, jP j = P , so we can exponentiate both sides to conclude P = ekt+C = eC ekt: To solve for the integrating constants C, since P (0) = P0, we have C k·0 C P0 = P (0) = e e = e ) ln P0 = C: Therefore, the population is given by ln P0 kt kt P (t) = e e = P0e : To find the time for the population to double, we want to find the t such that P (t) = 2P0. That is, ln(2) 2P = P ekt ) ekt = 2 ) kt = ln(2) ) t = : 0 0 k In particular, if k < 0 (we have a decreasing population) then we have our population will never double. Page 4 of 10 November 16, 2019 MAT186 { Week 7 Justin Ko Problem 2.4. (???) A pizza is put in a 200◦C oven and heats up according to the differential equation dH = −k(H − 200); where k > 0: dt The pizza is put in the oven at 20◦C and is removed 30 minutes later at a temperature of 120◦C. Find the proportionality constant k. Solution 2.4. The general solution to the differential equation is dH = −k(H − 200); dt can be solved using separation of variables, dH Z dH Z = −k(H − 200) ) = −kt dt ) ln jH − 200j = −kt + C dt (H − 200) Solving for H and using the fact H − 200 < 0 in a reasonable model, we see that ln jH − 200j = −kt + C ) ln(200 − H) = −kt + C ) H = 200 − De−kt where D = eC > 0. To find D, we can use the fact that at H(0) = 20, 20 = H(0) = 200 − D =) D = 180: Since H(30) = 120, we have 1 80 120 = H(30) = 200 − 180 · e−30k =) k = − · ln ≈ 0:027: 30 180 Problem 2.5. (???) Find the general form of a function y(x) such that dy = yx: dx Solution 2.5. Separating variables and integrating, we have dy Z 1 Z x2 = yx ) dy = x dx ) ln jyj = + A; dx y 2 for some constant A. Solving for y we get 2 2 x +A A x jyj = e 2 ) y = ±e e 2 : If we define the non-zero constant B = ±eA, then we have x2 y = Be 2 ; where B is some non-zero constant: dy However, notice that y ≡ 0 also satisfies dx = yx, so y = 0 is also a solution.
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