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Math 1432 (Calculus Ii) Lecture Notes Invitation-Only Section MATH 1432 (CALCULUS II) LECTURE NOTES INVITATION-ONLY SECTION VAUGHN CLIMENHAGA Contents I Integration 1 1 Integration by parts 1 2 More on integration by parts 2 3 Trigonometric integrals 5 4 More trigonometric integrals 8 5 Trigonometric substitutions 13 6 General quadratics, and rational functions 16 7 General partial fraction decompositions 21 8 Overview of integration strategies 27 9 Two more numerical strategies 30 10 Improper integrals 32 11 Another type of improper integral 34 II Applications of integration 38 12 Arc length and the catenary 38 13 Examples of arc length 41 14 Surface area 43 15 More applications 45 16 Center of mass 49 17 Fluid flow; probability 53 III Differential equations 58 18 Ideas and examples 58 19 Approximate solutions; separable DEs 61 20 Other population models 64 21 More DEs, separable and otherwise 67 22 Linear DEs, coupled DEs 70 IV Parametric curves, polar coordinates, and conic sections 75 23 Parametric curves 75 24 Calculus with parametrizations 78 25 Geometry of parametric curves 80 26 Polar coordinates 83 Date: April 26, 2019. i ii VAUGHN CLIMENHAGA 27 Calculus with polar coordinates 87 28 Arc length and introduction to conic sections 89 29 Equivalence of the parabola's definitions 92 30 Dandelin spheres 94 31 Ellipses (and hyperbolas) 97 32 Kepler and Newton 102 33 Completion of proofs of Kepler's laws 105 V Sequences and series 108 34 More on sequences, and the start of series 109 35 Examples and basic results 113 36 The integral test 117 37 Comparison tests and alternating series 119 38 Absolute convergence, ratio and root tests 122 39 Review of convergence; power series 125 40 More on power series 128 41 More with derivatives and integrals 130 42 Convergence of Taylor series 133 1 Part I. Integration Lecture 1 Integration by parts Date: Monday, January 14 Now that we have developed the basic theory of integration, especially the funda- mental theorem of calculus, we know that we can compute definite integrals by finding antiderivatives, or indefinite integrals as we tend to call them now. But so far our toolkit for finding these is rather small: we can look at a list of known examples, we can try various u-substitutions, and . that's about it. If that doesn't work { and we saw several examples for which it does not { then we are stuck. We found the substitution rule for integrals by looking at the chain rule for derivatives, and exploring its consequences for integrals. We can also do this with the product rule, which says that if f; g are differentiable functions, then d f(x)g(x) = f(x)g0(x) + g(x)f 0(x): dx This can be rewritten as Z d d f(x)g0(x) = f(x)g(x) g(x)f 0(x) = f(x)g(x) g(x)f 0(x) dx ; dx − dx − and we conclude that Z Z (1.1) f(x)g0(x) dx = f(x)g(x) g(x)f 0(x) dx: − We do not write a constant of integration because the right-hand side still contains an indefinite integral. The relationship (1.1) is called integration by parts and is a powerful tool for evaluating many integrals, especially when f; g can be chosen so that gf 0 is easier to integrate than fg0. Example 1.1. Suppose we want to evaluate R x cos x dx. Then we might try f(x) = x and g0(x) = cos x; to get this, we should put g(x) = sin x, and then (1.1) gives Z Z x cos x dx = x sin x (sin x) 1 dx = x sin x ( cos x)+C = x sin x+cos x+C: − | {z } · |{z}0 − − g(x) f (x) And indeed, we can verify this by differentiating and using the product rule: d (x sin x + cos x) = (sin x + x cos x) sin x = x cos x: dx − Remark 1.2. Since antiderivatives are only determined up to a constant, the fact that g0(x) = cos x actually only tells us that g(x) = sin x + C for some C. You can check that using this g(x) still gives us the same answer. Because we can choose g(x) to be any antiderivative of g0(x), we may as well choose it to be the antiderivative that is the simplest to write down, which usually happens when we put C = 0. 2 Remark 1.3. As with the substitution rule, not all choices are helpful! For example, if 1 2 we put f(x) = cos x and g0(x) = x in the example above, we would get g(x) = 2 x and f 0(x) = sin x, so − Z 1 Z 1 1 Z x cos x dx = x2 cos x x2( sin x) dx = x2 cos x + x2 sin x dx : 2 − 2 − 2 We have done nothing wrong { the equation we derived is true { but we have not done anything helpful, either, since we do not know how to evaluate R x2 sin x dx. We pause a moment to recall that we can write the chain rule in an alternate form by writing u = g(x) and y = f(u) = f(g(x)), so that we have the following diagram: f g ◦ x u = g(x) y = f(u) = f(g(x)) g f dy dy du Then the chain rule becomes the very sensible-looking equation dx = du dx . Remark 1.4. It looks like we are simply cancelling the two appearances of the term du, but this is not quite right; we have not given any independent meaning to the symbols dy R dy, du, and dx outside of a derivative like dx , or an integral like f(x) dx. Thus this should be regarded as a bookkeeping tool more than anything else; however, it is in some ways easier to remember, and the fact that it conforms to our expectation of how dy fractions should behave suggests that the notation dx is appropriate to use. A similar bookkeeping tool is useful for integration by parts. Using the notation u = f(x) and v = g(x), we write du = f 0(x) dx and dv = g0(x) dx (despite the fact that dx, du, and dv have no independent meaning in their own right!) and rewrite (1.1) in the following form, which is easier to remember: Z Z (1.2) u dv = uv v du: − In Example 1.1 we would put u = x, du = dx, dv = cos x dx, and v = sin x, obtaining the same result as before. R 1 Example 1.5. To evaluate ln x dx, put u = ln x, dv = dx, du = x dx, and v = x: Z Z 1 Z ln x dx = x ln x x dx = x ln x 1 dx = x ln x x + C: |{z} |{z} |{z} |{z} − |{z} x − − u dv v u v |{z} du Lecture 2 More on integration by parts Date: Wednesday, January 16 3 2.1. Iterated integration by parts Example 2.1. To evaluate R t2et dt, we can put u = t2 and dv = et dt, so du = 2t dt and v = et, giving Z Z (2.1) t2et dt = t2et 2tet dt : |{z} − uv | {z } R v du To evaluate the last integral we use integration by parts a second time; bring out the factor of 2 and compute R tet dt by putting u = t, dv = et dt, du = dt, v = et, giving Z Z tet dt = tet et dt = tet et: − − Using this in (2.1) gives Z Z t2et dt = t2et 2 tet dt = t2et 2(tet et) + C = t2et 2tet + 2et + C: − − − − Exercise 2.2. Follow this same approach to show that if f(t) is a polynomial of degree n, then using integration by parts n times gives Z t n (n) t f(t)e dt = f(t) f 0(t) + f 00(t) + ( 1) f (t) e + C: − − · · · − Sometimes by using integration by parts multiple times, we end up with an expression that does not yield the integral directly, but which gives an equation that can be solved for it. This is best illustrated with an example. Example 2.3. To evaluate R ex sin x dx, we integrate by parts twice: Z Z ex sin x dx = ex cos x + ex cos x dx (u = ex and dv = sin x dx) − Z = ex cos x + ex sin x ex sin x dx (u = ex and dv = cos x dx): − − Since this last expression contains the original integral, one might at first think that we have gotten nowhere. But in fact, we are nearly done! Adding R ex sin x dx to both sides of the equation gives Z Z 1 2 ex sin x dx = ex(sin x cos x) ex sin x dx = ex(sin x cos x) + C; − ) 2 − where we add a constant of integration to get the most general antiderivative. 2.2. Definite integrals By the FTC, (1.1) has a counterpart for definite integrals: Z b Z b b f(x)g0(x) dx = f(x)g(x) g(x)f 0(x) dx a − (2.2) a a Z b = f(b)g(b) f(a)g(a) g(x)f 0(x) dx: − − a 4 R 1 1 1 Example 2.4. To evaluate 0 tan− x dx, we put f(x) = tan− x and g0(x) = 1, so 1 g(x) = x and f 0(x) = 1+x2 , giving Z 1 Z 1 1 1 1 x 2 − − tan x dx = x tan x 0 2 dx (then substitute u = 1 + x ) 0 − 0 1 + x π 1 Z 2 1 = du (using du = 2x dx) 4 − 2 1 u π ln 2 = : 4 − 2 R 1 1 Remark 2.5.
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