# week1.pdf - 1 Week 1 Lecture Notes 1.1 u2022 Review of...

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Unformatted text preview: 1 Week 1 Lecture Notes 1.1 • Review of Integration We start our course with a nice fresher on calculating integrals including the method of substitution   It is assumed that you are familair sections 5.1, 5.2, 5.3, 5.5 from MATA30 Even though its from MATA30, make sure to ask us or the TAs if there is anything from this content you have questions about! • After the review we will begin our study of more advanced techniques of integration • Let • We dene f (x) such that be a continuous function R f (x) dx to be F (x) = f (x) the 0 indenite integral of f (x) and it is dened as F (x)  F (x) is also called the antiderivative of f (x) R d (F (x) + C) =  Generally we say f (x) dx = F (x)+C for a constant C because dx F 0 (x) also  f (x) is called the integrand of the integral • We also dene Rb a f (x) dx to be the denite integral of f (x) over the closed interval [a, b]   The fundamental theorem of calculus says that F (a) if F 0 (x) = f (x) We can interpret a denite integral as the Example 1. Evaluate R x2 dx and R2 0 Rb a f (x) dx = F (x)|ba = F (b) − area under the curve of a function x2 dx Solution 1. • Since d dx therefore • Z  1 3 x 3  = x2 1 x2 dx = x3 + C 3 Then, Z 0 2 x2 dx =  23 1 3 2 1 3 x 0= 2 − 03 = 3 3 3 1 • If we wanted to compute R f (x) dx, we could think hard to nd an antiderivative, but in general this is not feasible and we need some techniques to help us • QUESTION:  • What is one technique you already know? Substitution Let's review the substitution rule  u = g (x) for some f (g (x)) g (x) dx equals f (x) dx R R  Then f (u) du = f (x) dx  We substitute 0 We think of ∗ Example 2. u = g (x) function du = g 0 (x) dx implying g (x) we pick such that f (u) du = by dierentiating both sides This is informal but that's OK Evaluate R x3 cos (x4 + 2) dx Solution 2. u = x4 + 2 • Let • Then 3 • Therefore be the inside function d (x4 + 2) dx = 4x3 dx du = dx x cos (x4 + 2) dx = 41 cos udu Z Example 3. 1 x cos x + 2 dx = 4 3 4 Evaluate  R π/2 0 so that cos udu = cos (x4 + 2) · 4x3 dx Z cos udu = and hence  1 1 sin u + C = sin x4 + 2 + C 4 4 sin x cos xdx Solution 3. • Let • Since there are integration limits and we made a substitution, we have to update them:  • u = sin x so that Thinking of du = cos xdx and so udu = sin x cos xdx u as a function of x, get u (0) = sin 0 = 0 and u (π/2) = sin (π/2) = 1 So the above integral equals Z 1 udu = 0 Example 4. Evaluate R π/3 0 1 2 1 1 u 0= 2 2 sec3 x tan xdx Solution 4. • Here we have to be careful with our choice of substitution 2 • Recall that d dx • So if we let u = sec x tan x = sec2 x we get and d dx sec x = d 1 dx cos x du = sec x tan x = sec x tan x and then sec3 x tan xdx = sec2 x sec x tan xdx = u2 du • Since u (0) = 1/ cos 0 = 1 • QUESTION:  1.2 u (π/3) = 1/ (1/2) = 2 therefore 2 Z 2  1 3 1 3 2 u du = u = 2 −1 3 1 3 1 and Why wouldn't the substitution Would get du = sec2 xdx u = tan x get work? but then there is one remaining sec x, not good! Techniques of Integration: Integration by Parts • Do the examples in Section 7.1 of the book • Recall the product rule for dierentiation: • Integrating this yields • If we move some things around, we get what is called the (f (x) g (x)) = f 0 (x) g (x) + f (x) g 0 (x) R R f (x) g (x) = f 0 (x) g (x) dx + f (x) g 0 (x) dx d dx integration by parts formula Theorem 1. (Integration by parts, indenite) 1. f (x) g 0 (x) dx = f (x) g (x) − R R 2. udv = uv − vdu • R R g (x) f 0 (x) dx The idea of integration by parts:   We use it on integrals of the form We choose u, dv from R f (x) g (x) dx f (x) g (x) dx so that udv = f (x) g (x) dx ∗ Either we choose u rst and then dv is the remaining terms ∗ Or we choose dv rst and u is the remaining terms ∗ dv must have dx R  Then we compute du = du dx and v = dv and apply the formula dx R  The resulting vdu from the formula should help us in solving our integral integral R Example 5. Evaluate xex dx Solution 5. • Step 1: Identify u, dv 3  Let's try u = x, dv = ex dx (remaining terms) so that udv = xex dx  NOTE: We are choosing dv and not v when applying the formula • Step 2: Apply integration by parts   Get d dx du = (x) dx = 1dx = dx v= R dv = R ex dx = ex Then Z Z x xe dx = • and Z udv = uv − Z x ex dx = xex − ex + C vdu = xe − The following examples will cover the basic tricks you need to know Example 6. (Repeated integration by parts) Evaluate R t2 sin tdt Solution 6. u = t2 , dv = sin tdt (remaining terms) so that udv = t2 sin tdt R v = sin tdt = − cos t and du = dtd (t2 ) dt = 2tdt • Let's try • Then • Get Z Z 2 t sin tdt = • Note that  • For • Get R Z Z Z 2 2 udv = uv− vdu = t (− cos t)− (− cos t) (2t) dt = −t cos t+2 t cos tdt t cos tdt is similar to our original integral, but it is actually simpler In situations like these, we just use integration by parts again! R t cos tdt, choose Z Z t cos tdt = • u = t, dv = cos tdt, so v = sin t Z udv = uv − and du = d dt Z vdu = t sin t − sin tdt = t sin t + cos t Therefore, Z Example 7. t2 sin tdt = −t2 cos t + 2 (t sin t + cos t) + C (Getting back where you started) Evaluate Solution 7. u = ex , dv = cos xdx (remaining terms) R v = cos xdx = sin x and du = ex dx • Let's try • Then (t) dt = dt 4 R ex cos xdx • Get Z Z x Z udv = uv − e cos xdx = • vdu = e sin x − so Get v = − cos x Z and x Z x  Is this bad or good? It's good! Substituting yields Z • ex cos xdx The last integral is the original integral that we started with! • QUESTION: • sin xex dx du = ex dx e sin xdx = −e cos x + • Z Let's try repreated integration by parts • u = ex , dv = sin xdx, • x   Z x x e cos xdx = e sin x − −e cos x + e cos xdx x x Isolating yields Z 2 x Z x e cos xdx = e (sin x + cos x) =⇒ Example 8. (dv = 1dx) Evaluate R ex (sin x + cos x) + C e cos xdx = 2 x log xdx Solution 8. • Here there is only one function • The trick is to rewrite it as • Then we let • Then • Get u = log x du = 1/xdx and R log xdx = R dv = 1dx so that udv = log xdx R R v = dv = 1dx = x and Z Z log xdx = x log x − • log x · 1dx x (1/x) dx = x log x − x + C In addition to the indenite form, there is the denite form: Theorem 2. (Integration by parts, denite) Z b udv = (uv)|ba a Z − vdu a 5 b Example 9. (Harder) Evaluate R1 0 arcsin xdx Solution 9. • Let's use the  dv = 1 that du = √ dx 1−x2 • Recalling 1dx arcsin 1 = π/2, trick and let dv = 1dx arcsin xdx = (x arcsin x)|10 Z 0 Then t (0) = 1 Z 1 0 • so that u = arcsin x so x π √ dx = − 2 1 − x2 Z 1 0 x √ dx 1 − x2 and dt = −2xdx t (1) = 0, so get 1 x √ dx = − 2 1 − x2 0 Z 1 1 1 √ dt = 2 t 1 Z t−1/2 dt = 0  1 1 2t1/2 0 = 1 2 So, Z 1 arcsin xdx = 0 • and What substitution should we do for the new integral?  t = 1 − x2 • 1 − 0 • QUESTION: v = x, get 1 Z so that π −1 2 We've seen an example where applying integration by parts resulted in R vdu as an easier version of our original integral • This is a general concept and it is called Example 10. reduction formulas Derive the reduction formula Z n−1 1 cos xdx = sin x cosn−1 x + n n n Z cosn−2 xdx Solution 10. R • If we compare the intergration by parts formula udv n−1 n−1 formula, we might guess u = cos x or v = cos x • Let's try • Since • Then R vdu with the above u = cosn−1 x cosn xdx = cosn−1 x cos xdx, therefore dv = cos xdx (the remaining terms)  d cosn−1 x dx = (n − 1) cosn−2 x (− sin x) dx dx R v = cos xdx = sin x du = and = uv − 6 • Get Z Z udv = uv − n−1 vdu = sin x cos n−1 = sin x cos • QUESTION: Z x− Z x + (n − 1) sin x (n − 1) cosn−2 x (− sin x) dx sin2 x cosn−2 xdx What should we do next?  sin2 x = 1 − cos2 x from cos2 x + sin2 x = 1 • Get n−1 sin x cos n−1 = sin x cos • QUESTION:  • Z  1 − cos2 x cosn−2 xdx x + (n − 1) Z x + (n − 1) n−2 cos Z xdx − (n − 1) cosn xdx What happened? We ended up with our original integral R cosn xdx again So we isolate to get Z (1 + (n − 1)) and hence Example 11. n−1 cos xdx = sin x cos Z x + (n − 1) cosn−2 xdx Z 1 n−1 n−1 cos xdx = sin x cos x+ cosn−2 xdx n n R π/2 above reduction formula to evaluate cos7 xdx 0 Z Use the n n Solution 11. • Since the reduction formula keeps subtracting o 2 from the power of cos x, we keep applying it until we can't apply it anymore • Get Z π/2 0 • π/2 Z Z 1 6 π/2 6 π/2 6 5 cos xdx = sin x cos x + cos xdx = 0 + cos5 xdx 7 7 7 0 0 0 7 Applying it again gives Z 0 π/2 π/2 Z Z 1 4 π/2 4 π/2 4 3 cos xdx = sin x cos x + cos xdx = 0 + cos3 xdx 5 5 5 0 0 0 5 7 • Applying it again gives Z π/2 0 • π/2 π/2 Z 2 π/2 2 1 2 2 cos xdx = sin x cos x + cos xdx = 0 + sin x = 3 3 0 3 3 0 0 3 Hence Z π/2 cos7 xdx = 0 • QUESTION:  6 4 2 16 · · = 7 5 3 35 Do you recognize this pattern? The pattern is seen inside the identity π 2 2 4 4 6 6 = · · · · · ··· 2 1 3 3 5 5 7   • This is known as the Wallis product You don't need to know this for this course Some history about the mathematician Wallis:   He worked as a crytographer for the British government At that time (1600s) underyling principles of cryptography were not well-understood like they are today   Wallis actually was a proponent for key based ciphers A modern example is the RSA public key system that works on the principle that factoring very large numbers into primes is computationally very hard  You can actually take MATD16 to learn about cryptography! • Let's do a few more examples to conclude our lecture • Next week we will start with the new topic of trigonometric integrals Example 12. (Harder) Evaluate R arcsin2 xdx Solution 12. • First we recall that • Let's try • Then d dx u = arcsin2 x √ arcsin x = 1/ 1 − x2 and du = and v= R dv = 1dx so that udv = arcsin2 xdx √  d arcsin2 x dx = 2 arcsin x/ 1 − x2 dx dx 1dx = x 8 • Get Z Z 2 arcsin xdx = • QUESTION: Z udv = uv − 2 Z 2x arcsin x √ dx 1 − x2 vdu = x arcsin x − This last integral looks harder than our original, but what's actually nice about it?  • We see the √ 2x/ 1 − x2 which we can integrate by substitution Rewrite the last integral as Z • arcsin x √ So let's do integration by parts with 2x dx 1 − x2 u = arcsin x and dv = √ 2x dx so that 1−x2 udv = 2x√arcsin x dx 1−x2 √ du = 1/ 1 − x2 dx and using the substitution t = 1 − x2 , dt = −2xdx Z Z √ 1 2x √ dx = − √ dt = −2t1/2 = −2 1 − x2 v= t 1 − x2 • Then • Hence Z 2x arcsin x √ dx = uv − 1 − x2 √ Z vdu = −2 arcsin x 1 − x2 + Z get √ 2 1 − x2 √ dx 1 − x2 √ = −2 arcsin x 1 − x2 + 2x • Hence Z   √ arcsin2 xdx = x arcsin2 x − −2 arcsin x 1 − x2 + 2x + C • SUMMAY:  Integration by parts formula useful for evaluating integrals that are products of dierent functions  There are some tricks: started,  dv  repeated integration by parts, getting back where you = 1dx There are reduction formulas that simplify integrals so that we can solve them 9 ...
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