Lesson 27: Integration by Substitution, part II (Section 10 version)
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Lesson 27: Integration by Substitution, part II (Section 10 version)

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Lesson 27: Integration by Substitution, part II (Section 10 version) Lesson 27: Integration by Substitution, part II (Section 10 version) Presentation Transcript

  • Section 5.5 Integration by Substitution, Part Deux V63.0121, Calculus I April 29, 2009 Announcements Class on Monday will be review . . . . . .
  • Yes, there is class on Monday No new material We will review the course We will answer questions, so bring some . . . . . .
  • Final stuff Old finals online, including Fall 2008 Review sessions: May 5 and 6, 6:00–8:00pm, SILV 703 Final is May 8, 2:00–3:50pm in CANT 101/200 . . Image credit: Pragmagraphr . . . . . .
  • Resurrection Policy If your final score beats your midterm score, we will add 10% to its weight, and subtract 10% from the midterm weight. . . Image credit: Scott Beale / Laughing Squid . . . . . .
  • Outline Recall: The method of substitution Multiple substitutions Odd and even functions Examples More examples and advice Course Evaluations . . . . . .
  • Last Time: The Substitution Rule Theorem If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then ∫ ∫ ′ f(g(x))g (x) dx = f(u) du or ∫ ∫ du f(u) dx = f(u) du dx . . . . . .
  • Last Time: The Substitution Rule for Definite Integrals Theorem If g′ is continuous and f is continuous on the range of u = g(x), then ∫ ∫ b g(b) f(g(x))g′ (x) dx = f(u) du. a g(a) . . . . . .
  • Last Time: The Substitution Rule for Definite Integrals Theorem If g′ is continuous and f is continuous on the range of u = g(x), then ∫ ∫ b g(b) f(g(x))g′ (x) dx = f(u) du. a g(a) The integral on the left happens in “x-land”, so its limits are values of x The integral on the right happens in “u-land”, so its limits need to be values of u To convert x to u, simply apply the substitution u = g(x). . . . . . .
  • Outline Recall: The method of substitution Multiple substitutions Odd and even functions Examples More examples and advice Course Evaluations . . . . . .
  • An exponential example Example√ ∫ √ ln 8 e2x e2x + 1 dx Find √ ln 3 . . . . . .
  • An exponential example Example√ ∫ √ ln 8 e2x e2x + 1 dx Find √ ln 3 Solution Let u = e2x , so du = 2e2x dx. We have √ ∫ ∫ √ 8√ ln 8 1 e2x e2x + 1 dx = u + 1 du √ 2 ln 3 3 . . . . . .
  • An exponential example Example√ ∫ √ ln 8 e2x e2x + 1 dx Find √ ln 3 Solution Let u = e2x , so du = 2e2x dx. We have √ ∫ ∫ √ 8√ ln 8 1 e2x e2x + 1 dx = u + 1 du √ 2 ln 3 3 Now let y = u + 1, dy = du. So ∫ ∫ ∫ 8√ 9 9 √ 1 1 1 y1/2 dy u + 1 du = y dy = 2 2 2 3 4 4 9 12 1 19 = · y3/2 = (27 − 8) = 23 3 3 4 . . . . . .
  • Another way to skin that cat Example√ ∫ √ ln 8 e2x e2x + 1 dx Find √ ln 3 Solution Let u = e2x + 1 . . . . . .
  • Another way to skin that cat Example√ ∫ √ ln 8 e2x e2x + 1 dx Find √ ln 3 Solution Let u = e2x + 1, so that du = 2e2x dx. . . . . . .
  • Another way to skin that cat Example√ ∫ √ ln 8 e2x e2x + 1 dx Find √ ln 3 Solution Let u = e2x + 1, so that du = 2e2x dx. Then √ ∫ ∫ √ ln 8 9√ 1 2x e2x e + 1 dx = u du √ 2 ln 3 4 . . . . . .
  • Another way to skin that cat Example√ ∫ √ ln 8 e2x e2x + 1 dx Find √ ln 3 Solution Let u = e2x + 1, so that du = 2e2x dx. Then √ ∫ ∫ √ ln 8 9√ 1 2x e2x e + 1 dx = u du √ 2 ln 3 4 9 1 3/2 u = 3 4 . . . . . .
  • Another way to skin that cat Example√ ∫ √ ln 8 e2x e2x + 1 dx Find √ ln 3 Solution Let u = e2x + 1, so that du = 2e2x dx. Then √ ∫ ∫ √ ln 8 9√ 1 2x e2x e + 1 dx = u du √ 2 ln 3 4 1 3/2 9 u = 3 4 1 19 = (27 − 8) = 3 3 . . . . . .
  • A third skinned cat Example√ ∫ √ ln 8 e2x e2x + 1 dx Find √ ln 3 Solution √ e2x + 1, so that Let u = u2 = e2x + 1 . . . . . .
  • A third skinned cat Example√ ∫ √ ln 8 e2x e2x + 1 dx Find √ ln 3 Solution √ e2x + 1, so that Let u = u2 = e2x + 1 =⇒ 2u du = 2e2x dx . . . . . .
  • A third skinned cat Example√ ∫ √ ln 8 e2x e2x + 1 dx Find √ ln 3 Solution √ e2x + 1, so that Let u = u2 = e2x + 1 =⇒ 2u du = 2e2x dx Thus √ ∫ ∫ √ 3 ln 8 3 13 19 2x e2x u · u du = e + 1 dx = u = √ 3 3 ln 3 2 2 . . . . . .
  • Outline Recall: The method of substitution Multiple substitutions Odd and even functions Examples More examples and advice Course Evaluations . . . . . .
  • Example ∫ π sin(x) dx Find −π . . . . . .
  • Example ∫ π sin(x) dx Find −π Solution ∫ π sin(x) = − cos(x)|π = cos(x)|−π = cos(−π) − cos(π) = 0 −π π −π . . . . . .
  • Example ∫ π sin(x) dx Find −π Solution ∫ π sin(x) = − cos(x)|π = cos(x)|−π = cos(−π) − cos(π) = 0 −π π −π This is obvious from the graph: y . . x . . . . . . .
  • Even and Odd Functions Definition A function f is even if for all x, f(−x) = f(x) A function f is odd if for all x, f(−x) = −f(x). . . . . . .
  • Even and Odd Functions Definition A function f is even if for all x, f(−x) = f(x) A function f is odd if for all x, f(−x) = −f(x). . . . . . .
  • Even and Odd Functions Definition A function f is even if for all x, f(−x) = f(x) A function f is odd if for all x, f(−x) = −f(x). These properties are revealed in the graph. . . . . . .
  • Even and Odd Functions Definition A function f is even if for all x, f(−x) = f(x) A function f is odd if for all x, f(−x) = −f(x). These properties are revealed in the graph. An odd function has rotational symmetry about the origin. . . . . . .
  • Even and Odd Functions Definition A function f is even if for all x, f(−x) = f(x) A function f is odd if for all x, f(−x) = −f(x). These properties are revealed in the graph. An odd function has rotational symmetry about the origin. An even function has reflective symmetry in the y-axis . . . . . .
  • Even and Odd functions pictured y . o . dd . x . y . e . ven x . . . . . . .
  • Examples of symmetric functions Even and odd functions abound. x → xn is odd when n is odd and even when n is even. Funny, that! sin is odd and cos is even. . . . . . .
  • Combining symmetric functions Theorem (a) The sum of even functions is even. The sum of odd functions is odd. (b) The product of even functions is even. The product of odd functions is even. The product of an odd function and an even function is an odd function. (c) If g is even, then f ◦ g is even. The composition of two odd functions is odd. The composition of an even function and an odd function is even. . . . . . .
  • Integrating symmetric functions Theorem Let a be any number. (a) If f is odd, then ∫ a f(x) dx = 0. −a . . . . . .
  • Integrating symmetric functions Theorem Let a be any number. (a) If f is odd, then ∫ a f(x) dx = 0. −a (b) If f is even, then ∫ ∫ a a f(x) dx = 2 f(x) dx. −a 0 . . . . . .
  • Proof (odd f). ∫ a f(x) dx, let u = −x. Then du = −dx and we have To compute −a ∫ ∫ −a a f(x) dx = − f(−u) du −a a ∫ −a f(u) du = a ∫ a =− f(u) du. −a The only number which is equal to its own negative is zero. . . . . . .
  • Proof (even f). With the same substitution we have ∫0 ∫ 0 f(x) dx = − f(−u) du −a a ∫0 =− f(u) du a ∫a f(u) du. = 0 So ∫ ∫ ∫ ∫ a a a 0 f(x) dx = f(x) dx + f(x) dx = 2 f(x) dx. −a −a 0 0 . . . . . .
  • Example Compute ∫ √ √ e π +1 sin(x) 1 + cos3 (x) dx. √ −e π −1 . . . . . .
  • Example Compute ∫ √ √ e π +1 sin(x) 1 + cos3 (x) dx. √ −e π −1 Solution The integrand is odd! So the answer is zero. . . . . . .
  • Example Compute ∫ ( ) 2 x4 + x2 + 3 dx. −2 . . . . . .
  • Solution Because the integrand is even we can simplify our arithmetic. It’s especially nice to plug in zero since the result is often zero. ∫ ∫ ( ) 2( ) 2 4 2 x4 + x2 + 3 dx x + x + 3 dx = 2 −2 0 [ ]2 x5 x3 =2 + 3x + 5 3 0 [5 ] 23 2 =2 + 3(2) + 5 3 [ ] 32 8 =2 + +6 5 3 2 (32 · 3 + 8 · 5 + 6 · 15) = 15 2 · 226 = 15 . . . . . .
  • Outline Recall: The method of substitution Multiple substitutions Odd and even functions Examples More examples and advice Course Evaluations . . . . . .
  • Example ∫ x3 dx (5x4 + 2)2 . . . . . .
  • Example ∫ x3 dx (5x4 + 2)2 Solution Let u = 5x4 + 2, so du = 20x3 dx. Then ∫ ∫ x3 1 1 dx = du 4 + 2)2 u2 20 (5x 11 =− · +C 20 u 1 =− +C 20(5x4 + 2) . . . . . .
  • Example ∫ sin(sin(θ)) cos(θ) dθ . . . . . .
  • Example ∫ sin(sin(θ)) cos(θ) dθ Solution Let u = sin(θ), so du = cos(θ) dθ. Then ∫ ∫ sin(sin(θ)) cos(θ) dθ = sin(u) du = − cos(u) + C = − cos(sin(θ)) + C . . . . . .
  • Example ∫ ex + e−x dx ex − e−x . . . . . .
  • Example ∫ ex + e−x dx ex − e−x Solution The numerator is the derivative of the denominator! Let ( ) u = ex − e−x , so du = ex + e−x dx. Then ∫ ∫ ex + e−x 1 dx = du x − e−x e u = ln |u| + C = ln ex − e−x + C . . . . . .
  • Example ∫ 3x dx 1 + 9x . . . . . .
  • Example ∫ 3x dx 1 + 9x Solution Notice 9x = (32 )x = 32x = (3x )2 . So let u = 3x , du = (ln 3) · 3x dx. Then ∫ ∫ 3x 1 1 x dx = du 1 + u2 (1 + 9 ) ln 3 1 arctan(u) + C = ln 3 1 arctan(3x ) + C = ln 3 . . . . . .
  • Example √ ∫ sec2 x √ dx x . . . . . .
  • Example √ ∫ sec2 x √ dx x Solution √ 1 x, so du = √ du. Then Let u = 2x √ ∫ ∫ sec2 x dx = 2 sec2 (u) du √ x = 2 tan(u) + C (√ ) = 2 tan x + C . . . . . .
  • Example ∫ dx x ln x . . . . . .
  • Example ∫ dx x ln x Solution 1 Let u = ln x, so du = dx. Then x ∫ ∫ dx 1 du = x ln x u = ln |u| + C = ln |ln x| + C . . . . . .
  • What do we substitute? Linear factors (ax + b) are easy substitutions: u = ax + b, du = a dx Look for function/derivative pairs in the integrand, one to make u and one to make du: xn and xn−1 (fudge the coefficient) sine and cosine (fudge the minus sign) ex and ex ax and ax (fudge the coefficient) √ 1 x and √ (fudge the factor of 2) x 1 ln x and x . . . . . .
  • Outline Recall: The method of substitution Multiple substitutions Odd and even functions Examples More examples and advice Course Evaluations . . . . . .
  • Course Evaluations Please fill out CAS and departmental evaluations CAS goes to SILV 909 (need a volunteer) departmental goes to WWH 627 (need another volunteer) Thank you for your input! . . . . . .