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# Lesson 24: Evaluating Definite Integrals (slides)

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Computing integrals with Riemann sums is like computing derivatives with limits. The calculus of integrals turns out to come from antidifferentiation. This startling fact is the Second Fundamental Theorem of Calculus!

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### Lesson 24: Evaluating Definite Integrals (slides)

1. 1. Section 5.3 Evaluating Definite Integrals V63.0121.006/016, Calculus I New York University April 20, 2010 Announcements April 16: Quiz 4 on §§4.1–4.4 April 29: Movie Day!! April 30: Quiz 5 on §§5.1–5.4 Monday, May 10, 12:00noon (not 10:00am as previously announced) Final Exam . Image credit: docman . . . . . . .
2. 2. Announcements April 16: Quiz 4 on §§4.1–4.4 April 29: Movie Day!! April 30: Quiz 5 on §§5.1–5.4 Monday, May 10, 12:00noon (not 10:00am as previously announced) Final Exam . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 2 / 48
3. 3. Homework: The Good Most got problems 1 and 3 right. . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 3 / 48
4. 4. Homework: The Bad (steel pipe) Problem A steel pipe is being carried down a hallway 9 ft wide. At the end of the hall there is a right-aangled turn into a narrower hallway 6 ft wide. What is the length of the longest pipe that can be carried horizontally around the corner? . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 4 / 48
5. 5. Homework: The Bad (steel pipe) Problem A steel pipe is being carried down a hallway 9 ft wide. At the end of the hall there is a right-aangled turn into a narrower hallway 6 ft wide. What is the length of the longest pipe that can be carried horizontally around the corner? . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 4 / 48
6. 6. Homework: The Bad (steel pipe) Problem A steel pipe is being carried down a hallway 9 ft wide. At the end of the hall there is a right-aangled turn into a narrower hallway 6 ft wide. What is the length of the longest pipe that can be carried horizontally around the corner? . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 4 / 48
7. 7. Homework: The Bad (steel pipe) Problem A steel pipe is being carried down a hallway 9 ft wide. At the end of the hall there is a right-aangled turn into a narrower hallway 6 ft wide. What is the length of the longest pipe that can be carried horizontally around the corner? 6 . . . 9 . . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 4 / 48
8. 8. Homework: The Bad (steel pipe) Problem A steel pipe is being carried down a hallway 9 ft wide. At the end of the hall there is a right-aangled turn into a narrower hallway 6 ft wide. What is the length of the longest pipe that can be carried horizontally around the corner? 6 . . θ . . 9 . . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 4 / 48
9. 9. Homework: The Bad (steel pipe) Problem A steel pipe is being carried down a hallway 9 ft wide. At the end of the hall there is a right-aangled turn into a narrower hallway 6 ft wide. What is the length of the longest pipe that can be carried horizontally around the corner? . . θ 6 . . θ . . 9 . . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 4 / 48
10. 10. Homework: The Bad (steel pipe) Problem A steel pipe is being carried down a hallway 9 ft wide. At the end of the hall there is a right-aangled turn into a narrower hallway 6 ft wide. What is the length of the longest pipe that can be carried horizontally around the corner? . . θ 6 . 6 . 9 . . θ . . 9 . . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 4 / 48
11. 11. Homework: The Bad (steel pipe) Problem A steel pipe is being carried down a hallway 9 ft wide. At the end of the hall there is a right-aangled turn into a narrower hallway 6 ft wide. What is the length of the longest pipe that can be carried horizontally around the corner? . . θ 6 . 6 . 9 . . θ cθ . . cs 9 . 9 . . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 4 / 48
12. 12. Homework: The Bad (steel pipe) Problem A steel pipe is being carried down a hallway 9 ft wide. At the end of the hall there is a right-aangled turn into a narrower hallway 6 ft wide. What is the length of the longest pipe that can be carried horizontally around the corner? . . θ 6 . 6 . cθ . se 6 9 . . θ cθ . . cs 9 . 9 . . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 4 / 48
13. 13. Solution Solution The longest pipe that barely fits is the smallest pipe that almost doesn’t fit. We want to find the minimum value of f(θ) = a sec θ + b csc θ on the interval 0 < θ < π/2. (a = 9 and b = 6 in our problem.) f′ (θ) = a sec θ tan θ − b csc θ cot θ sin θ cos θ a sin3 θ − b cos3 θ =a −b 2 = cos2 θ sin θ sin2 θ cos2 θ So the critical point is when b a sin3 θ = b cos3 θ =⇒ tan3 θ = a . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 5 / 48
14. 14. Finding the minimum If f′ (θ) = a sec θ tan θ − b csc θ cot θ, then f′′ (θ) = a sec θ tan2 θ + a sec3 θ + b csc θ cot2 θ + b csc3 θ which is positive on 0 < θ < π/2. So the minimum value is f(θmin ) = a sec θmin + b csc θmin ( )1/3 b b where tan3 θmin = =⇒ tan θmin = . a a Using 1 + tan2 θ = sec2 θ 1 + cot2 θ = csc2 θ We get the minimum value is √ √ ( )2/3 ( a )2/3 b min = a 1+ +b 1+ a b . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 6 / 48
15. 15. Simplifying √ √ ( )2/3 ( a )2/3 b min = a 1 + +b 1+ a b √ √ b2/3 a2/3 a2/3 b2/3 =b + 2/3 + a + b2/3 b a2/3 a2/3 √ √ b 2/3 a = 1/3 b + a 2/3 + 1/3 a2/3 + b2/3 b √ a √ 2/3 2/3 2/3 + a2/3 a2/3 + b2/3 =b b +a √ = (b + a ) b2/3 + a2/3 2/3 2/3 = (a2/3 + b2/3 )3/2 If a = 9 and b = 6, then min ≈ 21.070. . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 7 / 48
16. 16. Homework: The Bad (Diving Board) Problem If a diver of mass m stands at the end of a diving board with length L and linear density ρ, then the board takes on the shape of a curve y = f(x), where EIy′′ = mg(L − x) + 1 ρg(L − x)2 2 E and I are positive constants that depend of the material of the board and g < 0 is the acceleration due to gravity. (a) Find an expression for the shape of the curve. (b) Use f(L) to estimate the distance below the horizontal at the end of the board. . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 8 / 48
17. 17. Solution We have EIy′′ (x) = mg(L − x) + 1 ρg(L − x)2 2 Antidifferentiating once gives EIy′ (x) = − 2 mg(L − x)2 − 1 ρg(L − x)3 + C 1 6 Once more: EIy(x) = 1 mg(L − x)3 + 6 1 24 ρg(L − x)4 + Cx + D where C and D are constants. . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 9 / 48
18. 18. Don't stop there! Plugging y(0) = 0 into EIy′ (x) = 1 mg(L − x)3 + 6 1 24 ρg(L − x)4 + Cx + D gives 1 1 0 = 1 mgL3 + 6 1 24 ρgL 4 + D =⇒ D = − mgL3 − ρgL4 6 24 Plugging y′ (0) = 0 into EIy′ (x) = − 2 mg(L − x)2 − 1 ρg(L − x)3 + C 1 6 gives 1 1 0 = − 1 mgL2 − 1 ρgL3 + C =⇒ C = 2 6 mgL2 + ρgL3 2 6 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 10 / 48
19. 19. Solution completed So EIy(x) = 1 mg(L − x)3 + 6 1 24 ρg(L − x)4 ( ) 1 1 3 1 1 + mgL + L x − mgL3 − 2 ρgL4 2 6 6 24 which means ( ) 1 1 1 1 EIy(L) = mgL2 + L3 L − mgL3 − ρgL4 2 6 6 24 1 1 1 1 = mgL3 + L4 − mgL3 − ρgL4 2 6 6 24 1 1 = mgL3 + ρgL4 3 8 3 ( ) gL m ρL =⇒ y(L) = + EI 3 8 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 11 / 48
20. 20. Homework: The Ugly Some students have gotten their hands on a solution manual and are copying answers word for word. This is very easy to catch: the graders are following the same solution manual. This is not very productive: the best you will do is ace 10% of your course grade. This is a violation of academic integrity. I do not take it lightly. . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 12 / 48
21. 21. Objectives Use the Evaluation Theorem to evaluate definite integrals. Write antiderivatives as indefinite integrals. Interpret definite integrals as “net change” of a function over an interval. . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 13 / 48
22. 22. Outline Last time: The Definite Integral The definite integral as a limit Properties of the integral Comparison Properties of the Integral Evaluating Definite Integrals The Theorem of the Day Examples The Integral as Total Change Indefinite Integrals My first table of integrals Computing Area with integrals . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 14 / 48
23. 23. The definite integral as a limit Definition If f is a function defined on [a, b], the definite integral of f from a to b is the number ∫ b ∑n f(x) dx = lim f(ci ) ∆x a n→∞ i=1 b−a where ∆x = , and for each i, xi = a + i∆x, and ci is a point in n [xi−1 , xi ]. Theorem If f is continuous on [a, b] or if f has only finitely many jump discontinuities, then f is integrable on [a, b]; that is, the definite integral ∫ b f(x) dx exists and is the same for any choice of ci . a . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 15 / 48
24. 24. Notation/Terminology ∫ b ∑ n f(x) dx = lim f(ci ) ∆x a n→∞ i=1 ∫ — integral sign (swoopy S) f(x) — integrand a and b — limits of integration (a is the lower limit and b the upper limit) dx — ??? (a parenthesis? an infinitesimal? a variable?) The process of computing an integral is called integration . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 16 / 48
25. 25. Properties of the integral Theorem (Additive Properties of the Integral) Let f and g be integrable functions on [a, b] and c a constant. Then ∫ b 1. c dx = c(b − a) a ∫ b ∫ b ∫ b 2. [f(x) + g(x)] dx = f(x) dx + g(x) dx. a a a ∫ b ∫ b 3. cf(x) dx = c f(x) dx. a a ∫ b ∫ b ∫ b 4. [f(x) − g(x)] dx = f(x) dx − g(x) dx. a a a . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 17 / 48
26. 26. More Properties of the Integral Conventions: ∫ ∫ a b f(x) dx = − f(x) dx b a ∫ a f(x) dx = 0 a This allows us to have ∫ c ∫ b ∫ c 5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c. a a b . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 18 / 48
27. 27. Definite Integrals We Know So Far If the integral computes an area and we know the area, we can use that. For instance, y . ∫ 1√ π 1 − x2 dx = 0 4 By brute force we . computed x . ∫ 1 ∫ 1 1 1 x2 dx = x3 dx = 0 3 0 4 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 19 / 48
28. 28. Comparison Properties of the Integral Theorem Let f and g be integrable functions on [a, b]. ∫ b 6. If f(x) ≥ 0 for all x in [a, b], then f(x) dx ≥ 0 a ∫ b ∫ b 7. If f(x) ≥ g(x) for all x in [a, b], then f(x) dx ≥ g(x) dx a a 8. If m ≤ f(x) ≤ M for all x in [a, b], then ∫ b m(b − a) ≤ f(x) dx ≤ M(b − a) a . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 21 / 48
29. 29. Example ∫ 2 1 Estimate dx using Property ??. 1 x . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 22 / 48
30. 30. Example ∫ 2 1 Estimate dx using Property ??. 1 x Solution Since 1 1 1 1 ≤ x ≤ 2 =⇒ ≤ ≤ 2 x 1 y . we have ∫ 2 1 1 ·(2−1) ≤ dx ≤ 1·(2−1) 2 1 x or . ∫ 2 x . 1 1 ≤ dx ≤ 1 2 1 x (Not a very good estimate) . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 22 / 48
31. 31. Outline Last time: The Definite Integral The definite integral as a limit Properties of the integral Comparison Properties of the Integral Evaluating Definite Integrals The Theorem of the Day Examples The Integral as Total Change Indefinite Integrals My first table of integrals Computing Area with integrals . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 23 / 48
32. 32. Socratic dialogue The definite integral of velocity measures displacement (net distance) The derivative of displacement is velocity So we can compute displacement with the definite integral or an antiderivative of velocity But any function can be a velocity function, so . . . . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 24 / 48
33. 33. Theorem of the Day Theorem (The Second Fundamental Theorem of Calculus) Suppose f is integrable on [a, b] and f = F′ for another function F, then ∫ b f(x) dx = F(b) − F(a). a . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 25 / 48
34. 34. Theorem of the Day Theorem (The Second Fundamental Theorem of Calculus) Suppose f is integrable on [a, b] and f = F′ for another function F, then ∫ b f(x) dx = F(b) − F(a). a Note In Section 5.3, this theorem is called “The Evaluation Theorem”. Nobody else in the world calls it that. . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 25 / 48
35. 35. Proving the Second FTC Proof. b−a Divide up [a, b] into n pieces of equal width ∆x = as usual. For n each i, F is continuous on [xi−1 , xi ] and differentiable on (xi−1 , xi ). So there is a point ci in (xi−1 , xi ) with F(xi ) − F(xi−1 ) = F′ (ci ) = f(ci ) xi − xi−1 Or f(ci )∆x = F(xi ) − F(xi−1 ) . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 26 / 48
36. 36. Proving the Second FTC Proof. b−a Divide up [a, b] into n pieces of equal width ∆x = as usual. For n each i, F is continuous on [xi−1 , xi ] and differentiable on (xi−1 , xi ). So there is a point ci in (xi−1 , xi ) with F(xi ) − F(xi−1 ) = F′ (ci ) = f(ci ) xi − xi−1 Or f(ci )∆x = F(xi ) − F(xi−1 ) See if you can spot the invocation of the Mean Value Theorem! . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 26 / 48
37. 37. Proving the Second FTC We have for each i f(ci )∆x = F(xi ) − F(xi−1 ) Form the Riemann Sum: ∑ n ∑ n Sn = f(ci )∆x = (F(xi ) − F(xi−1 )) i=1 i=1 = (F(x1 ) − F(x0 )) + (F(x2 ) − F(x1 )) + (F(x3 ) − F(x2 )) + · · · · · · + (F(xn−1 ) − F(xn−2 )) + (F(xn ) − F(xn−1 )) = F(xn ) − F(x0 ) = F(b) − F(a) . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 27 / 48
38. 38. Proving the Second FTC We have shown for each n, Sn = F(b) − F(a) so in the limit ∫ b f(x) dx = lim Sn = lim (F(b) − F(a)) = F(b) − F(a) a n→∞ n→∞ . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 28 / 48
39. 39. Verifying earlier computations Example Find the area between y = x3 the x-axis, x = 0 and x = 1. . . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 29 / 48
40. 40. Verifying earlier computations Example Find the area between y = x3 the x-axis, x = 0 and x = 1. . Solution ∫ 1 1 x4 1 A= x3 dx = = 0 4 0 4 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 29 / 48
41. 41. Verifying earlier computations Example Find the area between y = x3 the x-axis, x = 0 and x = 1. . Solution ∫ 1 1 x4 1 A= x3 dx = = 0 4 0 4 Here we use the notation F(x)|b or [F(x)]b to mean F(b) − F(a). a a . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 29 / 48
42. 42. Verifying Archimedes Example Find the area enclosed by the parabola y = x2 and y = 1. . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 30 / 48
43. 43. Verifying Archimedes Example Find the area enclosed by the parabola y = x2 and y = 1. . 1 . . . . − . 1 1 . . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 30 / 48
44. 44. Verifying Archimedes Example Find the area enclosed by the parabola y = x2 and y = 1. . 1 . . . . − . 1 1 . Solution ∫ 1 [ ]1 ( [ )] x3 1 1 4 A=2− x dx = 2 − 2 =2− − − = −1 3 −1 3 3 3 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 30 / 48
45. 45. Computing exactly what we earlier estimated Example ∫ 1 4 Evaluate the integral dx. 0 1 + x2 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 31 / 48
46. 46. Example ∫ 1 4 Estimate dx using the midpoint rule and four divisions. 0 1 + x2 Solution Dividing up [0, 1] into 4 pieces gives 1 2 3 4 x0 = 0, x1 = , x2 = , x3 = , x4 = 4 4 4 4 So the midpoint rule gives ( ) 1 4 4 4 4 M4 = + + + 4 1 + (1/8)2 1 + (3/8)2 1 + (5/8)2 1 + (7/8)2 ( ) 1 4 4 4 4 = + + + 4 65/64 73/64 89/64 113/64 150, 166, 784 = ≈ 3.1468 47, 720, 465 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 32 / 48
47. 47. Computing exactly what we earlier estimated Example ∫ 1 4 Evaluate the integral dx. 0 1 + x2 Solution ∫ 1 ∫ 1 4 1 dx = 4 dx 0 1 + x2 0 1 + x2 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 33 / 48
48. 48. Computing exactly what we earlier estimated Example ∫ 1 4 Evaluate the integral dx. 0 1 + x2 Solution ∫ 1 ∫ 1 4 1 dx = 4 dx 0 1 + x2 0 1 + x2 = 4 arctan(x)|1 0 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 33 / 48
49. 49. Computing exactly what we earlier estimated Example ∫ 1 4 Evaluate the integral dx. 0 1 + x2 Solution ∫ 1 ∫ 1 4 1 dx = 4 dx 0 1 + x2 0 1 + x2 = 4 arctan(x)|1 0 = 4 (arctan 1 − arctan 0) . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 33 / 48
50. 50. Computing exactly what we earlier estimated Example ∫ 1 4 Evaluate the integral dx. 0 1 + x2 Solution ∫ 1 ∫ 1 4 1 dx = 4 dx 0 1 + x2 0 1 + x2 = 4 arctan(x)|1 0 = 4 (arctan 1 − arctan 0) (π ) =4 −0 4 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 33 / 48
51. 51. Computing exactly what we earlier estimated Example ∫ 1 4 Evaluate the integral dx. 0 1 + x2 Solution ∫ 1 ∫ 1 4 1 dx = 4 dx 0 1 + x2 0 1 + x2 = 4 arctan(x)|1 0 = 4 (arctan 1 − arctan 0) (π ) =4 −0 =π 4 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 33 / 48
52. 52. Computing exactly what we earlier estimated Example ∫ 2 1 Evaluate dx. 1 x . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 34 / 48
53. 53. Example ∫ 2 1 Estimate dx using Property ??. 1 x Solution Since 1 1 1 1 ≤ x ≤ 2 =⇒ ≤ ≤ 2 x 1 y . we have ∫ 2 1 1 ·(2−1) ≤ dx ≤ 1·(2−1) 2 1 x or . ∫ 2 x . 1 1 ≤ dx ≤ 1 2 1 x (Not a very good estimate) . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 35 / 48
54. 54. Computing exactly what we earlier estimated Example ∫ 2 1 Evaluate dx. 1 x Solution ∫ 2 1 dx 1 x . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 36 / 48
55. 55. Computing exactly what we earlier estimated Example ∫ 2 1 Evaluate dx. 1 x Solution ∫ 2 1 dx = ln x|2 1 1 x . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 36 / 48
56. 56. Computing exactly what we earlier estimated Example ∫ 2 1 Evaluate dx. 1 x Solution ∫ 2 1 dx = ln x|2 1 1 x = ln 2 − ln 1 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 36 / 48
57. 57. Computing exactly what we earlier estimated Example ∫ 2 1 Evaluate dx. 1 x Solution ∫ 2 1 dx = ln x|2 1 1 x = ln 2 − ln 1 = ln 2 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 36 / 48
58. 58. Outline Last time: The Definite Integral The definite integral as a limit Properties of the integral Comparison Properties of the Integral Evaluating Definite Integrals The Theorem of the Day Examples The Integral as Total Change Indefinite Integrals My first table of integrals Computing Area with integrals . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 37 / 48
59. 59. The Integral as Total Change Another way to state this theorem is: ∫ b F′ (x) dx = F(b) − F(a), a or, the integral of a derivative along an interval is the total change over that interval. This has many ramifications: . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 38 / 48
60. 60. The Integral as Total Change Another way to state this theorem is: ∫ b F′ (x) dx = F(b) − F(a), a or, the integral of a derivative along an interval is the total change over that interval. This has many ramifications: Theorem If v(t) represents the velocity of a particle moving rectilinearly, then ∫ t1 v(t) dt = s(t1 ) − s(t0 ). t0 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 38 / 48
61. 61. The Integral as Total Change Another way to state this theorem is: ∫ b F′ (x) dx = F(b) − F(a), a or, the integral of a derivative along an interval is the total change over that interval. This has many ramifications: Theorem If MC(x) represents the marginal cost of making x units of a product, then ∫ x C(x) = C(0) + MC(q) dq. 0 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 38 / 48
62. 62. The Integral as Total Change Another way to state this theorem is: ∫ b F′ (x) dx = F(b) − F(a), a or, the integral of a derivative along an interval is the total change over that interval. This has many ramifications: Theorem If ρ(x) represents the density of a thin rod at a distance of x from its end, then the mass of the rod up to x is ∫ x m(x) = ρ(s) ds. 0 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 38 / 48
63. 63. Outline Last time: The Definite Integral The definite integral as a limit Properties of the integral Comparison Properties of the Integral Evaluating Definite Integrals The Theorem of the Day Examples The Integral as Total Change Indefinite Integrals My first table of integrals Computing Area with integrals . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 39 / 48
64. 64. A new notation for antiderivatives To emphasize the relationship between antidifferentiation and integration, we use the indefinite integral notation ∫ f(x) dx for any function whose derivative is f(x). . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 40 / 48
65. 65. A new notation for antiderivatives To emphasize the relationship between antidifferentiation and integration, we use the indefinite integral notation ∫ f(x) dx for any function whose derivative is f(x). Thus ∫ x2 dx = 1 x3 + C. 3 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 40 / 48
66. 66. My first table of integrals . ∫ ∫ ∫ [f(x) + g(x)] dx = f(x) dx + g(x) dx ∫ ∫ ∫ xn+1 xn dx = + C (n ̸= −1) cf(x) dx = c f(x) dx n+1 ∫ ∫ 1 ex dx = ex + C dx = ln |x| + C x ∫ ∫ ax sin x dx = − cos x + C ax dx = +C ln a . ∫ ∫ cos x dx = sin x + C csc2 x dx = − cot x + C ∫ ∫ sec2 x dx = tan x + C csc x cot x dx = − csc x + C ∫ ∫ 1 sec x tan x dx = sec x + C √ dx = arcsin x + C ∫ 1 − x2 1 dx = arctan x + C 1 + x2 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 41 / 48
67. 67. Outline Last time: The Definite Integral The definite integral as a limit Properties of the integral Comparison Properties of the Integral Evaluating Definite Integrals The Theorem of the Day Examples The Integral as Total Change Indefinite Integrals My first table of integrals Computing Area with integrals . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 42 / 48
68. 68. Example Find the area between the graph of y = (x − 1)(x − 2), the x-axis, and the vertical lines x = 0 and x = 3. . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 43 / 48
69. 69. Example Find the area between the graph of y = (x − 1)(x − 2), the x-axis, and the vertical lines x = 0 and x = 3. Solution ∫ 3 Consider (x − 1)(x − 2) dx. 0 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 43 / 48
70. 70. Graph y . . . . . x . 1 . 2 . 3 . . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 44 / 48
71. 71. Example Find the area between the graph of y = (x − 1)(x − 2), the x-axis, and the vertical lines x = 0 and x = 3. Solution ∫ 3 Consider (x − 1)(x − 2) dx. Notice the integrand is positive on [0, 1) 0 and (2, 3], and negative on (1, 2). . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 45 / 48
72. 72. Example Find the area between the graph of y = (x − 1)(x − 2), the x-axis, and the vertical lines x = 0 and x = 3. Solution ∫ 3 Consider (x − 1)(x − 2) dx. Notice the integrand is positive on [0, 1) 0 and (2, 3], and negative on (1, 2). If we want the area of the region, we have to do ∫ 1 ∫ 2 ∫ 3 A= (x − 1)(x − 2) dx − (x − 1)(x − 2) dx + (x − 1)(x − 2) dx 0 1 2 [ ]1 [ ]2 [ ]3 = − 1 3 3x 3 2 2x + 2x − 1 x3 − 3 x2 + 2x + 1 x3 − 3 x2 + 2x 3 2 3 2 ( ) 0 1 2 5 1 5 11 = − − + = . 6 6 6 6 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 45 / 48
73. 73. Interpretation of “negative area" in motion There is an analog in rectlinear motion: ∫ t1 v(t) dt is net distance traveled. t0 ∫ t1 |v(t)| dt is total distance traveled. t0 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 46 / 48
74. 74. What about the constant? It seems we forgot about the +C when we say for instance ∫ 1 1 x4 1 1 3 x dx = = −0= 0 4 0 4 4 But notice [ 4 ]1 ( ) x 1 1 1 +C = + C − (0 + C) = + C − C = 4 0 4 4 4 no matter what C is. So in antidifferentiation for definite integrals, the constant is immaterial. . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 47 / 48
75. 75. Summary Second FTC: ∫ b b f(x) dx = F(x) a a where F is an antiderivative of f. Computes any “net change” over an interval Proving the FTC requires the MVT . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 48 / 48