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# Lesson 27: Evaluating Definite Integrals

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The second Fundamental Theorem of Calculus makes calculating definite integrals a problem of antidifferentiation! …

The second Fundamental Theorem of Calculus makes calculating definite integrals a problem of antidifferentiation!

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• 1. Section 5.3 Evaluating Deﬁnite Integrals V63.0121.027, Calculus I December 3, 2009 Announcements Final Exam is Friday, December 18, 2:00–3:50pm Final is cumulative; topics will be represented roughly according to time spent on them . . Image credit: docman . . . . . .
• 2. Outline Last time: The Deﬁnite Integral The deﬁnite integral as a limit Properties of the integral Estimating the Deﬁnite Integral The Midpoint Rule Comparison Properties of the Integral Evaluating Deﬁnite Integrals Examples The Integral as Total Change Indeﬁnite Integrals My ﬁrst table of integrals Computing Area with integrals . . . . . .
• 3. 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 (c i ) ∆x a n→∞ i=1 b−a where ∆x = , and for each i, xi = a + i∆x, and ci is a point n in [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 ∫ b integral f(x) dx exists and is the same for any choice of ci . a . . . . . .
• 4. Notation/Terminology ∫ b f(x) dx a ∫ — 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 . . . . . .
• 5. 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 . . . . . .
• 6. 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 . . . . . .
• 7. 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 2 By brute force we . computed x . ∫ 1 ∫ 1 1 1 x2 dx = x3 dx = 0 3 0 4 . . . . . .
• 8. Outline Last time: The Deﬁnite Integral The deﬁnite integral as a limit Properties of the integral Estimating the Deﬁnite Integral The Midpoint Rule Comparison Properties of the Integral Evaluating Deﬁnite Integrals Examples The Integral as Total Change Indeﬁnite Integrals My ﬁrst table of integrals Computing Area with integrals . . . . . .
• 9. The Midpoint Rule Given a partition of [a, b] into n pieces, let ¯i be the midpoint of x [xi−1 , xi ]. Deﬁne ∑n Mn = f(¯i ) ∆x. x i =1 . . . . . .
• 10. The Midpoint Rule Given a partition of [a, b] into n pieces, let ¯i be the midpoint of x [xi−1 , xi ]. Deﬁne ∑n Mn = f(¯i ) ∆x. x i =1 y . . = x2 y Example ∫ 1 Cmpute M2 for x2 dx. 0 . Solution ( ) ( ) 1 1 2 1 3 2 5 . M2 = · + · = . . 2 4 2 4 16 x . 1 . 2 . . . . . .
• 11. Why are midpoints often better? ∫ 1 1 Compare L2 , R2 , and M2 for x2 dx = : 0 3 y . . = x2 y . . x . 1 . 2 . . . . . .
• 12. Why are midpoints often better? ∫ 1 1 Compare L2 , R2 , and M2 for x2 dx = : 0 3 y . ( )2 . = x2 y 1 1 1 1 L2 = · (0)2 + · = = 0.125 2 2 2 8 . . . x . 1 . 2 . . . . . .
• 13. Why are midpoints often better? ∫ 1 1 Compare L2 , R2 , and M2 for x2 dx = : 0 3 y . ( )2 . = x2 y 1 2 1 1 1 L2 = · (0) + · = = 0.125 2 2 2 8 ( )2 1 1 1 5 R2 = · + · (1)2 = = 0.625 2 2 2 8 . . . x . 1 . 2 . . . . . .
• 14. Why are midpoints often better? ∫ 1 1 Compare L2 , R2 , and M2 for x2 dx = : 0 3 y . ( )2 . = x2 y 1 1 1 1 L2 = · (0)2 + · = = 0.125 2 2 2 8 ( )2 1 1 1 5 R2 = · + · (1)2 = = 0.625 2 2 2 8 . ( )2 ( )2 1 1 1 3 5 M2 = · + · = = 0.3125 . . 2 4 2 4 16 x . 1 . 2 . . . . . .
• 15. Why are midpoints often better? ∫ 1 1 Compare L2 , R2 , and M2 for x2 dx = : 0 3 y . ( )2 . = x2 y 1 1 1 1 L2 = · (0)2 + · = = 0.125 2 2 2 8 ( )2 1 1 1 5 R2 = · + · (1)2 = = 0.625 2 2 2 8 . ( )2 ( )2 1 1 1 3 5 M2 = · + · = = 0.3125 .. 2 4 2 4 16 x . 1 . 2 Where f is monotone, one of Ln and Rn will be too much, and the other two little. But Mn allows overestimates and underestimates to counteract. . . . . . .
• 16. Example ∫ 1 4 Estimate dx using the midpoint rule and four divisions. 0 1 + x2 . . . . . .
• 17. 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 x 0 = 0 , x1 = , x 2 = , 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 . . . . . .
• 18. 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 x 0 = 0 , x1 = , x 2 = , 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 . . . . . .
• 19. 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 x 0 = 0 , x1 = , x 2 = , 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 . . . . . .
• 20. Comparison Properties of the Integral Theorem Let f and g be integrable functions on [a, b]. 6. If f(x) ≥ 0 for all x in [a, b], then ∫ b f(x) dx ≥ 0 a . . . . . .
• 21. The integral of a nonnegative function is nonnegative Proof. If f(x) ≥ 0 for all x in [a, b], then for any number of divisions n and choice of sample points {ci }: n ∑ n ∑ Sn = f(ci ) ∆x ≥ 0 · ∆x = 0 i=1 ≥0 i =1 Since Sn ≥ 0 for all n, the limit of {Sn } is nonnegative, too: ∫ b f(x) dx = lim Sn ≥ 0 a n→∞ ≥0 . . . . . .
• 22. Comparison Properties of the Integral Theorem Let f and g be integrable functions on [a, b]. 6. If f(x) ≥ 0 for all x in [a, b], then ∫ b f(x) dx ≥ 0 a 7. If f(x) ≥ g(x) for all x in [a, b], then ∫ b ∫ b f(x) dx ≥ g(x) dx a a . . . . . .
• 23. The deﬁnite integral is “increasing” Proof. Let h(x) = f(x) − g(x). If f(x) ≥ g(x) for all x in [a, b], then h(x) ≥ 0 for all x in [a, b]. So by the previous property ∫ b h(x) dx ≥ 0 a This means that ∫ b ∫ b ∫ b ∫ b f(x) dx − g(x) dx = (f(x) − g(x)) dx = h(x) dx ≥ 0 a a a a So ∫ ∫ b b f(x) dx ≥ g(x) dx a a . . . . . .
• 24. Comparison Properties of the Integral Theorem Let f and g be integrable functions on [a, b]. 6. If f(x) ≥ 0 for all x in [a, b], then ∫ b f(x) dx ≥ 0 a 7. If f(x) ≥ g(x) for all x in [a, b], then ∫ b ∫ b 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 . . . . . .
• 25. Bounding the integral using bounds of the function Proof. If m ≤ f(x) ≤ M on for all x in [a, b], then by the previous property ∫ b ∫ b ∫ b m dx ≤ f(x) dx ≤ M dx a a a By Property 1, the integral of a constant function is the product of the constant and the width of the interval. So: ∫ b m(b − a) ≤ f(x) dx ≤ M(b − a) a . . . . . .
• 26. Example ∫ 2 1 Estimate dx using Property 8. 1 x . . . . . .
• 27. Example ∫ 2 1 Estimate dx using Property 8. 1 x Solution Since 1 1 1 1 ≤ x ≤ 2 =⇒ ≤ ≤ 2 x 1 we have ∫ 2 1 1 · (2 − 1) ≤ dx ≤ 1 · (2 − 1) 2 1 x or ∫ 2 1 1 ≤ dx ≤ 1 2 1 x . . . . . .
• 28. Outline Last time: The Deﬁnite Integral The deﬁnite integral as a limit Properties of the integral Estimating the Deﬁnite Integral The Midpoint Rule Comparison Properties of the Integral Evaluating Deﬁnite Integrals Examples The Integral as Total Change Indeﬁnite Integrals My ﬁrst table of integrals Computing Area with integrals . . . . . .
• 29. Socratic proof 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 the antiderivative of velocity But any function can be a velocity function, so ... . . . . . .
• 30. 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 . . . . . .
• 31. 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. . . . . . .
• 32. Proving the Second FTC b−a Divide up [a, b] into n pieces of equal width ∆x = as n usual. For 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 ) x i − x i −1 Or f(ci )∆x = F(xi ) − F(xi−1 ) . . . . . .
• 33. 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) . . . . . .
• 34. 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) See if you can spot the invocation of the Mean Value Theorem! . . . . . .
• 35. 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→∞ . . . . . .
• 36. Example Find the area between y = x3 and the x-axis, between x = 0 and x = 1. . . . . . . .
• 37. Example Find the area between y = x3 and the x-axis, between x = 0 and x = 1. Solution ∫ 1 1 x4 1 A= x3 dx = = 0 4 0 4 . . . . . . .
• 38. Example Find the area between y = x3 and the x-axis, between 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 . . . . . .
• 39. Example Find the area enclosed by the parabola y = x2 and y = 1. . . . . . .
• 40. Example Find the area enclosed by the parabola y = x2 and y = 1. . 1 . . . . − . 1 1 . . . . . . .
• 41. Example Find the area enclosed by the parabola y = x2 and y = 1. . 1 . . . . − . 1 1 . Solution ∫ 1 [ ]1 [ ( )] 2 x3 1 1 4 A=2− x dx = 2 − =2− − − = −1 3 −1 3 3 3 . . . . . .
• 42. Example ∫ 1 4 Evaluate the integral dx. 0 1 + x2 . . . . . .
• 43. 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 x 0 = 0 , x1 = , x 2 = , 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 . . . . . .
• 44. Example ∫ 1 4 Evaluate the integral dx. 0 1 + x2 Solution ∫ 1 ∫ 1 4 1 dx = 4 dx 0 1 + x2 0 1 + x2 . . . . . .
• 45. 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 . . . . . .
• 46. 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) . . . . . .
• 47. 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 . . . . . .
• 48. 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 . . . . . .
• 49. Example ∫ 2 1 Evaluate dx. 1 x . . . . . .
• 50. Example ∫ 2 1 Estimate dx using Property 8. 1 x Solution Since 1 1 1 1 ≤ x ≤ 2 =⇒ ≤ ≤ 2 x 1 we have ∫ 2 1 1 · (2 − 1) ≤ dx ≤ 1 · (2 − 1) 2 1 x or ∫ 2 1 1 ≤ dx ≤ 1 2 1 x . . . . . .
• 51. Example ∫ 2 1 Evaluate dx. 1 x Solution ∫ 2 1 dx 1 x . . . . . .
• 52. Example ∫ 2 1 Evaluate dx. 1 x Solution ∫ 2 1 dx = ln x|2 1 1 x . . . . . .
• 53. Example ∫ 2 1 Evaluate dx. 1 x Solution ∫ 2 1 dx = ln x|2 1 1 x = ln 2 − ln 1 . . . . . .
• 54. Example ∫ 2 1 Evaluate dx. 1 x Solution ∫ 2 1 dx = ln x|2 1 1 x = ln 2 − ln 1 = ln 2 . . . . . .
• 55. Outline Last time: The Deﬁnite Integral The deﬁnite integral as a limit Properties of the integral Estimating the Deﬁnite Integral The Midpoint Rule Comparison Properties of the Integral Evaluating Deﬁnite Integrals Examples The Integral as Total Change Indeﬁnite Integrals My ﬁrst table of integrals Computing Area with integrals . . . . . .
• 56. 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 between the sides of that interval. This has many ramiﬁcations: . . . . . .
• 57. 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 between the sides of 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 . . . . . .
• 58. 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 between the sides of 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 . . . . . .
• 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 between the sides of 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 . . . . . .
• 60. Outline Last time: The Deﬁnite Integral The deﬁnite integral as a limit Properties of the integral Estimating the Deﬁnite Integral The Midpoint Rule Comparison Properties of the Integral Evaluating Deﬁnite Integrals Examples The Integral as Total Change Indeﬁnite Integrals My ﬁrst table of integrals Computing Area with integrals . . . . . .
• 61. 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). . . . . . .
• 62. 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 . . . . . .
• 63. My ﬁrst table of integrals ∫ ∫ ∫ [f(x) + g(x)] dx = f(x) dx + g(x) dx ∫ ∫ ∫ x n +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 . . . . . .
• 64. Outline Last time: The Deﬁnite Integral The deﬁnite integral as a limit Properties of the integral Estimating the Deﬁnite Integral The Midpoint Rule Comparison Properties of the Integral Evaluating Deﬁnite Integrals Examples The Integral as Total Change Indeﬁnite Integrals My ﬁrst table of integrals Computing Area with integrals . . . . . .
• 65. 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. . . . . . .
• 66. 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 . . . . . .
• 67. Graph y . . . . . x . 1 . 2 . 3 . . . . . . .
• 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. Solution ∫ 3 Consider (x − 1)(x − 2) dx. Notice the integrand is positive on 0 [0, 1) and (2, 3], and negative on (1, 2). . . . . . .
• 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. Notice the integrand is positive on 0 [0, 1) 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 ]1 [1 3 ]2 [1 ]3 = 3 x3 − 3 x2 + 2x 0 − 3x − 3 x2 + 2x + 3 3 2 3 x − 2 x + 2x ( 2 ) 2 1 2 5 1 5 11 = − − + = . 6 6 6 6 . . . . . .
• 70. 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 . . . . . .
• 71. What about the constant? It seems we forgot about the +C when we say for instance ∫ 1 1 3 x4 1 1 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. . . . . . .
• 72. What have we learned today? The second Fundamental Theorem of Calculus: ∫ b f(x) dx = F(b) − F(a) a where F′ = f. Deﬁnite integrals represent net change of a function over an interval. ∫ We write antiderivatives as indeﬁnite integrals f(x) dx . . . . . .