Upcoming SlideShare
×

# Lesson 25: Evaluating Definite Integrals (Section 041 slides)

604 views
544 views

Published on

A remarkable theorem about definite integrals is that they can be calculated with antiderivatives.

Published in: Technology, Education
1 Like
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

Views
Total views
604
On SlideShare
0
From Embeds
0
Number of Embeds
3
Actions
Shares
0
37
0
Likes
1
Embeds 0
No embeds

No notes for slide

### Lesson 25: Evaluating Definite Integrals (Section 041 slides)

1. 1. . Section 5.3 Evaluating Definite Integrals V63.0121.041, Calculus I New York University December 6, 2010 Announcements Today: Section 5.3 Wednesday: Section 5.4 Monday, December 13: Section 5.5 ”Monday,” December 15: Review and Movie Day! Monday, December 20, 12:00–1:50pm: Final Exam . . . . . .
2. 2. Announcements Today: Section 5.3 Wednesday: Section 5.4 Monday, December 13: Section 5.5 ”Monday,” December 15: Review and Movie Day! Monday, December 20, 12:00–1:50pm: Final Exam . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 2 / 41
3. 3. 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. . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 3 / 41
4. 4. OutlineLast time: The Definite Integral The definite integral as a limit Properties of the integral Comparison Properties of the IntegralEvaluating Definite Integrals ExamplesThe Integral as Total ChangeIndefinite Integrals My first table of integralsComputing Area with integrals . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 4 / 41
5. 5. The definite integral as a limitDefinitionIf f is a function defined on [a, b], the definite integral of f from a to bis the number ∫ b ∑n f(x) dx = lim f(ci ) ∆x a n→∞ i=1 b−awhere ∆x = , and for each i, xi = a + i∆x, and ci is a point in n[xi−1 , xi ].TheoremIf f is continuous on [a, b] or if f has only finitely many jumpdiscontinuities, 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 5 / 41
6. 6. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 6 / 41
7. 7. Example ∫ 1 4Estimate dx using the midpoint rule and four divisions. 0 1 + x2 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 7 / 41
8. 8. Example ∫ 1 4Estimate dx using the midpoint rule and four divisions. 0 1 + x2SolutionDividing up [0, 1] into 4 pieces gives 1 2 3 4 x0 = 0, x1 = , x2 = , x3 = , x4 = 4 4 4 4So 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 7 / 41
9. 9. Example ∫ 1 4Estimate dx using the midpoint rule and four divisions. 0 1 + x2SolutionDividing up [0, 1] into 4 pieces gives 1 2 3 4 x0 = 0, x1 = , x2 = , x3 = , x4 = 4 4 4 4So 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 7 / 41
10. 10. Example ∫ 1 4Estimate dx using the midpoint rule and four divisions. 0 1 + x2SolutionDividing up [0, 1] into 4 pieces gives 1 2 3 4 x0 = 0, x1 = , x2 = , x3 = , x4 = 4 4 4 4So 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 7 / 41
11. 11. Properties of the integralTheorem (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 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 8 / 41
12. 12. More Properties of the IntegralConventions: ∫ ∫ a b f(x) dx = − f(x) dx b a ∫ a f(x) dx = 0 aThis allows us to haveTheorem ∫ c ∫ b ∫ c 5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c. a a b . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 9 / 41
13. 13. Illustrating Property 5Theorem ∫ c ∫ b ∫ c 5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c. a a b y . a b c x . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
14. 14. Illustrating Property 5Theorem ∫ c ∫ b ∫ c 5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c. a a b y ∫ b f(x) dx a . a b c x . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
15. 15. Illustrating Property 5Theorem ∫ c ∫ b ∫ c 5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c. a a b y ∫ b ∫ c f(x) dx f(x) dx a b . a b c x . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
16. 16. Illustrating Property 5Theorem ∫ c ∫ b ∫ c 5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c. a a b y ∫ b ∫ c ∫ c f(x) dxf(x) dxf(x) dx a a b . a b c x . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
17. 17. Illustrating Property 5Theorem ∫ c ∫ b ∫ c 5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c. a a b y . a c x b . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
18. 18. Illustrating Property 5Theorem ∫ c ∫ b ∫ c 5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c. a a b y ∫ b f(x) dx a . a c x b . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
19. 19. Illustrating Property 5Theorem ∫ c ∫ b ∫ c 5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c. a a b y ∫ c f(x) dx = b∫ b − f(x) dx c . a c x b . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
20. 20. Illustrating Property 5Theorem ∫ c ∫ b ∫ c 5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c. a a b y ∫ c ∫ c f(x) dx f(x) dx = a b∫ b − f(x) dx c . a c x b . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
21. 21. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 11 / 41
22. 22. Comparison Properties of the IntegralTheoremLet 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 12 / 41
23. 23. Comparison Properties of the IntegralTheoremLet 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 7. If f(x) ≥ g(x) for all x in [a, b], then ∫ b ∫ b f(x) dx ≥ g(x) dx a a . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 12 / 41
24. 24. Comparison Properties of the IntegralTheoremLet 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 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 12 / 41
25. 25. Estimating an integral with inequalitiesExample ∫ 2 1Estimate dx using Property 8. 1 x . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 13 / 41
26. 26. Estimating an integral with inequalitiesExample ∫ 2 1Estimate dx using Property 8. 1 xSolutionSince 1 1 1 1 ≤ x ≤ 2 =⇒ ≤ ≤ 2 x 1we have ∫ 2 1 1 · (2 − 1) ≤ dx ≤ 1 · (2 − 1) 2 1 xor ∫ 2 1 1 ≤ dx ≤ 1 2 1 x . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 13 / 41
27. 27. OutlineLast time: The Definite Integral The definite integral as a limit Properties of the integral Comparison Properties of the IntegralEvaluating Definite Integrals ExamplesThe Integral as Total ChangeIndefinite Integrals My first table of integralsComputing Area with integrals . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 14 / 41
28. 28. 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 . . . . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 15 / 41
29. 29. Theorem of the DayTheorem (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 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 16 / 41
30. 30. Theorem of the DayTheorem (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). aNoteIn Section 5.3, this theorem is called “The Evaluation Theorem”.Nobody else in the world calls it that. . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 16 / 41
31. 31. Proving the Second FTC b−a Divide up [a, b] into n pieces of equal width ∆x = as usual. n . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 17 / 41
32. 32. Proving the Second FTC b−a Divide up [a, b] into n pieces of equal width ∆x = as usual. n 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 ) xi − xi−1 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 17 / 41
33. 33. Proving the Second FTC b−a Divide up [a, b] into n pieces of equal width ∆x = as usual. n 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 ) xi − xi−1 Or f(ci )∆x = F(xi ) − F(xi−1 ) . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 17 / 41
34. 34. Proof continued We have for each i f(ci )∆x = F(xi ) − F(xi−1 ) . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
35. 35. Proof continued We have for each i f(ci )∆x = F(xi ) − F(xi−1 ) Form the Riemann Sum: . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
36. 36. Proof continued 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
37. 37. Proof continued 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 )) . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
38. 38. Proof continued 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 )) . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
39. 39. Proof continued 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 )) . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
40. 40. Proof continued 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 )) . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
41. 41. Proof continued 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 )) . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
42. 42. Proof continued 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 )) . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
43. 43. Proof continued 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 )) . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
44. 44. Proof continued 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 )) . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
45. 45. Proof continued 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 )) . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
46. 46. Proof continued 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 )) . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
47. 47. Proof continued 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 )) . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
48. 48. Proof continued 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) . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
49. 49. Proof Completed We have shown for each n, Sn = F(b) − F(a) Which does not depend on n. . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 19 / 41
50. 50. Proof Completed We have shown for each n, Sn = F(b) − F(a) Which does not depend on n. So in the limit ∫ b f(x) dx = lim Sn = lim (F(b) − F(a)) = F(b) − F(a) a n→∞ n→∞ . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 19 / 41
51. 51. Computing area with the Second FTCExampleFind the area between y = x3 and the x-axis, between x = 0 and x = 1. . . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 20 / 41
52. 52. Computing area with the Second FTCExampleFind 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 . . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 20 / 41
53. 53. Computing area with the Second FTCExampleFind 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 20 / 41
54. 54. Computing area with the Second FTCExampleFind the area enclosed by the parabola y = x2 and the line y = 1. . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 21 / 41
55. 55. Computing area with the Second FTCExampleFind the area enclosed by the parabola y = x2 and the line y = 1. 1 . −1 1 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 21 / 41
56. 56. Computing area with the Second FTCExampleFind the area enclosed by the parabola y = x2 and the line y = 1. 1 . −1 1Solution ∫ 1 [ ]1 ( [ )] x3 1 1 4 A=2− x dx = 2 − 2 =2− − − = −1 3 −1 3 3 3 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 21 / 41
57. 57. Computing an integral we estimated beforeExample ∫ 1 4Evaluate the integral dx. 0 1 + x2 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 22 / 41
58. 58. Example ∫ 1 4Estimate dx using the midpoint rule and four divisions. 0 1 + x2SolutionDividing up [0, 1] into 4 pieces gives 1 2 3 4 x0 = 0, x1 = , x2 = , x3 = , x4 = 4 4 4 4So 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 23 / 41
59. 59. Computing an integral we estimated beforeExample ∫ 1 4Evaluate the integral dx. 0 1 + x2Solution ∫ 1 ∫ 1 4 1 dx = 4 dx 0 1 + x2 0 1 + x2 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 24 / 41
60. 60. Computing an integral we estimated beforeExample ∫ 1 4Evaluate the integral dx. 0 1 + x2Solution ∫ 1 ∫ 1 4 1 dx = 4 dx 0 1 + x2 0 1 + x2 = 4 arctan(x)|1 0 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 24 / 41
61. 61. Computing an integral we estimated beforeExample ∫ 1 4Evaluate the integral dx. 0 1 + x2Solution ∫ 1 ∫ 1 4 1 dx = 4 dx 0 1 + x2 0 1 + x2 = 4 arctan(x)|1 0 = 4 (arctan 1 − arctan 0) . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 24 / 41
62. 62. Computing an integral we estimated beforeExample ∫ 1 4Evaluate the integral dx. 0 1 + x2Solution ∫ 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 24 / 41
63. 63. Computing an integral we estimated beforeExample ∫ 1 4Evaluate the integral dx. 0 1 + x2Solution ∫ 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 24 / 41
64. 64. Computing an integral we estimated beforeExample ∫ 2 1Evaluate dx. 1 x . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 25 / 41
65. 65. Estimating an integral with inequalitiesExample ∫ 2 1Estimate dx using Property 8. 1 xSolutionSince 1 1 1 1 ≤ x ≤ 2 =⇒ ≤ ≤ 2 x 1we have ∫ 2 1 1 · (2 − 1) ≤ dx ≤ 1 · (2 − 1) 2 1 xor ∫ 2 1 1 ≤ dx ≤ 1 2 1 x . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 26 / 41
66. 66. Computing an integral we estimated beforeExample ∫ 2 1Evaluate dx. 1 xSolution ∫ 2 1 dx 1 x . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 27 / 41
67. 67. Computing an integral we estimated beforeExample ∫ 2 1Evaluate dx. 1 xSolution ∫ 2 1 dx = ln x|2 1 1 x . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 27 / 41
68. 68. Computing an integral we estimated beforeExample ∫ 2 1Evaluate dx. 1 xSolution ∫ 2 1 dx = ln x|2 1 1 x = ln 2 − ln 1 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 27 / 41
69. 69. Computing an integral we estimated beforeExample ∫ 2 1Evaluate dx. 1 xSolution ∫ 2 1 dx = ln x|2 1 1 x = ln 2 − ln 1 = ln 2 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 27 / 41
70. 70. OutlineLast time: The Definite Integral The definite integral as a limit Properties of the integral Comparison Properties of the IntegralEvaluating Definite Integrals ExamplesThe Integral as Total ChangeIndefinite Integrals My first table of integralsComputing Area with integrals . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 28 / 41
71. 71. The Integral as Total ChangeAnother way to state this theorem is: ∫ b F′ (x) dx = F(b) − F(a), aor the integral of a derivative along an interval is the total changebetween the sides of that interval. This has many ramifications: . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 29 / 41
72. 72. The Integral as Total ChangeAnother way to state this theorem is: ∫ b F′ (x) dx = F(b) − F(a), aor the integral of a derivative along an interval is the total changebetween the sides of that interval. This has many ramifications:TheoremIf v(t) represents the velocity of a particle moving rectilinearly, then ∫ t1 v(t) dt = s(t1 ) − s(t0 ). t0 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 29 / 41
73. 73. The Integral as Total ChangeAnother way to state this theorem is: ∫ b F′ (x) dx = F(b) − F(a), aor the integral of a derivative along an interval is the total changebetween the sides of that interval. This has many ramifications:TheoremIf MC(x) represents the marginal cost of making x units of a product,then ∫ x C(x) = C(0) + MC(q) dq. 0 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 29 / 41
74. 74. The Integral as Total ChangeAnother way to state this theorem is: ∫ b F′ (x) dx = F(b) − F(a), aor the integral of a derivative along an interval is the total changebetween the sides of that interval. This has many ramifications:TheoremIf ρ(x) represents the density of a thin rod at a distance of x from itsend, then the mass of the rod up to x is ∫ x m(x) = ρ(s) ds. 0 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 29 / 41
75. 75. OutlineLast time: The Definite Integral The definite integral as a limit Properties of the integral Comparison Properties of the IntegralEvaluating Definite Integrals ExamplesThe Integral as Total ChangeIndefinite Integrals My first table of integralsComputing Area with integrals . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 30 / 41
76. 76. A new notation for antiderivativesTo emphasize the relationship between antidifferentiation andintegration, we use the indefinite integral notation ∫ f(x) dxfor any function whose derivative is f(x). . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 31 / 41
77. 77. A new notation for antiderivativesTo emphasize the relationship between antidifferentiation andintegration, we use the indefinite integral notation ∫ f(x) dxfor any function whose derivative is f(x). Thus ∫ x2 dx = 1 x3 + C. 3 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 31 / 41
78. 78. My first table of integrals. ∫ ∫ ∫ [f(x) + g(x)] dx = f(x) dx + g(x) dx ∫ ∫ ∫ xn+1 n x 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 32 / 41
79. 79. OutlineLast time: The Definite Integral The definite integral as a limit Properties of the integral Comparison Properties of the IntegralEvaluating Definite Integrals ExamplesThe Integral as Total ChangeIndefinite Integrals My first table of integralsComputing Area with integrals . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 33 / 41
80. 80. Computing Area with integralsExampleFind the area of the region bounded by the lines x = 1, x = 4, thex-axis, and the curve y = ex . . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 34 / 41
81. 81. Computing Area with integralsExampleFind the area of the region bounded by the lines x = 1, x = 4, thex-axis, and the curve y = ex .SolutionThe answer is ∫ 4 4 ex dx = ex |1 = e4 − e. 1 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 34 / 41
82. 82. Computing Area with integralsExampleFind the area of the region bounded by the curve y = arcsin x, thex-axis, and the line x = 1. . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 35 / 41
83. 83. Computing Area with integralsExampleFind the area of the region bounded by the curve y = arcsin x, thex-axis, and the line x = 1.Solution ∫ 1 The answer is arcsin x dx, but we y 0 do not know an antiderivative for π/2 arcsin. . x 1 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 35 / 41
84. 84. Computing Area with integralsExampleFind the area of the region bounded by the curve y = arcsin x, thex-axis, and the line x = 1.Solution ∫ 1 The answer is arcsin x dx, but we y 0 do not know an antiderivative for π/2 arcsin. Instead compute the area as ∫ π/2 π − sin y dy . 2 0 x 1 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 35 / 41
85. 85. Computing Area with integralsExampleFind the area of the region bounded by the curve y = arcsin x, thex-axis, and the line x = 1.Solution ∫ 1 The answer is arcsin x dx, but we y 0 do not know an antiderivative for π/2 arcsin. Instead compute the area as ∫ π/2 π π π/2 − sin y dy = −[− cos x]0 . 2 0 2 x 1 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 35 / 41
86. 86. Computing Area with integralsExampleFind the area of the region bounded by the curve y = arcsin x, thex-axis, and the line x = 1.Solution ∫ 1 The answer is arcsin x dx, but we y 0 do not know an antiderivative for π/2 arcsin. Instead compute the area as ∫ π/2 π π π/2 π − sin y dy = −[− cos x]0 = −1 . 2 0 2 2 x 1 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 35 / 41
87. 87. ExampleFind the area between the graph of y = (x − 1)(x − 2), the x-axis, andthe vertical lines x = 0 and x = 3. . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 36 / 41
88. 88. ExampleFind the area between the graph of y = (x − 1)(x − 2), the x-axis, andthe vertical lines x = 0 and x = 3.Solution ∫ 3Consider (x − 1)(x − 2) dx. 0 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 36 / 41
89. 89. Graph y . x 1 2 3 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 37 / 41
90. 90. ExampleFind the area between the graph of y = (x − 1)(x − 2), the x-axis, andthe vertical lines x = 0 and x = 3.Solution ∫ 3Consider (x − 1)(x − 2) dx. Notice the integrand is positive on [0, 1) 0and (2, 3], and negative on (1, 2). . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 38 / 41
91. 91. ExampleFind the area between the graph of y = (x − 1)(x − 2), the x-axis, andthe vertical lines x = 0 and x = 3.Solution ∫ 3Consider (x − 1)(x − 2) dx. Notice the integrand is positive on [0, 1) 0and (2, 3], and negative on (1, 2). If we want the area of the region, wehave to do ∫ 1 ∫ 2 ∫ 3A= (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 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 38 / 41
92. 92. Interpretation of “negative area" in motionThere is an analog in rectlinear motion: ∫ t1 v(t) dt is net distance traveled. t0 ∫ t1 |v(t)| dt is total distance traveled. t0 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 39 / 41
93. 93. 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. . . . . . . V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 40 / 41
94. 94. Summary 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 . . . . . .V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 41 / 41