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# Lesson 25: Evaluating Definite Integrals (Section 4 version)

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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 25: Evaluating Definite Integrals (Section 4 version)

1. 1. Section 5.3 Evaluating Deﬁnite Integrals V63.0121, Calculus I April 21, 2009 Announcements Final Exam is Friday, May 8, 2:00–3:50pm Final is cumulative; topics will be represented roughly according to time spent on them . Image credit: docman . . . . . . .
2. 2. Outline . . . . . .
3. 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(ci ) ∆x n→∞ a i=1 b−a , and for each i, xi = a + i∆x, and ci is a point where ∆x = n in [xi−1 , xi ]. . . . . . .
4. 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. 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 c dx = c(b − a) 1. a ∫ ∫ ∫ b b b [f(x) + g(x)] dx = f(x) dx + g(x) dx. 2. a a a ∫ ∫ b b cf(x) dx = c f(x) dx. 3. a a ∫ ∫ ∫ b b b [f(x) − g(x)] dx = f(x) dx − g(x) dx. 4. a a a . . . . . .
6. 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 f(x) dx = f(x) dx + f(x) dx for all a, b, and c. 5. a a b . . . . . .
7. 7. 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 . . . . . .
8. 8. Outline . . . . . .
9. 9. Socratic proof The deﬁnite integral of velocity measures displacement (net distance) The derivative of displacement is velocity So we can compute displacement with the antiderivative of velocity? . . . . . .
10. 10. 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 . . . . . .
11. 11. 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. . . . . . .
12. 12. Proving 2FTC 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 ) xi − xi−1 Or f(ci )∆x = F(xi ) − F(xi−1 ) . . . . . .
13. 13. We have for each i f(ci )∆x = F(xi ) − F(xi−1 ) Form the Riemann Sum: n n ∑ ∑ (F(xi ) − F(xi−1 )) Sn = f(ci )∆x = 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) . . . . . .
14. 14. 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) n→∞ n→∞ a . . . . . .
15. 15. Example Find the area between y = x3 and the x-axis, between x = 0 and x = 1. . . . . . . .
16. 16. Example Find the area between y = x3 and the x-axis, between x = 0 and x = 1. Solution ∫ 1 1 x4 1 x3 dx = A= = 4 4 . 0 0 . . . . . .
17. 17. Example Find the area between y = x3 and the x-axis, between x = 0 and x = 1. Solution ∫ 1 1 x4 1 x3 dx = A= = 4 4 . 0 0 Here we use the notation F(x)|b or [F(x)]b to mean F(b) − F(a). a a . . . . . .
18. 18. Example Find the area enclosed by the parabola y = x2 and y = 1. . . . . . .
19. 19. Example Find the area enclosed by the parabola y = x2 and y = 1. . . . . . . .
20. 20. Example Find the area enclosed by the parabola y = x2 and y = 1. . Solution [ ]1 [ ( )] ∫ 1 x3 −1 1 4 2 A=2− x dx = 2 − =2− −− = 3 3 3 3 −1 −1 . . . . . .
21. 21. Outline . . . . . .
22. 22. 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: . . . . . .
23. 23. 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 . . . . . .
24. 24. 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 . . . . . .
25. 25. 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 . . . . . .
26. 26. Outline . . . . . .
27. 27. 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). . . . . . .
28. 28. 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 . . . . . .
29. 29. My ﬁrst table of integrals ∫ ∫ ∫ [f(x) + g(x)] dx = f(x) dx + g(x) dx ∫ ∫ ∫ xn+1 xn dx = cf(x) dx = c f(x) dx + C (n ̸= −1) n+1 ∫ ∫ 1 ex dx = ex + C dx = ln |x| + C x ∫ ∫ ax ax dx = +C sin x dx = − cos x + C ln a ∫ ∫ csc2 x dx = − cot x + C cos x dx = sin x + C ∫ ∫ sec2 x dx = tan x + C csc x cot x dx = − csc x + C ∫ ∫ 1 √ dx = arcsin x + C sec x tan x dx = sec x + C 1 − x2 ∫ 1 dx = arctan x + C 1 + x2 . . . . . .
30. 30. Outline . . . . . .
31. 31. 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. . . . . . .
32. 32. 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 (x − 1)(x − 2) dx. Notice the integrand is positive on Consider 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 (x − 1)(x − 2) dx − (x − 1)(x − 2) dx + (x − 1)(x − 2) dx A= 0 1 2 [1 ]1 [1 3 ]2 [1 ]3 x3 − 3 x2 + 2x 0 − − 3 x2 + 2x 3 32 3 x − 2 x + 2x 3x = + 3 (2) 2 1 2 5 1 5 11 = −− += . 6 6 6 6 . . . . . .
33. 33. Graph from previous example y . . . . . x . 2 . 3 . 1 . . . . . . .
34. 34. Summary integrals can be computed with antidifferentiation integral of instantaneous rate of change is total net change The second Funamental Theorem of Calculus requires the Mean Value Theorem . . . . . .