lecture_notes3

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Unformatted text preview: 1 preview : integrals (theory/applications), differential equations, series 1st main goal : prepare for FTC Appendix E. : sigma notation n 1 Tues 9/7 def : i=1 ai = a1 + a2 + a3 + · · · + an−1 + an : sum , series i : index , ai : terms , 1, n : limits ex 5 i = 1 + 2 + 3 + 4 + 5 = 15 i=1 5 i=1 i2 = 1 + 4 + 9 + 16 + 25 = 55 note : A series can be written in different ways. ex 5 4 i= i=1 (j + 1) = 1 + 2 + 3 + 4 + 5 = 15 j =0 ok set j = i − 1 thm n n 1. i=1 n cai = c · i=1 ai , where c is any constant n n 2. n (ai + bi ) = i=1 n i=1 ai + i=1 n bi 3. i=1 ai b i = i=1 n ai · i=1 bi n pf 1. i=1 cai = ca1 + ca2 + · · · + can = c(a1 + a2 + · · · + an ) = c · i=1 ai ok 2. hw 3. example (n = 2) 2 ai bi = a1 b1 + a2 b2 i=1 2 2 ai · i=1 i=1 bi = (a1 + a2 ) · (b1 + b2 ) = a1 b1 + a1 b2 + a2 b1 + a2 b2 ok 2 thm n (ai+1 − ai ) = an+1 − a1 : telescoping series i=1 pf n (ai+1 − ai ) i=1 = (a2 − a1 ) + (a3 − 2 ) + (a4 − 3 ) + · · · + (an − an−1 ) + (an+1 − n ) a a a = −a1 + an+1 thm n ok 1. i=1 n 1 = 1 + 1 + 1 + ··· + 1 = n i = 1 + 2 + 3 + ··· + n = i=1 n 2. 3. i=1 n(n + 1) 2 implies ↓ (n = 5 ⇒ S = 15) (n = 5 ⇒ S = 55) i2 = 1 + 4 + 9 + · · · + n2 = n(n + 1)(2n + 1) 6 pf 1. ok 2. (i + 1)2 − i2 = i2 + 2i + 1 − i2 = 2i + 1 n i=1 n ((i + 1) − i ) = 2 2 n n n n (2i + 1) = 2 i=1 i=1 i+ i=1 1 = 2S + n , where S = i=1 i ((i + 1)2 − i2 ) = (22 − 12 ) + (32 − 22 ) + (42 − 32 ) + · · · + ((n + 1)2 − n2 ) = (n + 1)2 − 12 = n2 + 2n n(n + 1) 2 i=1 ⇒ 2S + n = n2 + 2n ⇒ 2S = n2 + n = n(n + 1) ⇒ S = 3. hw , (i + 1)3 − i3 = · · · ok 3 5.1 area Given f (x) ≥ 0 , a ≤ x ≤ b. 2 Wed 9/8 y f(x) R a x0 x1 x2 b xn x R = region in the xy -plane between y = 0 and y = f (x) for a ≤ x ≤ b = {(x, y ) : a ≤ x ≤ b, 0 ≤ y ≤ f (x)} problem : find the area of R solution : approximate by rectangles choose n ≥ 1 , set ∆x = x0 = a x1 = a + ∆x x2 = a + 2∆x ... xn = a + n∆x = a + b − a = b area of ith rectangle = f (xi )∆x n b−a , xi = a + i∆x , i = 0, . . . , n n area of region R ≈ i=1 f (xi )∆x : Riemann sum ↑ approximately n area of region R = nlim →∞ f (xi )∆x i=1 4 ex f (x) = x , 0 ≤ x ≤ 1 y f(x) R x 0 1 R = {(x, y ) : 0 ≤ x ≤ 1, 0 ≤ y ≤ x} of course , area = 1 1 · base · height = 2 2 1 b−a = n n a = 0 , b = 1 , ∆x = xi = a + i∆x = f (xi ) = xi = n i n i n i1 1n 1 n(n + 1) n+1 · =2 i = 2· = n n i=1 n 2 2n i=1 n n f (xi )∆x = i=1 n area = nlim →∞ i=1 f (xi )∆x = nlim →∞ 1 1 n+1 = nlim + →∞ 2 2n 2n = 1 2 ok 5 5.2 definite integral As before, given f (x) , a ≤ x ≤ b , set ∆x = Let x∗ be any point such that xi−1 ≤ x∗ ≤ xi . i i n b−a , xi = a + i∆x. n Then i=1 f (x∗ )∆x is a Riemann sum. i ex x∗ = xi : right-hand RS i x∗ = xi−1 : left-hand RS i xi−1 + xi x∗ = : midpoint RS i 2 def : ex 1 0 n b a n f (x) dx = nlim →∞ f (x∗ )∆x : definite integral = i area work probability i=1 ··· dx = nlim →∞ 1· i=1 1 = nlim 1 = 1 →∞ n b−a 1 i = , xi = a + i∆x = , f (x) = 1 n n n a = 0 , b = 1 , ∆x = 1 0 n x dx = nlim →∞ i1 1 n(n + 1) 1 · = nlim 2 · = →∞ n n 2 2 i=1 n i n f (x) = x , f (xi ) = xi = 12 x dx 0 n = nlim →∞ 2 i=1 i21 1 n(n + 1)(2n + 1) 1 · = nlim 3 · = →∞ n n n 6 3 x2 i i = n 2 f (x) = x , f (xi ) = 1x e dx 0 n n 1 i1 = nlim = ··· = e − 1 e1/n · →∞ n n i=1 i=1 ↑ f (x) = ex , f (xi ) = exi = ei/n geometric series (hw2) = nlim →∞ ei/n · thm : If f (x) is continuous, then b a f (x) dx exists. pf : MATH 451 (advanced calculus) 3 Fri 9/10 6 def n Rn = i=1 n f (xi )∆x : right-hand RS , f i=1 lim R = n→∞ n b a f (x) dx b a Mn = xi−1 + xi ∆x : midpoint RS , 2 lim Mn = n→∞ f (x) dx question : For a given value of n, which approximation is more accurate? ex : 1x e dx 0 = e − 1 = 1.71828183 = I Mn |I − Mn | 1.6487 0.0696 1.7005 0.0178 1.7138 0.0045 1.7172 0.0011 ↓ ↓ I 0 Rn |I − Rn | n ∆x 11 2.7183 1.0000 2.1835 0.4652 2 0.5 4 0.25 1.9420 0.2237 8 0.125 1.8279 0.1096 ↓ ↓ ↓ ↓ ∞0 I 0 conclusions 1. The midpoint RS is more accurate than the right-hand RS. 2. If ∆x decreases by a factor of 1 , then the error in the right-hand RS decreases 2 1 by a factor of approximately 2 and the error in the midpoint RS decreases by a factor of approximately 1 . 4 Why? Consider n = 2. y y x x0 x1 x2 x0 x1 x2 ex (group work) x Draw the region R, express the area of R as a limit of Riemann sums, and evaluate the limit (if possible). 1. R = {(x, y ) : 1 ≤ x ≤ 2, 0 ≤ y ≤ x} 2. R = {(x, y ) : 0 ≤ x ≤ 1, 0 ≤ y ≤ x2 } √ 3. R = {(x, y ) : −1 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x2 } 7 properties of the definite integral 1. 2. 3. b a b a c · f (x) dx = c · b a f (x) dx b a 4 Mon 9/13 b a (f (x) + g (x)) dx = f (x) dx = c a f (x) dx + b c g (x) dx b a f (x) dx + f (x) dx b a 4. If f (x) ≤ g (x) for a ≤ x ≤ b, then pf 1. b a n f (x) dx ≤ n b a g (x) dx. b a c · f (x) dx = nlim →∞ 1x e dx 0 i=1 c · f (xi )∆x = c · nlim →∞ f (xi )∆x = c · i=1 f (x) dx ok 2. hw2 , 3. , 4. omit ex : y e 1 x 0 1 1x e dx 0 ≤ 1 (e + 1) = 1.8591 : upper bound (exact value = 1.718) 2 ex (e − 1)x + 1 e ex ≤ e for 0 ≤ x ≤ 1 ⇒ ≤ 1 0 e dx = e = 2.7183 : too big , try again ex ≤ (e − 1)x + 1 for 0 ≤ x ≤ 1 ⇒ 1x e dx 0 ≤ 1 0 ((e − 1)x + 1) dx = (e − 1) 1 0 1 2 x dx + 1 0 dx ok 1 = (e − 1) · + 1 = 2 (e + 1) = 1.8591 question : can you find a better upper bound? a lower bound? note : If f (x) changes sign, then ex : f (x) = sin x ⇒ y 0 π 2π x π 0 b a f (x) dx = signed area. 2π π sin x dx > 0 , sin x dx < 0 , 2π 0 sin x dx = 0 8 5.3 FTC , 5.4 antiderivatives thm (FTC, part 1) If F (x) = pf x a f (t) dt, then F (x) = f (x). f a recall : F (x) = lim F (x + h) = x +h a x x+h F (x + h) − F (x) h→0 h f (t) dt = x x a t f (t) dt + x +h x f (t) dt = F (x) + x+h x f (t) dt ⇒ F (x + h) − F (x) = ⇒ x+h f (t) dt ≈ f (t∗ ) · h , where x ≤ t∗ ≤ x + h F (x + h) − F (x) ≈ f (t∗ ) h F (x + h) − F (x) = lim f (t∗ ) = f (x) h→0 h→0 h x a ⇒ F (x) = lim ex ok (if f is continuous) f (x) = x , a = 0 , F (x) = check F (x) = ∆t = x 0 n f (t) dt = x 0 t dt , FTC ⇒ F (x) = x 5 Tues 9/14 t dt = nlim →∞ ti ∆t i=1 b−a x x = , ti = a + i∆t = i · n n n n F (x) = nlim →∞ ⇒ F (x) = i· i=1 x2 xx x2 n x2 n(n + 1) x2 · = nlim 2 i = nlim 2 · = →∞ n →∞ n nn 2 2 i=1 =x ok 2 9 thm (FTC, part 2) If f (x) = F (x), then pf ∆x = b a f (x) dx = b a F (x) dx = F (b) − F (a). b−a , xi = a + i∆x n n n xi + ∆x = a + i∆x + ∆x = a + (i + 1)∆x = xi+1 F (xi + ∆x) − F (xi ) f (xi )∆x = F (xi )∆x ≈ ∆x ∆x i=1 i=1 i=1 n n = x1 = a + ∆ x (F (xi+1 ) − F (xi )) = F (xn+1 ) − F (x1 ) : telescoping sum i=1 xn+1 = a + (n + 1)∆x = a + n∆x + ∆x = b + ∆x b a n f (x) dx = nlim →∞ i=1 f (xi )∆x = nlim (F (b + ∆x) − F (a + ∆x)) = F (b) − F (a) →∞ ok notation We say that F (x) is an antiderivative of f (x) and we write f (x)dx = F (x) + c. We also write note We previously showed that 1 0 b a b F (x) dx = F (x) = F (b) − F (a). a x dx = nlim →∞ 1 0 i1 1 n(n + 1) 1 = nlim 2 =. →∞ n 2 2 i=1 n n 1 0 n Now we can use the FTC instead : ex x3 1. x dx = −1 3 1 2 1 −1 x dx = 2 3 x2 x2 dx = 2 2 1 0 = 1 . 2 = 1 −1 − 3 3 = 2. dx −1 = −1 x2 x 1 1 −1 = −1 −1 − 1 −1 = −2 1 dx 1 is positive, so why is negative? −1 x2 x2 1 answer : FTC doesn’t apply, because 2 is not continuous at x = 0. x question : The function 10 f (x) x n F (x) xn+1 , n=1 n+1 ln x x ln x − x ex − cos x sin x cosh x sinh x tan−1 x x t2 e a x−1 ln x ex sin x cos x ex − e−x sinh x = 2 x e + e−x cosh x = 2 1 x2 + 1 ex 2 more later dt 6 Wed 9/15 summary of FTC d 1. dx 2. b a x a f (t)dt = f (x) ⇒ f (x)dx = f (b) − f (a) d dx g (x) a “integration and differentiation are inverse operations” note : f (t)dt = f (g (x)) · g (x) x a pf : Set F (x) = ⇒ d dx d dx g ( x) a f (t)dt. Then F (x) = f (x) and g (x) a f (t)dt = F (g (x)). ex : a d F (g (x)) = F (g (x)) · g (x) = f (g (x)) · g (x) ok dx ↑ chain rule x2 t 2 e dt = ex · 2x f (t)dt = =e t x2 a check : x2 t e dt a = e −e x2 a d ⇒ dx x2 t e dt a = d x2 2 e − e a = e x · 2x dx ok 11 6. applications of integration 6.1 area 6.2, 6.3 volume 6.4 work ← 6.5 average value of a function 6.4 work force = mass × acceleration work = force × distance units metric British mass kilogram: kg slug : slug distance meter : m foot : ft time second : s second : s m force Newton : N = kg · s2 pound : lb work Joule : J = N·m foot-pound : ft-lb conversion : 1 m = 3.28 ft , 1 N = 0.225 lb , 1 J = 0.738 ft-lb ex How much work is done in lifting a 1 kg book to a height of 1 m off the ground? W = force × distance = m g × d = 1 kg × 9.8 m × 1 m = 9.8 J s2 ex : A water tank has the shape of an inverted cone. x R H R : base radius of cone H : height of cone h h : water level ¢x xi 0 Find the work done in pumping the water out the top of the tank. 12 strategy : think of the water volume as a stack of books h : width of a water layer n xi = i∆x : height of ith layer ∆x = ri = · · · ? : radius of ith layer 7 Fri 9/17 R ri R R = ⇒ ri = xi H xi H H ri xi check : xi = 0 ⇒ ri = 0 xi = H ⇒ ri = R ok work = force × distance = mass × acceleration × distance 2 mass of water in ith layer = density × volume = ρ · πri ∆x 2 force acting on ith layer = ρπri ∆x · g 2 work done in raising ith layer = ρgπri ∆x · (H − xi ) work done in raising entire water volume n = nlim →∞ h2 x (H 0 2 ρgπri (H h − xi )∆x = 2 3 h 0 ρgπx 2R 2 i=1 H2 (H − x)dx h3 h4 h3 − x)dx = (Hx − x )dx = H − = (4H − 3h) 0 3 4 12 R2 h3 W = ρgπ 2 · (4H − 3h) , check : h = 0 ⇒ W = 0 ok H 12 plug in some numbers R = 4 m , H = 10 m , h = 8 m , ρ = 1000 kg/m3 , g = 9.8 m/s2 ¡ 16 83 kg m m2 ¡ 3 ¡ W = 10 (9.8) (3.14) 2 · (40 − 24) 3 2 2 m · m ¡ ¡ 10 12 msm¡ ¡ ¡ 4 ≈ 10 · 10 · ¡¡ · · 5 · 102 · 16 J ≈ 3.2 · 106 J 3 ¡ 3 ¡ 3 13 note In the previous examples, the force (due to gravity) is assumed to be constant as the mass is raised (book, water layer). However in general, the force may depend on the displacement. ex Compute the work done in moving an object from x = a to x = b, subject to a force f (x). b−a x , ∆x = , xi = a + i∆x n a b work done in moving object from xi−1 to xi ≈ f (xi )∆x n · · · · · · · · · · · · · “ · · · · · · · · · · · · from a to b ≈ n i=1 f (xi )∆x ⇒ W = nlim →∞ ex f (xi )∆x = i=1 b a f (x)dx 8 Mon 9/20 A spring is stretched x units from its natural length. : natural length 0 : stretched length 0 x Hooke’s law : The force needed to maintain a spring in its stretched state is proportional to the displacement x from its natural length. force = f (x) = kx , k > 0 : spring constant ex A spring has natural length 10 cm and a force of 40 N is needed to maintain the spring when it is stretched to a length of 15 cm. How much work is done in stretching the spring from 15 cm to 20 cm. 40 N = k · (15 cm − 10 cm) = k · 5 cm ⇒ k = 8 N/cm W= b a f (x)dx = 10 5 x2 kxdx = k 2 10 =8 5 100 25 N − = 300 cm2 = 3 N m = 3 J 2 2 cm 14 ex Find the work done in moving a particle from the Earth’s surface to ∞. 0 R r R : radius of Earth r : distance from particle to center of Earth , R ≤ r < ∞ f (r) = GM m : force on particle due to gravity r2 G : gravitational constant M : mass of Earth m : mass of particle W= ∞ R f (r)dr = GM m ∞ R dr −1 = GM m · r2 r ∞ R = GM m · 0 − −1 GM m = R R ↑ improper integral , more later note : We can compute the escape velocity of the particle. 2 mvesc GM m 2GM initial kinetic energy = work ⇒ = ⇒ vesc = 2 R R /2 1 1/2 vesc 6.67 · 10−11 N · m2 /kg2 × 5.98 · 1024 kg km = 2 · ≈ 11 ≈ 30 · sound speed (air) 6.37 · 106 m s note : black hole ⇒ M → ∞ , R → 0 ⇒ vesc → ∞ : impossible ⇒ nothing can escape the gravitational field of a black hole 15 8.8 improper integrals def b a and f (x) is a bounded function for a ≤ x ≤ b a = −∞ or b = ∞ b otherwise, f (x) dx is an improper integral, i.e. if or a f (x) → ±∞ in (a, b) ex 1 0 f (x) dx is a proper integral if (a, b) is a bounded interval 9 Tues 9/21 x dx : proper x dx : improper (because b = ∞) ∞ 1 ∞ dx : improper 1 x 1 dx : improper (because 0x 2 dx : proper 1x 1 x → ∞ as x → 0) note : An improper integral is evaluated by taking a limit of proper integrals. If the limit is finite, the integral converges; otherwise, it diverges. dx 1b 1 = lim − = lim − + 1 = 1 : converges 1 x2 1 b→∞ x 1 b→∞ b 1 ⇒ the area under the graph of y = 2 from x = 1 to x = ∞ is finite x ∞ dx 1∞ 1 short cut : =− = − +1=1 1 x2 x1 ∞ ex : ∞ dx = lim b→∞ x2 b ex : ∞ 1 dx = ln x x ∞ 1 = ln ∞ − ln 1 = ∞ : diverges 1 from x = 1 to x = ∞ is infinite x ⇒ the area under the graph of y = ex : ∞ dx √ : diverges 1 x √ √∞ ∞ dx √ = 2 x = 2( ∞ − 1) = ∞ : diverges pf 1 : 1 1 x ∞ dx ∞ dx 1 1 √≥ pf 2 : √ ≥ for x ≥ 1 ⇒ : diverges (comparison test . . .) 1 1 xx x x 16 4 3.5 3 2.5 2 1.5 1 0.5 0 111 comparison of y = √ , , 2 xxx 0 1 2 3 4 5 6 √ 1/ x , p = 1 2 1/x , p = 1 1/x2 , p = 2 10 Wed 9/22 note So far we considered Now consider ex 1 0 1 0 1 0 1 0 ∞ 1 f (x) dx, where f (x) is bounded for 1 ≤ x < ∞. f (x) dx, where f (x) is unbounded for 0 ≤ x ≤ 1. √ dx √ =2 x x dx = ln x x dx 1 =− x2 x 1 0 1 0 1 0 = 2 − 0 = 2 : converges , p = 1 2 = ln 1 − ln 0 = 0 − (−∞) = ∞ : diverges , p = 1 1 1 =− − − = −1 + ∞ = ∞ : diverges , p = 2 1 0 note : The comparison test can also be used in these examples. summary (p-test) ∞ 1 dx : xp 3 2 converges if p > 1 diverges if p ≤ 1 ∞ 1 , 1 0 dx : xp 1 0 converges if p < 1 diverges if p ≥ 1 pf : omit ex : p = ⇒ dx : converges , x3/2 dx : diverges x3/2 17 note : An improper integral the integrals ex : 2 0 c a b a f (x)dx converges if and only if (⇔) f (x)dx and 1 0 b c f (x)dx converge for all c such that a ≤ c ≤ b. 2 1 dx = 1−x dx + 1−x dx : diverges 1−x y 1 2 0 1 x 1 0 1 dx = − ln(1 − x) = − ln 0 + ln 1 = ∞ : diverges 0 1−x ∞ −x e dx 0 ex : y 1 = −e−x ∞ 0 = −e−∞ − (−e0 ) = 0 − (−1) = 1 : converges e −x x xe−x dx = 1 : converges 0 ∞ x 1 lim xe−x = ∞ · 0 = xlim x = = xlim x = 0 (l’Hopital’s rule, pf later) x→∞ →∞ e →∞ e ∞ y ex : e−x xe−x x 1 ∞ 0 b a ∞ xe−x dx = ? : integration by parts (pf soon) b b a a udv = uv − ∞ 0 vdu , choose : u = x , dv = e−x dx ⇒ du = dx , v = −e−x ¨ ¨∞ ∞ −x e dx 0 ⇒ − xe−x dx = −xe¨x 0 + ¨¨ = 0+1 = 1 question : What happens if we choose u = e−x , dv = xdx? . . . 18 integration by parts (u(x)v (x)) = u(x)v (x) + u (x)v (x) (uv ) = uv + u v ⇒ ⇒ b a b a 11 Mon 9/27 (uv ) dx = b a b uv dx + b a b a u v dx = uv b a uv dx = uv − a vu dx v dx = ⇒ ex b a du dv · dx = dv , u dx = · dx = du dx dx b b a a udv = uv − vdu ok ∞ −x 2 e dx 0 : converges e−x 2 y 1 e−1 e−x x 1 The antiderivative is not an elementary function and integration by parts doesn’t help in this case, so we will use a different approach. ∞ −x 2 e dx 0 1 −x2 e dx 0 = 1 −x 2 e dx 0 + ∞ −x 2 e dx 1 : converges (proper integral) 2 x ≥ 1 ⇒ x2 ≥ x ⇒ −x2 ≤ −x ⇒ e−x ≤ e−x for x ≥ 1 (this is because ex is an increasing function, i.e. a ≤ b ⇒ ea eb ) ⇒ ∞ −x 2 e dx 1 ≤ ∞ −x e dx 1 : converges (comparison test) ok note : ∞ −x 2 e dx 0 √ = π : MATH 255 2 19 ex ∞ 0 dx : converges x2 + 1 y 1 1 x2 + 1 x note : 1 1 ≤ 2⇒ x2 + 1 x ∞ 0 dx ≤ x2 + 1 ∞ 0 1 dx =− x2 x ∞ 0 =− 1 1 + = ∞ : diverges ∞0 Hence the comparison test yields no information. 1 1 But if x → ∞, then 2 ∼ 2 , so we suspect that the given integral converges. x +1 x ∞ 0 dx = x2 + 1 ∞ dx dx + 0 x2 + 1 1 x2 + 1 ↑ ↑ proper converges because 1 ∞ 1 dx ≤ x2 + 1 ∞ 1 dx : converges x2 ok 12 Tues 9/28 alternative : x dx = tan−1 x = arctan x 2+1 x √ x = tan θ : trigonometric substitution x2 + 1 θ 1 dx = sec2 θ dθ √ sec θ = x2 + 1 ¨ dx = x2 + 1 ¨ sec2¨ dθ ¨θ = dθ = θ = tan−1 x ¨ 2¨ sec θ ¨¨ tan x ok tan−1 x π 2 π − 32 −π − π 2 π 2 π 3π 2 x −π 2 x ∞ 0 ∞ dx π π = tan−1 x = tan−1 ∞ − tan−1 0 = − 0 = : converges 0 x2 + 1 2 2 ok 20 ex ∞ 0 √ y 1 dx : diverges x2 + 1 1 x2 + 1 √ x idea : √ 1 x2 + 1 ∼ 1 for x → ∞ , so we expect that the integral diverges x Let’s turn this idea into a proof. ∞ ∞ ∞ dx 1 1 dx √ √ ≤ ⇒ ≤ = ln x = ln ∞ − ln 0 = ∞ : diverges 0 0 0 x x x2 + 1 x2 + 1 However, this yields no information about the given integral. To proceed, we need a reverse inequality and there is a trick for that. √ 1 1 1 √ ⇔ x2 + 1 ≤ 2x ⇔ x2 + 1 ≤ 4x2 ⇔ 1 ≤ 3x2 ⇔ x ≥ √ ≥ 2x 3 x2 + 1 ∞ 0 √ dx = x2 + 1 1 √ 3 0 √ dx + x2 + 1 ∞ 1 √ 3 √ dx x2 + 1 ∞ 1 √ 3 1st integral : proper 2nd integral : alternative √ x θ 1 √ dx = x2 + 1 x2 + 1 x = tan θ dx = sec2 θ dθ √ sec θ = x2 + 1 ∞ 1 √ 3 √ dx ≥ x2 + 1 ∞ 1 √ 3 dx 1 = 2x 2 dx : diverges x ok sec θdθ = sec2 θ dθ = sec θdθ sec θ 1 cos θ dθ = dθ = cos θ cos2 θ cos θ dθ = 1 − sin2 θ du =? 1 − u2 u = sin θ ⇒ du = cos θdθ 21 1 0.8 comparison of y = √ 0.6 1 2 , e −x x2 + 1 x2 + 1 1 , 0.4 0.2 √ 1/ x 2 + 1 e−x 2 1/(x2 + 1) 2.5 3 3.5 4 4.5 5 0 0 0.5 1 1.5 2 3 10 partial fractions , idea : = 3 2·5 = 1 2 − 1 5 1 1 a b a(1 − u) + b(1 + u) = = + = 1 − u2 (1 + u)(1 − u) 1 + u 1 − u (1 + u)(1 − u) (a + b) + u(−a + b) a+b=1 ⇒ = −a + b = 0 1 − u2 ⇒ 1 1 1 = + 1 − u2 2(1 + u) 2(1 − u) ⇒ a=b= 1 2 13 Wed 9/29 , check . . . du = 1 − u2 du du 1 1 1 1+u + = ln(1 + u) − ln(1 − u) = ln 2(1 + u) 2(1 − u) 2 2 2 1−u 1 1 + sin θ 1 1 + sin θ 1 + sin θ 1 (1 + sin θ)2 sec θdθ = ln = ln · = ln 2 1 − sin θ 2 1 − sin θ 1 + sin θ 2 cos2 θ = ln 1 + sin θ = ln(sec θ + tan θ) cos θ sec θdθ = ln(sec θ + tan θ) , check . . . √ dx √ = ln( x2 + 1 + x) , check . . . x2 + 1 √ ∞ ∞ dx √ = ln( x2 + 1 + x) = ln ∞ − ln 1 : diverges 0 0 x2 + 1 ok 22 9.1 arclength problem : compute the length of the graph of a function y f (x) x a 1st approximation : b (b − a)2 + (f (b) − f (a))2 a+b , x2 = b 2 2nd approximation : set x0 = a , x1 = (x1 − x0 )2 + (f (x1 ) − f (x0 ))2 + (x2 − x1 )2 + (f (x2 ) − f (x1 ))2 nth approximation : ∆x = n b−a , xi = a + i∆x , xi − xi−1 = ∆x n n (xi − xi−1 i=1 n )2 + (f (xi ) − f (xi−1 ))2 b a = i=1 (∆x)2 f (xi ) − f (xi−1 ) 1+ xi − xi−1 2 L = nlim →∞ 1 + (f (x))2 · ∆x = i=1 1 + (f (x))2 dx ex : straight line y (x1 , y1 ) (x0 , y0 ) x y = mx + b : point-slope formula y = m(x − x0 ) + y0 = y1 − y 0 · (x − x0 ) + y0 : alternative x1 − x0 ok check : x = x0 ⇒ y = y0 , x = x1 ⇒ y = y1 L= = b a 1 + (f (x))2 dx = x1 x0 1+ y1 − y0 2 dx x1 − x0 (x1 − x0 )2 + (y1 − y0 )2 ok 1+ y1 − y0 2 · (x1 − x0 ) = x1 − x0 23 ex : circumference of a circle of radius r, L = 2πr y −r L=2 b a r 1 + (f (x))2 dx x x2 + y 2 = r2 ⇒ f (x) = (r2 − x2 )1/2 1 −x f (x) = (r2 − x2 )−1/2 · (−2x) = √ 2 2 r − x2 r2 − x2 + x2 r2 x2 = =2 1 + (f (x)) = 1 + 2 r − x2 r2 − x2 r x2 b r r √ L=2 1 + (f (x))2 dx = 2 dx : improper a −r r 2 − x2 2 y 14 Fri 10/1 x √ r θ r2 − x2 √ r 2 − x2 r −r r x x dx sin θ = ⇒ cos θ dθ = , cos θ = r r sin θ 1 −π −π 2 π 2 π θ −1 x = r ⇒ sin θ = 1 ⇒ θ = π 2 x = −r ⇒ sin θ = −1 ⇒ θ = − π 2 L=2 π /2 −π/2 sec · r cos θ dθ = 2r θ π /2 −π/2 dθ = 2r · π = 2πr ok 24 ex : find the length of the parabola y = x2 for 0 ≤ x ≤ 1 y 1 note : L > x 1 f (x) = x2 ⇒ f (x) = 2x ⇒ L = √ 2x 1 √ 1 + 4x2 dx = 1 sec θ · 2 sec2 θdθ = 1 2 b a √ 2 = 1.4142 1 + (f (x))2 dx = 1√ 0 1 + 4x2 dx 1 + 4x2 θ tan θ = 2x ⇒ sec2 θdθ = 2dx √ 1 + 4x2 = sec θ sec3 θdθ du (1 − u2 )2 sec3 θdθ = dθ = cos3 θ cos θ dθ = cos4 θ cos θdθ = (1 − sin2 θ)2 u = sin θ ⇒ du = cos θdθ (1 − u2 )2 = ((1 − u)(1 + u))2 = (1 − u)2 (1 + u)2 1 a b c d = + + + = ··· (1 − u2 )2 1 − u (1 − u)2 1 + u (1 + u)2 alternative sec3 θdθ = sec θ · sec2 θdθ u = sec θ , dv = sec2 θdθ ⇒ du = sec θ tan θdθ , v = tan θ , check . . . sin2 θ dθ sec θdθ = sec θ tan θ − sec θ tan θdθ = sec θ tan θ − sec θ · cos2 θ 3 2 = sec θ tan θ − sec3 θ(1 − cos2 θ)dθ = sec θ tan θ − sec3 θ + sec θdθ 1 ⇒ sec3 θdθ = 2 (sec θ tan θ + ln(sec θ + tan θ)) √ √ √ 1 1 + 4x2 dx = 4 2x 1 + 4x2 + ln(2x + 1 + 4x2 ) 1√ 0 1 + 4x2 dx = √ √ 2x 1 + 4x2 + ln(2x + 1 + 4x2 ) √ √ 1 = 4 (2 5 + ln(2 + 5)) = 1.4789 1 4 1 0 15 Mon 10/4 25 ex : find the arclength of the curve y = y 1 √ x for 0 ≤ x ≤ 1 x 1 f (x) = L= b a √ 1 x ⇒ f (x) = √ 2x 1 0 1 + (f (x))2 dx = √ 1+ 1 dx : improper , converges 4x substitute : y = L= 1 0 dx dx x ⇒ dy = √ = 2x 2y 1 0 1+ 1 · 2y dy = 4y 2 4y 2 + 1 dy = arclength of a parabola ok 9.2 surface area def : A surface of revolution is formed by rotating a curve about an axis. ex : cylinder y l r x z l : length r : radius S : surface area To find S , cut the cylinder and spread it flat to form a rectangle. (show using paper) l 2¼r ) S = 2¼rl 26 ex : cone y l r l : slant length x z r : radius S : surface area To find S , cut the cone and spread it flat to form a circular sector. l : radius of circle θ : angle of sector l µ 2πr : length of edge of sector S : area of sector = area of cone 2¼r claim a) 2πr = lθ b) S = 1 l2 θ 2 pf a) 2πr = 2πl · b) S = πl2 · note Another proof is given on hw5 using A = f (x)dx , L = a b b a ⇒ S = 1 l · lθ = 2 1 2l · 2πr = πrl θ = lθ 2π ok θ = 1 l2 θ 2 2π 1 + (f (x))2 dx. 27 ex : truncated cone y r1 l 16 Tues 10/5 r2 x z l1 l, r1 , r2 = slant length and radii of truncated cone S = surface area of truncated cone S1 = · · · · · · ” · · · · · · small cone = πr1 l1 S2 = · · · · · · ” · · · · · · large cone = πr2 (l1 + l) S = S2 − S1 = πr2 (l1 + l) − πr1 l1 = πl1 (r2 − r1 ) + πr2 l we can eliminate l1 l l1 r1 r2 l1 l1 + l = ⇒ l1 r2 = r1 (l1 + l) ⇒ l1 (r2 − r1 ) = r1 l r1 r2 ⇒ S = πr1 l + πr2 l = π (r1 + r2 )l note 1. S = 2π r1 + r2 r1 + r2 l = 2π rl , where r = = average radius 2 2 2. r1 = r2 ⇒ truncated cone reduces to cylinder , S = 2πrl 3. r1 → 0 ⇒ truncated cone reduces to cone , S = πrl 28 general case y ¢li f(x) a z ∆x = xi{1 xi b x b−a , xi = a + i∆x n (xi − xi−1 )2 + (f (xi ) − f (xi−1 ))2 ≈ n surface area of ith slice ≈ π (f (xi−1 ) + f (xi ))∆li ∆li ≈ 1 + (f (xi ))2 · ∆x b a S = nlim →∞ π (f (xi−1 ) + f (xi )) 1 + (f (xi ))2 · ∆x = i=1 2πf (x) 1 + (f (x))2 dx check 1 : cylinder y r f(x) = r x l S= b a 2πf (x) 1 + (f (x))2 dx = l 0 2πrdx = 2πrl ok check 2 : cone y r l h r f(x) = | x h x S= h 0 √ r r2 r r2 h2 2π x 1 + 2 dx = 2π 1 + 2 · = πr h2 + r2 = πrl h h h h2 ok check 3 : truncated cone , see review sheet 29 ex : sphere y 17 Wed 10/6 x −r r √ −x 1 f (x) = r2 − x2 = (r2 − x2 )1/2 ⇒ f (x) = 2 (r2 − x2 )1/2 · −2x = √ 2 r − x2 x2 r2 − x2 + x2 r2 2 1 + (f (x)) = 1 + 2 = =2 r − x2 r2 − x2 r − x2 ¨ r√ b r r 2 −¨ r¨¨ x2 · √ 2¨¨¨ dx = 2πr dx = 2πr · 2r S = 2π f (x) 1 + (f (x))2 dx = 2π −r ¨ a −r r − x2 ¨¨ S = 4πr2 ex : torus ok y a x {r r √ assume a ≥ r , equation of circle : x2 + (y − a)2 = r2 ⇒ y = a ± r2 − x2 √ upper semicircle : f+ (x) = a + r2 − x2 √ lower semicircle : f− (x) = a − r2 − x2 S = S+ + S− = 1 + (f+ (x))2 = r −r 2πf+ (x) 1 + (f+ (x))2 dx + 2 r −r 2πf− (x) 1 + (f− (x))2 dx r = 1 + (f− (x))2 , because f− (x) = −f+ (x) 2 − x2 r √ √ r r S = 2π (a + r2 − x2 + a − r2 − x2 ) √ 2 dx = 2π · 2a · r −r r − x2 x dx ⇒ cos θdθ = sin θ = r r r x √ θ r 2 − x2 √ cos θ = r r2 − x2 ⇒ S = 4πar π 2 r −r √ dx r2 − x2 −π 2 r cos θ dθ = 4πar · π = 4π 2 ar r cos θ 30 note 1. Storus = 4π 2 ar = 2πa · 2πr = product of circumferences of two circles 2. For the cylinder and cone, we can find the surface area by cutting the surface, spreading it flat, and computing the area of the resulting shape. Does this work for the sphere and torus? No. (Math 433 , differential geometry) summary y f(x) a b x b a b a b a b a area under graph of y = f (x) on a ≤ x ≤ b : A = arclength · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · : L = surface area · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·· : S = volume · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · : V = ex 1 f (x) = , 1 ≤ x < ∞ x A= L= ∞ 1 ∞ 1 ∞ 1 ∞ 1 f (x)dx 1 + (f (x))2 dx 2πf (x) 1 + (f (x))2 dx πf (x)2 dx y dx : diverges, p = 1 x 1+ x 1 1+ 1 x 1 x4 1 dx : diverges , comparison test , x4 ≥1 1 x4 1 x S= V= 2π 1 1 1 + 4 dx : diverges , comparison test , x x ∞ 1 1+ ≥ π π dx = − x2 x = π : converges , p = 2 This shape is called Gabriel’s horn; it has finite volume and infinite surface area. (see 595/25) 31 9.3 center of mass problem : Find the point at which a thin plate balances. 18 Fri 10/8 ex : two masses m1 , m2 are connected by a rod of negligible mass m1 m2 d1 d2 balance of moments (prevents tipping) : m1 d1 = m2 d2 m1 m2 x 0 x1 ¹ x2 x x1 , x2 , x : coordinates of m1 , m2 , CM on x-axis ¯ m1 (¯ − x1 ) = m2 (x2 − x) ⇒ (m1 + m2 )¯ = m1 x1 + m2 x2 ⇒ x = x ¯ x ¯ m1 x1 + m2 x2 m1 + m2 mi xi : moment of mass i about x = 0 , units are mass × distance note : The balance of moments can be written as m1 (¯ − x1 ) + m2 (¯ − x2 ) = 0. x x n n n ex : n masses m1 , . . . , mn are connected by a rod of negligible mass balance of moments ⇒ i=1 n n mi (¯ − xi ) = 0 ⇒ x i=1 mi x = ¯ i=1 mi xi ⇒ mx = M ¯ M m M= i=1 mi xi : total moment , m = i=1 mi : total mass ⇒ x = ¯ note : In the case n = 2, this agrees with the formula derived above. ex : m1 = 2, m2 = 3, m3 = 1, x1 = −10, x2 = 6, x3 = −4 : where is the CM ? m1=2 m3=1 m2=3 x ¹0 {10 {4 x 6 M −6 M = m1 x1 + m2 x2 + m3 x3 = −20 + 18 − 4 = −6 ⇒ x= ¯ = = −1 m = m1 + m2 + m3 = 2 + 3 + 1 = 6 m 6 note : M = m1 x1 + · · · + mn xn = mx ⇒ The total moment of n masses mi ¯ located at xi is equal to the moment of one mass m located at x. ¯ 32 two-dimensional case y y3 m1 y1 x2 x1 y2 x3 x m3 19 Mon 10/11 m2 center of mass : (¯, y ) = ? x¯ n n balance of moments : i=1 n mi (¯ − xi ) = 0 , x i=1 mi (¯ − yi ) = 0 y mi xi ⇒ x= ¯ i=1 n = mi My , My : moment about y −axis ⇒ My = mx ¯ m i=1 n mi yi ⇒ y= ¯ i=1 n = mi Mx , Mx : moment about x−axis ⇒ Mx = my ¯ m i=1 ex : m1 = m2 = m3 = 1 y m3 (0;1) m2 ({1;0) m1 (1;0) ¹¹ (x;y)=(0;1=3) x x= ¯ y= ¯ m1 x1 + m2 x2 + m3 x3 −1 + 1 + 0 = =0 m1 + m2 + m3 3 m1 y1 + m2 y2 + m3 y3 0+0+1 1 = = m1 + m2 + m3 3 3 33 continuous mass distribution Consider a region of uniform density in the xy -plane. y CM = (¯, y ) = ? x¯ x symmetry principle If a region is symmetric about a line l, then the CM is on l. ex 1 : isoceles triangle , CM is on l l ex 2 : rectangle , CM is on l1 and l2 ⇒ CM is at center of rectangle l2 l1 general case : R = {(x, y ) : a ≤ x ≤ b , 0 ≤ y ≤ f (x)} y f(x) a xi{1 xi b x We define x = ¯ My Mx , y= ¯ , but how do we find Mx , My ? m m 34 moment principles 1. The moments of a region are unchanged if all the mass is concentrated at the CM of the region. m1 m m2 ex : x ⇒ M = m1 x1 + m2 x2 = mx ¯ x2 0 x1 ¹ x 2. Let R1 and R2 be two disjoint regions. Then MR1 ∪R2 = MR1 + MR2 , i.e. the moments of a union of two regions are equal to the sums of the moments of the two regions. b−a , xi = a + i∆x , x∗ = 1 (xi−1 + xi ). In general, set ∆x = i 2 n 1 CM of ith rectangle = (x∗ , 2 f (x∗ )) i i mass of ith rectangle = density × area = ρf (x∗ )∆x i moment of ith rectangle about y −axis = mass × distance = ρf (x∗ )∆x · x∗ i i 1 moment of ith rectangle about x−axis = mass × distance = ρf (x∗ )∆x · 2 f (x∗ ) i i n My = nlim →∞ Mx = nlim →∞ ρx∗ f (x∗ )∆x = i i 1 ρ 2 f (x∗ )2 ∆x = i b ¡ ρxf (x)dx ¡ ρf (x)dx b a ρxf (x)dx = mx ¯ = my ¯ b 1 ¡ ρf 2 2 a ¡ (x) dx b ¡ ρf (x)dx a¡ i=1 n i=1 b 2 1 2 a ρf (x) dx My ⇒x= ¯ = m a¡ b Mx , y= ¯ = m a¡ ex : rectangle (take ρ = 1) , CM = y h f(x) = h lh 2, 2 x l x2 My = xf (x)dx = x · hdx = h a 0 2 b l l 0 hl2 = 2 b a Mx = b 2 1 2 a f (x) dx = l2 1 2 0 h dx h2 l = , m= 2 f (x)dx = ok l 0 hdx = hl My hl2 1 l Mx h2 l 1 h x= ¯ = · = , y= ¯ = · = m 2 hl 2 m 2 hl 2 35 ex : triangular plate y 1 20 Tues 10/12 f (x) = 1 − |x| x {1 1 b 2 1 2 a f (x) dx b a 1 symmetry ⇒ x = 0 ¯ y= ¯ = 0 (1 − x)2 dx 1 2 f (x)dx ·2·1 = − 1 (1 − x)3 = − 1 (0 − 1) = 3 3 0 1 1 3 ⇒ CM = (0, 1 ) 3 ex : another triangular plate y 1 f (x) = 1 − x x x ¯ b 1 symmetry ⇒ CM lies on the line y = x ⇒ y = x ¯¯ My x= ¯ = m a xf (x)dx b 1 = 0 x(1 − x)dx 1 2 a f (x)dx ·1·1 = 12 2x 1 − 3 x3 1 2 1 0 = 1 6 1 2 = 1 3 ⇒ CM = ( 1 , 1 ) 33 question : The line x = x divides the triangle into two parts. Which part has ¯ greater area, the left or the right? left : right : 1/3 0 (1 − x)dx = − 1 (1 − x)2 2 1/3 0 1 1 14 = − 2 (( 2 )2 − 1) = − 2 ( 9 − 1) = 3 1 4 = − 2 (0 − ( 2 )2 ) = − 1 (− 9 ) = 3 2 5 18 1 1/3 1 (1 − x)dx = − 2 (1 − x)2 1/3 4 18 So the left part has 25% more area than the right part. Nonetheless, the CM lies in the right part. 36 ex : half disk y f (x) = x √ 1 − x2 21 Wed 10/20 {1 1 1 y= ¯ b 2 1 2 a f (x) dx b a = 0 (1 − x2 )dx π/2 = x − 1 x2 3 π/2 1 0 = 2 3 f (x)dx π/2 = 4 = 0.4244 3π Hence the CM lies higher in this case than in the case of the triangular plate. more general case y f(x) R = f(x;y) : a · x · b; g(x) · y · f(x)g g(x) a b xi{1 xi x(f (x) − g (x))dx (f (x) − g (x))dx b b x b 2 1 2 a (f (x) b a My = x= ¯ m a Mx , y= ¯ = m − g (x)2 )dx a (f (x) − g (x))dx ex : f (x) = x , g (x) = x2 , 0 ≤ x ≤ 1 y 1 1/2 2/5 1/4 1 2 x 1 1 0 m= Mx = My = b a (f (x) − g (x))dx = (x − x2 )dx = 12 1 2 0 (x 1 0 1 2 − 1 3 = 1 2 1 4 1 6 1 3 b 2 1 2 a (f (x) b a − g (x)2 )dx = − x4 )dx = 1 3 − 1 5 = 1 15 x(f (x) − g (x))dx = 1 12 1 6 (x2 − x3 )dx = 1 15 1 6 − = 1 12 My x= ¯ = m = 1 2 Mx , y= ¯ = m = 2 5 37 ex : f (x) = xm , g (x) = xn y 1 x 1 For some choice of m, n, the CM lies outside the region. Theorem of Pappus Let R be a region that lies on one side of a line l. l R CM (hw7) A = area of R V = volume obtained by rotating R about l d = distance traveled by CM when R is rotated about l pf : (special case) y CM ⇒ V =A·d f(x) R : rotated about x{axis g(x) a A= b a b b a x (f (x) − g (x))dx , V = 1 2a b π (f (x)2 − g (x)2 )dx = V A ok Mx d = 2π y = 2π · ¯ = 2π · m ex : cylinder y (f (x)2 − g (x)2 )dx (f (x) − g (x))dx b a r l x r = πr 2 ⇒ V = A · d = rl πr = πr2 l A = rl , d = 2π · ok (hw7 : torus) 38 9.5 probability X : random variable ex X = velocity of gas molecules X = waiting time in the supermarket check-out line X = GPA of a college student def A random variable X has a probability density function f (x) with the property that ex X = velocity {20 {10 0 {10 f(x) {20 x meter=second b a 22 Fri 10/22 f (x) dx = prob(a ≤ X ≤ b) = probability that X lies between a and b. X = waiting time 0 1 2 3 f(x) 4 f(x) 0:0 1:0 2:0 3:0 4:0 x grade points x minutes X = GPA note large f (x) ⇒ high probability that X is close to x small f (x) ⇒ low probability that X is close to x In practice, the pdf can be obtained by direct measurement or as the solution of a differential equation. properties of a pdf 1. f (x) ≥ 0 2. ∞ −∞ f (x) dx = prob(−∞ < X < ∞) = 1 ∞ −∞ n n def : The mean value of a random variable X (average value, expected value) is µ = µ(X ) = xf (x) dx ≈ i=1 xi f (xi )∆x ≈ i=1 xi · prob(xi−1 ≤ X ≤ xi ), i.e. the values of xi are weighted by the probability that X is close to xi . 39 ex : Gaussian pdf 1 2 f (x) = √ e−x π 1. f (x) ≥ 0 2. ∞ y x ∞ 1 ∞ −x 2 1 1√ 2 √ e−x dx = √ e dx = √ · π = 1 −∞ −∞ π π −∞ π √ ∞ −x 2 π recall : e dx = (Math 255) 0 2 ∞ ∞ 1 2 3. µ = xf (x) dx = x √ e−x dx = 0 −∞ −∞ π f (x)dx = ok def : The median value of a random variable X is denoted by m = m(X ) and is defined by the relation prob(X ≤ m) = prob(X ≥ m) = 1 . 2 note 1. This is equivalent to the equation m −∞ f (x) dx = ∞ m f (x) dx = 1 2, i.e. half the area under the graph of f lies to the left of m and half lies to the right. 2. In general, m = µ. (more later) def The standard deviation of a random variable is denoted by σ = σ (X ) and is defined by σ = note small σ ⇒ X is more likely to be close to µ large σ ⇒ X is less likely to be close to µ ex 1 2 f (x) = √ e−x ⇒ µ = 0 , σ = ? π σ2 = 1 2 2 x2 √ e−x dx , u = x , dv = xe−x dx −∞ −∞ π 2 e−x du = dx , v = −2 √ 2 −x ∞ 1 e 1 ∞ −x 2 1 π 1 1 = √ x · + e dx = √ · = ⇒ σ = √ = 0.7071 π −2 −∞ 2 −∞ π2 2 2 ∞ ∞ −∞ (x − µ)2 f (x) dx 1/2 23 Mon 10/25 , where µ is the mean of X . (x − µ)2 f (x) dx = ∞ 40 −(x−µ)2 1 def : Let µ and σ > 0 be given and set f (x) = √ e 2σ2 . Then f (x) is a σ 2π pdf of a random variable, called a normal distribution, with mean µ and standard deviation σ . 1. µ shifts the pdf and σ scales the height and width of the pdf 2. The previous example corresponds to µ = 0 , σ = check (hw7) −(x−µ)2 1 √ e 2σ2 dx = 1 −∞ −∞ σ 2π −(x−µ)2 ∞ ∞ 1 xf (x) dx = x √ e 2σ2 dx = µ −∞ −∞ σ 2π −(x−µ)2 ∞ ∞ 1 (x − µ)2 f (x) dx = (x − µ)2 √ e 2σ2 dx = σ 2 −∞ −∞ σ 2π ex 1 √. 2 ∞ f (x) dx = ∞ 1. Find the probability that X lies within 1 standard deviation of µ. prob( µ − σ ≤ X ≤ µ + σ ) = = µ +σ µ −σ f (x) dx −(x−µ)2 1 x−µ dx √ e 2σ2 dx , u = √ , du = √ µ −σ σ 2π 2σ 2σ √ √ 1/ 2 1 2 √ e−u · 2σ du √ = −1/ 2 σ 2π √ √ 1/ 2 1 −x2 √√ e dx = erf(1/ 2) = 0.6827 = −1/ 2 π µ+σ 2. Find the probability that X is 3 standard deviations or more greater than µ. √ ∞ 1 −x2 1 √√ e prob(X ≥ µ + 3σ ) = . . . = dx = 2 (1 − erf(3/ 2)) = 0.001349 3/ 2 π ex : annual rainfall in a certain region is normally distributed with µ = 25 in and σ = 5 in. ⇒ 68% of years have rainfall between 20 in and 30 in 0.13% of years have rainfall greater than or equal to 40 in question : which of the following random variables are normally distributed? velocity of gas molecules - yes waiting time - no GPA - yes daily change in stock price - ? 41 −(x−µ)2 1 pdf of a normal distribution : f (x) = √ e 2σ2 σ 2π , µ=0 1.5 1.5 σ = 1/4 1 1 σ = 1/2 0.5 0.5 0 −5 0 5 0 −5 0 5 1.5 1.5 σ=1 1 1 σ=2 0.5 0.5 0 −5 0 5 0 −5 0 5 ...
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This note was uploaded on 01/28/2011 for the course CHEM 210 taught by Professor Kiste during the Spring '10 term at Michigan Flint.

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