Problem Set 3 Solutions: Integration

SOLUTIONS TO 18.01 EXERCISES Unit 3. Integration 3A. Differentials, indefinite integration 3A-1 a) 7x6 dx. (d(sin 1) = 0 because sin 1 is a constant.) b) (1/2)x−1/2 dx c) (10x9 − 8)dx d) (3e3x sin x + e3x cos x)dx √ √ e) (1/2 x)dx + (1/2 y)dy = 0 implies � � √ √ √ y 1/2 xdx 1− x 1 dy = − dx = 1 − √ dx √ = − √ dx = − √ 1/2 y x x x 3A-2 a) (2/5)x5 + x3 + x2 /2 + 8x + c b) (2/3)x3/2 + 2x1/2 + c � c) Method 1 (slow way) Substitute: u = 8 + 9x, du = 9dx. Therefore � √ 8 + 9xdx = u1/2 (1/9)du = (1/9)(2/3)u3/2 + c = (2/27)(8 + 9x)3/2 + c Method 2 (guess and check): It’s often faster to guess the form of the antideriv­ ative and work out the constant factor afterwards: Guess (8 + 9x)3/2 ; d 27 (8 + 9x)3/2 = (3/2)(9)(8 + 9x)1/2 = (8 + 9x)1/2 . dx 2 2 So multiply the guess by to make the derivative come out right; the answer is 27 then 2 (8 + 9x)3/2 + c 27 COPYRIGHT DAVID JERISON AND MIT 1996, 2003 1 E. Solutions to 18.01 Exercises 3. Integration d) Method 1 (slow way) Use the substitution: u = 1 − 12x4 , du = −48x3 dx. � � 1 1 x3 (1−12x4 )1/8 dx = u1/8 (−1/48)du = − (8/9)u9/8 +c = − (1−12x4 )9/8 +c 48 54 Method 2 (guess and check): guess (1 − 12x4 )9/8 ; d 9 (1 − 12x4 )9/8 = (−48x3 )(1 − 12x4 )1/8 = −54(1 − 12x4 )1/8 . dx 8 1 to make the derivative come out right, getting the So multiply the guess by − 54 previous answer. e) Method 1 (slow way): Use substitution: u = 8 − 2x2 , du = −4xdx. � � x 12 1 √ dx = u1/2 (−1/4)du = − u3/2 + c = − (8 − 2x2 )3/2 + c 43 6 8 − 2x2 Method 2 (guess and check): guess (8 − 2x2 )3/2 ; differentiating it: d 3 (8 − 2x2 )3/2 = (−4x2 )(8 − 2x2 )1/2 = −6(8 − 2x2 )1/2 ; dx 2 1 so multiply the guess by − to make the derivative come out right. 6 The next four questions you should try to do (by Method 2) in your head. Write down the correct form of the solution and correct the factor in front. f) (1/7)e7x + c 5 g) (7/5)ex + c √ h) 2e x +c i) (1/3) ln(3x + 2) + c. For comparison, let’s see how much slower substitution is: u = 3x + 2, � dx = 3x + 2 � du = 3dx, so (1/3)du = (1/3) ln u + c = (1/3) ln(3x + 2) + c u j) � x+5 dx = x � � � 5 1+ dx = x + 5 ln x + c x 2 3. Integration E. Solutions to 18.01 Exercises k) � x dx = x+5 � � 1− 5 x+5 � dx = x − 5 ln(x + 5) + c In Unit 5 this sort of algebraic trick will be explained in detail as part of a general method. What underlies the algebra in both (j) and (k) is the algorithm of long division for polynomials. l) u = ln x, du = dx/x, so � � ln x dx = udu = (1/2)u2 + c = (1/2)(ln x)2 + c x m) u = ln x, du = dx/x. � � dx du = = ln u + c = ln(ln x) + c x ln x u 3A-3 a) −(1/5) cos(5x) + c b) (1/2) sin2 x + c, coming from the substitution u = sin x or −(1/2) cos2 x + c, coming from the substitution u = cos x. The two functions (1/2) sin2 x and −(1/2) cos2 x are not the same. Nevertheless the two answers given are the same. Why? (See 1J-1(m).) c) −(1/3) cos3 x + c d) −(1/2)(sin x)−2 + c = −(1/2) csc2 x + c e) 5 tan(x/5) + c f) (1/7) tan7 x + c. g) u = sec x, du = sec x tan xdx, � � sec9 x tan xdx (sec x)8 sec x tan xdx = (1/9) sec9 x + c 3B. Definite Integrals 3B-1 a) 1 + 4 + 9 + 16 = 30 b) 2 + 4 + 8 + 16 + 32 + 64 = 126 c) −1 + 4 − 9 + 16 − 25 = −15 d) 1 + 1/2 + 1/3 + 1/4 = 25/12 3 E. Solutions to 18.01 Exercises 3B-2 a) 6 � (−1)n+1 (2n + 1) n=1 3. Integration b) n � 1/k 2 c) k=1 n � sin(kx/n) k=1 3B-3 a) upper sum = right sum = (1/4)[(1/4)3 +(2/4)3 +(3/4)3 +(4/4)3 ] = 15/128 lower sum = left sum = (1/4)[03 + (1/4)3 + (2/4)3 + (3/4)3 ] = 7/128 b) left sum = (−1)2 + 02 + 12 + 22 = 6; right sum = 02 + 12 + 22 + 32 = 14; upper sum = (−1)2 + 12 + 22 + 32 = 15; lower sum = 02 + 02 + 12 + 22 = 5. c) left sum = (π/2)[sin 0+sin(π/2)+sin(π)+sin(3π/2)] = (π/2)[0+1+0−1] = 0; right sum = (π/2)[sin(π/2)+sin(π)+sin(3π/2)+sin(2π)] = (π/2)[1+0−1+0] = 0; upper sum = (π/2)[sin(π/2)+sin(π/2)+sin(π)+sin(2π)] = (π/2)[1+1+0+0] = π; lower sum = (π/2)[sin(0)+sin(π)+sin(3π/2)+sin(3π/2)] = (π/2)[0+0−1−1] = −π. 3B-4 Both x2 and x3 are increasing functions on 0 ≤ x ≤ b, so the upper sum is the right sum and the lower sum is the left sum. The difference between the right and left Riemann sums is (b/n)[f (x1 + · · · + f (xn )] − (b/n)[f (x0 + · · · + f (xn−1 )] = (b/n)[f (xn ) − f (x0 )] In both cases xn = b and x0 = 0, so the formula is (b/n)(f (b) − f (0)) a) (b/n)(b2 − 0) = b3 /n. Yes, this tends to zero as n → ∞. b) (b/n)(b3 − 0) = b4 /n. Yes, this tends to zero as n → ∞. 3B-5 The expression is the right Riemann sum for the integral � 1 sin(bx)dx = −(1/b) cos(bx)|10 = (1 − cos b)/b 0 so this is the limit. 3C. Fundamental theorem of calculus 4 3. Integration E. Solutions to 18.01 Exercises 3C-1 � 6 �6 � (x − 2)−1/2 dx = 2(x − 2)1/2 � = 2[(4)1/2 − 11/2 ] = 2 3 3 �2 � 3C-2 a) (2/3)(1/3)(3x + 5)3/2 � = (2/9)(113/2 − 53/2 ) 0 b) If n �= −1, then �2 (1/(n + 1))(1/3)(3x + 5)n+1 �0 = (1/3(n + 1))((11n+1 − 5n+1 ) If n = −1, then the answer is (1/3) ln(11/5). √ �π c) (1/2)(cos x)−2 �3π/4 = (1/2)[(−1)−2 − (−1/ 2)−2 ] = −1/2 �2 3C-3 a) (1/2) ln(x2 + 1)�1 = (1/2)[ln 5 − ln 2] = (1/2) ln(5/2) �2b b) (1/2) ln(x2 + b2 )�b = (1/2)[ln(5b2 ) − ln(2b2 )] = (1/2) ln(5/2) 3C-4 As b → ∞, � b �b x−10 dx = −(1/9)x−9 �1 = −(1/9)(b−9 − 1) → −(1/9)(0 − 1) = 1/9. 1 This integral is the area of the infinite region between the curve y = x−10 and the x-axis for x > 0. π � 3C-5 a) 0 � b) 0 π sin xdx = − cos x|0 = −(cos π − cos 0) = 2 π/a π/a sin(ax)dx = −(1/a) cos(ax)|0 = −(1/a)(cos π − cos 0) = 2/a 3C-6 a) x2 − 4 = 0 implies x = ±2. So the area is �2 � 2 � 2 � x3 8 (x2 − 4)dx = 2 (x2 − 4)dx = − 4x�� = − 4 · 2 = −16/3 3 3 −2 0 0 (We changed to the interval (0, 2) and doubled the integral because x2 − 4 is even.) Notice that the integral gave the wrong answer! It’s negative. This is because the graph y = x2 − 4 is concave up and is below the x-axis in the interval −2 < x < 2. So the correct answer is 16/3. √ b) Following part (a), x2 − a = 0 implies x = ± a. The area is � √a � √a � √a � x3 �� a3/2 � 4 3/2 2 2 (a − x )dx = 2 (a − x )dx = 2ax − � = 2 a3/2 − = a √ 3 0 3 3 − a 0 5 E. Solutions to 18.01 Exercises 3. Integration 3D. Second fundamental theorem 3D-1 Differentiate both sides; left side L(x): L� (x) = d dx � 0 x √ dt 1 =√ , by FT2; a2 + x2 a2 + x2 � d R (x) = (ln(x + a2 + x2 ) − ln a) = dx � right side R(x): 1 √ a2 + x2 1 + √a2x+x2 √ x + a2 + x2 = Since L� (x) = R� (x), we have L(x) = R(x) + C for some constant C = L(x) − R(x). The constant C may be evaluated by assigning a value to x; the most convenient choice is x = 0, which gives 0 � L(0) = = 0; R(0) = ln(0 + � 0 + a2 ) − ln a = 0; therefore C = 0 and 0 L(x) = R(x). � b) Put x = c; the equation becomes√0 = ln(c + c2 + a2 ); solve this for c by first exponentiating both sides: 1 = c + c2 + a2 ; then subtract c and square both sides; after some algebra one gets c = 12 (1 − a2 ). 3D-3 Sketch y = 1 − t2 first, as shown at the right. 1 + t2 -t 2 y = 11 + t2 1 y -1 t 1 -1 � 3D-4 a) x 3 sin(t )dt, by the FT2. � b) 0 x 3 sin(t )dt + 2 0 3D-5 This problem reviews the idea of the proof of the FT2. t a) f (t) = √ 1 + t4 6 � c) 1 x sin(t3 )dt − 1 3. Integration E. Solutions to 18.01 Exercises 1+Δx � 1 Δx f (t)dt = 1 shaded area ≈ height width f(t) 2/2 1 1+∆x lim Δx→0 � 1 Δx 1+Δx f (t)dt = 1 . lim Δx→0 t shaded area 1 = height = f (1) = √ . width 2 b) By definition of derivative, � 1+Δx F (1 + Δx) − F (1) 1 F � (1) = lim = lim f (t)dt; Δx→0 Δx→0 Δx 1 Δx 1 by FT2, F � (1) = f (1) = √ . 2 � x 3D-6 a) If F1 (x) = � x dt, then F1 (x) = x − a1 and F2 (x) = dt and F2 (x) = a1 a2 x − a2 . Thus F1 (x) − F2 (x) = a2 − a1 , a constant. b) By the FT2, F1� (x) = f (x) and F2� (x) = f (x); therefore F1 = F2 + C, for some constant C. 3D-7 a) Using the FT2 and the chain rule, as in the Notes, d dx � x2 √ √ u sin udu = 0 b) = � d c) dx � x2 sin(x2 ) · d(x2 ) = 2x2 sin(x2 ) dx � 1 2 1 − sin x sin x · cos x = 1. (So 0 dt = x) 1 − t2 x2 tan udu = tan(x2 ) · 2x − tan x x 3D-8 a) Differentiate both sides using FT2, and substitute x = π/2: f (π/2) = 4. 7 E. Solutions to 18.01 Exercises 3. Integration b) Substitute x = 2u and follow the method of part (a); put u = π, get finally f (π/2) = 4 − 4π. 3E. Change of Variables; Estimating Integrals 1 3E-1 L( ) = a 1/a � dt 1 1 . Put t = , dt = − 2 du. Then t u u � 1/a � a dt u 1 dt du = − 2 du =⇒ L( ) = =− = −L(a) t u a t u 1 1 √ √ 3E-2 a) We want −t2 = −u2 /2, so u = t 2, du = 2dt. 1 � x/√2 � x/√2 2√ 2 1 1 e−u /2 du = √ e−t 2dt = √ e−t dt π 2π 0 0 0 √ √ 1 1 π 1 =⇒ E(x) = √ F (x/ 2) and = lim E(x) = · x→∞ π 2 2 π 1 √ 2π � x 2 b) The integrand is even, so � N � N 2 1 2 −u2 /2 √ e du = √ e−u /2 du = 2E(N ) −→ 1 2π −N 2π 0 lim E(x) = −1/2 as N → ∞ because E(x) is odd. x→−∞ � b 2 1 √ e−u /2 du = E(b)−E(a) by FT1 or by “interval addition” Notes PI (3). 2π a Commentary: The answer is consistent with the limit, � N 2 1 √ e−u /2 du = E(N ) − E(−N ) = 2E(N ) −→ 1 as N → ∞ 2π −N dx 3E-3 a) Using u = ln x, du = , x � e √ 1 2 . 3 ln x dx = x 1 � 0 √ udu = 2 3/2 ��1 u � = 3 0 b) Using u = cos x, du = − sin x, � 0 π sin x dx = (2 + cos x)3 � 1 −1 �−1 −du 1 1 1 1 4 � = = ( 2 − 2) = .. � 3 2 (2 + u) 2(2 + u) 1 2 1 3 9 � c) Using x = sin u, dx = cos udu, . 0 8 1 √ dx = 1 − x2 � 0 π 2 �π/2 cos u π � du = u� = cos u 2 0 3. Integration E. Solutions to 18.01 Exercises 3E-4 Substitute x = t/a; then x = ±1 ⇒ t = ±a. We then have � 1� � a� � a� π t2 dt 1 2 = 1 − x dx = 1− 2 = 2 a2 − t2 dt. Multiply­ 2 a a a −a −1 −a ing by a2 gives the value πa2 /2 for the integral, which checks, since the integral represents the area of the semicircle. a -a a 3E-5 One can use informal reasoning based on areas (as in Ex. 5, Notes FT), but it is better to use change of variable. a) Goal: F (−x) = −F (x). Let t = −u, dt = −du, then � −x � x F (−x) = f (t)dt = f (−u)(−du) 0 0 x � Since f is even (f (−u) = f (u)), F (−x) = − f (u)du = −F (x). 0 b) Goal: F (−x) = F (x). Let t = −u, dt = −du, then � −x � x F (−x) = f (t)dt = f (−u)(−du) 0 0 x � Since f is odd ((f (−u) = −f (u)), F (−x) = f (u)du = F (x). 0 1 1 > on (0,1); therefore 3 1+x 1+x 3E-6 a) x3 < x on (0,1) ⇒ � 1 0 dx > 1 + x3 1 � 0 �1 dx � = ln(1 + x)� = ln 2 = .69 1+x 0 b) 0 < sin x < 1 on (0, π) ⇒ sin2 x < sin x on (0, π); therefore � π 2 π � sin xdx < 0 0 �π � sin xdx = − cos x� = −(−1 − 1) = 2. 0 9 E. Solutions to 18.01 Exercises � 20 c) � � x2 + 1dx > 10 3. Integration 20 √ x2 ��20 1 � = (400 − 100) = 150 2 10 2 x2 dx = 10 �� � � N sin x � �N � � 3E-7 � dx� ≤ 1 2 � 1 � x | sin x| x2 dx ≤ �N 1 1 x2 dx �N � = − x1 �� = − N1 + 1 < 1. 1 3F. Differential Equations: Separation of Variables. Applications 3F-1 a) y = (1/10)(2x + 5)5 + c � � dx =⇒ (1/2)(y + 1)2 = x + c. You √ can leave this in implicit form or solve for y: y = −1 ± 2x + a for any constant a (a = 2c) b) (y + 1)dy = dx =⇒ (y + 1)dy = c) y 1/2 dy = 3dx =⇒ (2/3)y 3/2 = 3x + c =⇒ y = (9x/2 + a)2/3 , with a = (3/2)c. d) y −2 dy = xdx =⇒ −y −1 = x2 /2 + c =⇒ y = −1/(x2 /2 + c) 3F-2 a) Answer: 3e16 . y −1 dy = 4xdx =⇒ ln y = 2x2 + c y(1) = 3 =⇒ ln 3 = 2 + c =⇒ c = ln 3 − 2. Therefore ln y = 2x2 + (ln 3 − 2) 18+ln 3−2 At x = 3, y = e 16 = 3e √ b) Answer: y = 11/2 + 3 2. (y + 1)−1/2 dy = dx =⇒ 2(y + 1)1/2 = x + c √ y(0) = 1 =⇒ 2(1 + 1)1/2 = c =⇒ c = 2 2 At x = 3, √ √ √ 2(y + 1)1/2 = 3 + 2 2 =⇒ y + 1 = (3/2 + 2)2 = 13/2 + 3 2 √ Thus, y = 11/2 + 3 2. c) Answer: y = � 550/3 ydy = x2 dx =⇒ y 2 /2 = (1/3)x3 + c y(0) = 10 =⇒ c = 102 /2 = 50 Therefore, at x = 5, y 2 /2 = (1/3)53 + 50 =⇒ y = 10 � 550/3 3. Integration E. Solutions to 18.01 Exercises d) Answer: y = (2/3)(e24 − 1) (3y + 2)−1 dy = dx =⇒ (1/3) ln(3y + 2) = x + c y(0) = 0 =⇒ (1/3) ln 2 = c Therefore, at x = 8, (1/3) ln(3y + 2) = 8 + (1/3) ln 2 =⇒ ln(3y + 2) = 24 + ln 2 =⇒ (3y + 2) = 2e24 Therefore, y = (2e24 − 2)/3 e) Answer: y = − ln 4 at x = 0. Defined for −∞ < x < 4. e−y dy = dx =⇒ −e−y = x + c y(3) = 0 =⇒ −e0 = 3 + c =⇒ c = −4 Therefore, y = − ln(4 − x), The solution y is defined only if x < 4. y(0) = − ln 4 3F-3 a) Answers: y(1/2) = 2, y(−1) = 1/2, y(1) is undefined. y −2 dy = dx =⇒ −y −1 = x + c y(0) = 1 =⇒ −1 = 0 + c =⇒ c = −1 Therefore, −1/y = x − 1 and 1 y= 1−x The values are y(1/2) = 2, y(−1) = −1/2 and y is undefined at x = 1. b) Although the formula for y makes sense at x = 3/2, (y(3/2) = 1/(1 − 3/2) = −2), it is not consistent with the rate of change interpretation of the differential equation. The function is defined, continuous and differentiable for −∞ < x < 1. But at x = 1, y and dy/dx are undefined. Since y = 1/(1 − x) is the only solution to the differential equation in the interval (0, 1) that satisfies the initial condition y(0) = 1, it is impossible to define a function that has the initial condition y(0) = 1 and also satisfies the differential equation in any longer interval containing x = 1. To ask what happens to y after x = 1, say at x = 3/2, is something like asking what happened to a rocket ship after it fell into a black hole. There is no obvious reason why one has to choose the formula y = 1/(1 − x) after the “explosion.” For example, one could define y = 1/(2 − x) for 1 ≤ x < 2. In fact, any formula y = 1/(c − x) for c ≥ 1 satisfies the differential equation at every point x > 1. 3F-4 a) If the surrounding air is cooler (Te − T < 0), then the object will cool, so dT /dt < 0. Thus k > 0. b) Separate variables and integrate. (T − Te )−1 dT = −kdt =⇒ ln |T − Te | = −kt + c 11 E. Solutions to 18.01 Exercises 3. Integration Exponentiating, T − Te = ±ec e−kt = Ae−kt The initial condition T (0) = T0 implies A = T0 − Te . Thus T = Te + (T0 − Te )e−kt c) Since k > 0, e−kt → 0 as t → ∞. Therefore, T = Te + (T0 − Te )e−kt −→ Te as t → ∞ d) T − Te = (T0 − Te )e−kt The data are T0 = 680, Te = 40 and T (8) = 200. Therefore, 200 − 40 = (680 − 40)e−8k =⇒ e−8k = 160/640 = 1/4 =⇒ −8k = − ln 4. The number of hours t that it takes to cool to 50◦ satisfies the equation 50 − 40 = (640)e−kt =⇒ e−kt = 1/64 =⇒ −kt = −3 ln 4. To solve the two equations on the right above simultaneously for t, it is easiest just to divide the bottom equation by the top equation, which gives t = 3, t = 24. 8 e) T − Te = (T0 − Te )e−kt The data at t = 1 and t = 2 are 800 − Te = (1000 − Te )e−k and 700 − Te = (1000 − Te )e−2k Eliminating e−k from these two equations gives 700 − Te = 1000 − Te � 800 − Te 1000 − Te �2 (800 − Te )2 = (1000 − Te )(700 − Te ) 8002 − 1600Te + Te2 = (1000)(700) − 1700Te + Te2 100Te = (1000)(700) − 8002 Te = 7000 − 6400 = 600 f) To confirm the differential equation: y � (t) = T � (t − t0 ) = k(Te − T (t − t0 )) = k(Te − y(t)) The formula for y is y(t) = T (t − t0 ) = Te + (T0 − Te )e−k(t−t0 ) = a + (y(t0 ) − a)e−c(t−t0 ) with k = c, Te = a and T0 = T (0) = y(t0 ). 3F-6 y = cos3 u − 3 cos u, x = sin4 u 12 3. Integration E. Solutions to 18.01 Exercises dy = (3 cos2 u · (− sin u) + 3 sin u)du, dx = 4 sin3 u cos udu dy 3 sin u(1 − cos2 u) 3 = = 3 4 cos u dx 4 sin u cos u 3F-7 a) y � = −xy; y(0) = 1 dy 1 = −xdx =⇒ ln y = − x2 + c y 2 To find c, put x = 0, y = 1: ln 1 = 0 + c =⇒ c = 0. 2 1 =⇒ ln y = − x2 =⇒ y = e−x /2 2 b) cos x sin ydy = sin xdx; y(0) = 0 sin ydy = sin x dx =⇒ − cos y = − ln(cos x) + c cos x Find c: put x = 0, y = 0: − cos 0 = − ln(cos 0) + c =⇒ c = −1 =⇒ cos y = ln(cos x) + 1 3F-8 a) From the triangle, y � = slope tangent = =⇒ y 1 dy = dx =⇒ ln y = x + c1 =⇒ y = ex+c1 = Aex (A = ec1 ) y P 1 b) If P bisects tangent, then P0 bisects OQ (by euclidean geometry) So P0 Q = x ( since OP0 = x). 13 E. Solutions to 18.01 Exercises Slope tangent = y � = 3. Integration −y dy dx =⇒ =− x y x =⇒ ln y = − ln x + c1 P Exponentiate: y = 1 c1 c · e = ,c > 0 x x Ans: The hyperbolas y = c ,c > 0 x 3G. Numerical Integration 3G-1 Left Riemann sum: (Δx)(y0 + y1 + y2 + y3 ) Trapezoidal rule: (Δx)((1/2)y0 + y1 + y2 + y3 + (1/2)y4 ) Simpson’s rule: (Δx/3)(y0 + 4y1 + 2y2 + 4y3 + y4 ) a) Δx = 1/4 and √ √ y0 = 0, y1 = 1/2, y2 = 1/ 2, y3 = 3/2, y4 = 1. √ √ Left Riemann sum: (1/4)(0 + 1/2 + 1/ 2 + 3/2) ≈ .518 √ √ Trapezoidal rule: (1/4)((1/2) · 0 + 1/2 + 1/ 2 + 3/2 + (1/2)1) ≈ .643 √ √ Simpson’s rule: (1/12)(1 · 0 + 4(1/2) + 2(1/ 2) + 4( 3/2) + 1) ≈ .657 as compared to the exact answer .6666 . . . b) Δx = π/4 √ √ y0 = 0, y1 = 1/ 2, y2 = 1, y3 = 1/ 2, y4 = 0. 14 3. Integration E. Solutions to 18.01 Exercises √ √ Left Riemann sum: (π/4)(0 + 1/ 2 + 1 + 1/ 2) ≈ 1.896 √ √ Trapezoidal rule: (π/4)((1/2) · 0 + 1/ 2 + 1 + 1/ 2 + (1/2) · 0) ≈ 1.896 (same as Riemann sum) √ √ Simpson’s rule: (π/12)(1 · 0 + 4(1/ 2) + 2(1) + 4(1/ 2) + 1 · 0) ≈ 2.005 as compared to the exact answer 2 c) Δx = 1/4 y0 = 1, y1 = 16/17, y2 = 4/5, y3 = 16/25, y4 = 1/2. Left Riemann sum: (1/4)(1 + 16/17 + 4/5 + 16/25) ≈ .845 Trapezoidal rule: (1/4)((1/2) · 1 + 16/17 + 4/5 + 16/25 + (1/2)(1/2)) ≈ .8128 Simpson’s rule: (1/12)(1 · 1 + 4(16/17) + 2(4/5) + 4(16/25) + 1(1/2)) ≈ .785392 as compared to the exact answer π/4 ≈ .785398 (Multiplying the Simpson’s rule answer by 4 gives a passable approximation to π, of 3.14157, accurate to about 2 × 10−5 .) d) Δx = 1/4 y0 = 1, y1 = 4/5, y2 = 2/3, y3 = 4/7, y4 = 1/2. Left Riemann sum: (1/4)(1 + 4/5 + 2/3 + 4/7) ≈ .76 Trapezoidal rule: (1/4)((1/2) · 1 + 4/5 + 2/3 + 4/7(1/2)(1/2)) ≈ .697 Simpson’s rule: (1/12)(1 · 1 + 4(4/5) + 2(2/3) + 4(4/7) + 1(1/2)) ≈ .69325 Compared with the exact answer ln 2 ≈ .69315, Simpson’s rule is accurate to about 10−4 . 15 E. Solutions to 18.01 Exercises 3. Integration b b4 . Using Simpson’s rule with two subintervals, Δx = 4 0 b/2, so that we get the same answer as above:1 � � b b 3 3 b4 3 3 3 S(x ) = (0 + 4(b/2) + b ) = b = . 6 6 2 4 � 3G-2 We have x3 dx = 3G-3 The sum � √ √ 1 + 2 + ... + 10, 000 � 104 √ is related to the trapezoidal estimate of xdx : S= 0 y= x 1 1 � (1) 0 104 2 3 10,000 √ √ 1√ 1√ 4 1√ 4 xdx ≈ 0 + 1 + ... + 10 = S − 10 2 2 2 But � 0 104 �104 √ 2 3/2 �� 2 xdx = x � = · 106 � 3 3 0 From (1), (2) 2 · 106 ≈ S − 50 3 Hence S ≈ 666, 717 (3) In (1), we have >, as in the picture. Hence in (2), we have >, so in (3), we have <, Too high. 3G-4 As in Problem 3 above, let S= 1 1 1 + + ... + 1 2 n Then by trapezoidal rule, 1The fact that Simpson’s rule is exact on cubic polynomials is very significant to its effectiveness as a numerical approximation. It implies that the approximation converges at a rate proportional to the the fourth derivative of the function times (Δx)4 , which is fast enough for many practical purposes. 16 3. Integration E. Solutions to 18.01 Exercises 1 y= x 1 1 � � Since 1 n 2 3 n-1 n n dx 1 1 1 1 1 1 1 1 ≈ · + + + ... + · = S − − x 2 2 2 3 2 n 2 2n 1 dx 1 1 = ln n, we have S ≈ ln n + + . (Estimate is too low.) x 2 2n 3G-5 Referring to the two pictures above, one can see that if f (x) is concave down on [a, b], the trapezoidal rule gives too low an estimate; if f (x) is concave up, the trapezoidal rule gives too high an estimate.. 17