# [17-08-24]_[1910-17F]_HWSolutions01.pdf - H OMEWORK S...

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Unformatted text preview: H OMEWORK S OLUTIONS MATH 1910 Fall 2017 Sections 5.3, 5.1, 5.2 Problem 5.3.15 Z  18t5 − 10t4 − 28t dt. Evaluate the indefinite integral S OLUTION. Z Z Z Z  18t5 − 10t4 − 28t dt = 18 t5 dt − 10 t4 dt − 28 t dt t5 t2 t6 − 10 − 28 + C 6 5 2 = 3t6 − 2t5 − 14t2 + C = 18 Problem 5.3.20 Z Evaluate the indefinite integral S OLUTION. Z dx . x4/3 Z 3 dx x−1/3 + C = − 1/3 + C = x−4/3 dx = 4/3 −1/3 x x Problem 5.3.28 5.3.15 5.3.20 Z  θ + sec2 θ dθ. Evaluate the indefinite integral S OLUTION. Z Problem 5.3.35  θ2 θ + sec2 θ dθ = + tan θ + C 2 5.3.28 Z Evaluate the indefinite integral sec 12t tan 12t dt. S OLUTION. Recall that d dx sec x = sec x tan x. Z sec 12t tan 12t dt = 1 1 sec 12t + C 12 5.3.35 Problem 5.3.61   dy 1 Solve the initial value problem = cos 3π − θ , dθ 2 S OLUTION. We have Z y= y(3π) = 8.     1 1 cos 3π − θ dθ = −2 sin 3π − θ + C. 2 2 To find C, use the initial condition:     1 3π 8 = y(3π) = −2 sin 3π − · 3π + C = −2 sin + C = 2 + C. 2 2 It follows that C = 6, so y = −2 sin 3π − 1 2  · θ + 6. 5.3.61 Problem 5.3.75 A mass oscillates at the end of a spring. Let s(t) be the displacement of the mass from the equilibrium position at time t. Assuming that the mass is located at the origin at t = 0 and has velocity v(t) = sin(πt/2) m/s, state the differential equation satisfied by s(t), and find s(t). S OLUTION. Velocity is the derivative of position, so we have π  t , s(0) = 0. s 0 (t) = sin 2 To find s(t) we follow the same procedure as the previous problem. Z π  π  2 s(t) = sin t dt = − cos t +C 2 π 2 To find C, use the initial condition. 0 = s(0) = − π  2 2 cos ·0 +C=− +C π 2 π Thus C = 2/π, so we have s(t) = − π  2 2 cos t + . π 2 π 2 5.3.75 Problem 5.1.37 150 X Rewrite and evaluate the sum n2 . n=51 S OLUTION. 150 X n2 = n=51 150 X 50 X n2 − n=1 n2 n=1 150(150 + 1)(2 · 150 + 1) 50(50 + 1)(2 · 50 + 1) − = 6 6 = 1, 136, 275 − 42, 925 = 1, 093, 350 5.1.37 Problem 5.1.38 200 X Rewrite and evaluate the sum k3 . k=101 S OLUTION. 200 X k3 = k=101 200 X k=1  = k3 − 100 X k3 k=1 200(200 + 1) 2 2  − 100(100 + 1) 2 2 = 404, 010, 000 − 25, 502, 500 = 378, 507, 500 5.1.38 3 Problem 5.1.45 Evaluate the limit lim N 2 X i −i+1 N→∞ i=1 N . S OLUTION. lim N 2 X i −i+1 N→∞ i=1 N3 N X 1 = lim N→∞ N3 i=1 i2 − N X N X i+ i=1 ! 1 i=1   N(N + 1)(2N + 1) N(N + 1) 1 = lim − +N N→∞ N3 6 2   1 N(N + 1)(2N + 1) N(N + 1) − + 3 = lim N→∞ 6N3 2N3 N 1 = 3 5.1.45 Problem 5.1.53 Show, f(x) = 3x2 + 4x over [0, 2], that 2 X N N RN =  j=1 12j2 8j + N2 N  . Then evaluate lim RN . N→∞ S OLUTION. Recall that for a function f(x) on an interval [a, b], the right-endpoint approximation RN is given by N X RN = ∆x f(a + j∆x) j=1 2 where ∆x = (b − a)/N. Taking f(x) = 3x + 4x, a = 0, and b = 2, we have ∆x = 2/N and RN "  2  #  N N  2 X 12j2 8j 2 X 2 2 3 j· +4 j· = . = + N N N N N2 N j=1 j=1 4 We can now compute the limit. lim RN N→∞  N  2 X 12j2 8j = lim + N→∞ N N2 N j=1 N N X X 16 24 j2 + 2 j = lim 3 N→∞ N N j=1 j=1 S E C T I O N 5.2 The Definite Integral   16 N(N + 1) 24 N(N + 1)(2N + 1) = lim + 2 y N→∞ N3 6 N 2  1 N(N + 1) N(N + 1)(2N + 1) 0.5 +8 = lim 4 N→∞ N3 N2 x = 8 + 8− 0.5 = 16 1 2 3 4 533 5.1.53 −1 7. ! 0 5" 25 − x 2 dx Problem 5.1.73 √ solution The region s bounded by the graph of y = 25 − x 2 and the x-axis over the interval [0, 5] is  2 N 1 X one-quarter of a circle of radius 5. jHence, Evaluate lim by interpreting it as the area of part of a familiar geometric figure. 1− N→∞ N N! 5 " j=1 1 25π 25 − x 2 dx = π(5)2 = . 4 4 0 S OLUTION. Comparing the the formula for RN , we see that the limit is equal √ expression above y to the area under f(x) = 1 − x2 over [0, 1]. This region is the same as the quarter of the unit 5 disk x2 + y2 ≤ 1 that lies in the first quadrant. Thus the limit is a quarter the area of the unit 4 circle: s 3  2 N 1 X 2 j π 1 lim 1 − 5.1.73 = π · 12 = . 1 N→∞ N N 4 4 j=1 8. ! 1 2 3 4 5 x 3 |x| dx −2 Problem 5.2.8 Z 3 the interval [−2, 3] consists of solution The region bounded by the graph of y = |x| and the x-axis over 1 Draw a graph of the signed area represented by the definite integral andhas compute it using two right triangles, both above the axis. One triangle has area 2 (2)(2) = 2, and |x| thedx other area 12 (3)(3) = 92 . −2 Hence, geometry. ! 3 9 13 |x| dx = +2= . S OLUTION. The area consists of two right triangles: 2 2 −2 y 3 2 1 −2 9. ! −1 1 2 3 x 2 −2 (2 − |x|) dx solution The region bounded by the graph of y = 2 − |x| and the x-axis over the interval [−2, 2] is a 5 triangle above the axis with base 4 and height 2. Consequently, ! 2 1 (2 − |x|) dx = (2)(4) = 4. 2 −2 y 2 • The region bounded by the graph of y = 4x − 8 and the x-axis over the interval [−1, 4] consists of a triangle below the axis with base 3 and height 12 and a triangle above the axis with base 2 and height 8. Hence, ! 4 1 1 (4x − 8) dx = − (3)(12) + (2)(8) = −10. 2 2 −1 y Thus the integral is the sum of the areas of the triangles. Z3 5 −1 |x| dx = 1 −52 −2 · 2 ·1 2 +2 1 x 13 ·33 · 34 = 2 2 5.2.8 −10 InProblem Exercises 135.2.14 and 14, refer to Figure 14. Let f(x) be the function with the following graph: y y = f(x) 2 4 6 x FIGURE 14 The two parts of the graph are semicircles. Evaluate: Z4 ! 2 ! 6 a) f(x) dx 13. Evaluate: (a) f (x) dx (b) f (x) dx Z16 0 0 b) |f(x)| Let dx f (x) be given by Figure 14. solution 1 "2 (a) The definite integral 0 f (x) dx is the signed area of a semicircle of radius 1 which lies below the x-axis. S OLUTION. a) The integral computes the signed area of half of smaller semicircle and half of Therefore, the area of the larger semicircle. ! 2 1 π 2 Z4 f (x) dx 1 = −22 . 3 1 = − 22π (1) f(x)0 dx = − · π · 1 + · π · 2 = π 4 4 4 1 "6 (b) The definite integral 0 f (x) dx is the signed area of a semicircle of radius 1 which lies below the x-axis and semicircle ofcounts radius 2the which lies above x-axis. b)aThe integral unsigned areathe of half of Therefore, the smaller semicircle the entire larger semicircle. Z!6 6 3π 11 9 2 11 2 f (x) − ·ππ(1) f(x) dxdx== · π ·(2) 12 + · 22 = = π. 5.2.14 2 2 4 2 42 0 1 14. Evaluate: (a) ! 4 (b) f (x) dx 1 ! 6 1 |f (x)| dx solution Let f (x) be given by Figure 14. "4 (a) The definite integral 1 f (x) dx is the signed area of one-quarter of a circle of radius 1 which lies below the x-axis and one-quarter of a circle of radius 2 which lies above the x-axis. Therefore, ! 1 "6 4 f (x) dx = 1 1 3 π (2)2 − π (1)2 = π. 4 4 4 (b) The definite integral 1 |f (x)| dx is the signed area of one-quarter of a circle of radius 1 and a semicircle of radius 2, both of which lie above the x-axis. Therefore, ! 6 1 9π 1 |f (x)| dx = π (2)2 + π (1)2 = . 2 4 4 1 6 536 CHAPTER 5 THE INTEGRAL Problem 5.2.16 Za Zc InFind Exercises 15 and refer g(t) to Figure 15. g(t) dt are as large as possible. a, b, and c so16, that dt and 0 b y y = g(t) 2 1 1 2 3 4 5 t −1 −2 FIGURE 15 ! 5 as much of the positive area as possible. This is achieved by taking 3 want to take S OLUTION.! We 15. Evaluate g(t) dt and g(t) dt. a = 4 and b = 1, c = 4. 5.2.16 0 3 solution The region bounded by the curve y = g(x) and the x-axis over the interval [0, 3] is comprised of two 1 right triangles, 2 below the axis, and one with area 2 above the axis. The definite integral Problem 5.2.34one with area 1 is therefore equal to 2 − 2 = 32 . Z2 • The bounded the curve = g(x) and the x-axis over summary the intervalto[3, 5] is comprised Use the region properties of theby integral andy the formulas in the chapter evaluate (4x +of7)another dx. two right triangles, one with area 1 above the axis and one with area 1 below the −3 axis. The definite integral is therefore equal to 0. S OLUTION. ! a ! Z2 Z Z 2c Z2 16. Find a, b, and c such that g(t) dt and g(t) dt are as large as possible. (4x 0+ 7) dx = 4 b x dx + 7 dx −3 −3 −3 ! a ! Z Z2 solution To make the value of g(t) dt as0 large as possible, we want to include as much positive area =4 x dx + x dx + 7(2!− (−3)) 0 c −3 0 as possible. This happens when we take a = 4. Now, g(t) dt as large as possible, we ! of Z to make Zthe value • −3 2 b want to make sure to include all of the positive positive area. This happens when we take = 4 −area and x dxonly + the x dx + 25) 0 0 b = 1 and c = 4.   1 for the 1 points 2 Riemann sum shown in Figure 16. Compute 17. Describe the partition P and the set of=sample · 22 − C(−3) + 35 4 2 2 the value of the Riemann sum. = y 25. 5.2.34 34.25 20 15 8 0.5 1 2 2.5 3 3.2 4.5 5 x FIGURE 16 solution The partition P is defined by x0 = 0 < x1 = 1 < x2 = 2.5 < x3 = 3.2 < x4 = 5 The set of sample points is given by C = {c1 = 0.5, c2 = 2, c3 = 3, c4 = 4.5}. Finally, the value of the Riemann sum is 34.25(1 − 0) + 20(2.5 − 1) + 8(3.2 7 − 2.5) + 15(5 − 3.2) = 96.85. 18. Compute R(f, P , C) for f (x) = x 2 + x for the partition P and the set of sample points C in Figure 16. [The curve shown is not f (x) = x 2 + x.] solution Problem 5.2.37 Z1 (u2 − 2u) du. Use the properties of the integral and the formulas in the chapter summary to evaluate 0 S OLUTION. Z1 Z1 Z1 2 2 u du u du − 2 (u − 2u) du = 0 3 0 1 12 −2· 3 2 1 =− 3 0 = 8 5.2.37 ...
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