Section 7.4 #1, 5, 6, 8, 12, 13, 44, 53; Section 7.5 #7, 10, 11, 20, 22; Section 7.7 #1, 4, 10, 15, 22, 44

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1 Math B Prof. Audrey Terras HW #4 Solutions Due Tuesday, Oct. 9 Section 7.4 #, 5, 6, 8,, 3, 44, 53; Section 7.5 #7,,,, ; Section 7.7 #, 4,, 5,, Since 5 = 5 )5 + ), start with So, 5 = A 5 + B 5 +. = A5 + ) + B5 ). Equating coefficients gives = A B, = 5A + 5B. The solution to this system is A =, B =. Alternatively, we could choose = 5 above, which gives B = ; = 5 gives A =. Therefore 5 = Since s 4 = s )s + )s + ), start with So, s 4 = A s + B s + + Cs + D s +. = As + )s + ) + Bs )s + ) + Cs + D)s )s + ) = A + B + C)s 3 + A B + D)s + A + B C)s + A B D). Equating coefficients gives A + B + C = A B + D = A + B C = A B D =. From the first and third equations we find that A + B = and C =. From the second and fourth we find A B = and D =. So A = /, B = /, and s 4 = / s / s + s +.

2 7.4.6 Since y 3 y + y = y )y + ), start with So, Equating coefficients gives y y 3 y + y = A y + By + C y +. y = Ay + ) + By + C)y ) y = A + B)y + B + C)y + A C. A + B = B + C = A C =. The solution to this system is A =, B =, C =, so Using the answer from problem, y y 3 y + y = y + y + y +. 5 d = 5 d d = ln 5 + ln C. Note: this could be rewritten as ln C Using the answer from problem 5, / s 4 ds = s / s + ) s ds = + ln s ln s + arctan s + C Because 5 + y appears in the integrand, we should try the substitution y = 5 tan θ, dy = 5 sec θ dθ. So, y 5 + y dy = = 5 5 tan θ 5 + tan θ) 5 sec θ dθ = 5 sec θ ) dθ = 5 tan θ 5θ + C. tan θ dθ For the last step, we must get back to the original variable y. Recall that 5 tan θ = y, and therefore θ = arctany/5), so y = y 5 arctany/5) + C. 5 + y

3 The area of the indicated region is 3 4 d. We should try the substitution = sin θ, d = cos θ dθ because of the appearance of 4. If we change the limits as well, we won t have to back-substitute at the end. So, = becomes θ =, and = becomes θ = π/ d = π/4 8 sin 3 θ 4 sin θ) π/4 cos θ) dθ = 8 sin 3 θ dθ π/4 ) = 8 sin θ sin θ cos θ) dθ = 8 cos θ + cos3 θ π/4 3 = ).69 There are many equivalent ways to epress the final answer) Let f ) =. For each part, divide the interval 6 into two intervals of width 3. a) b) c) d) LEFT) = 3 f ) + f 3)) = 7 RIGHT) = 3 f 3) + f 6)) = 35 TRAP) = = 8 MID) = 3 f.5) + f 4.5)) =

4 7.5. Let f ) = sin. For each part, divide the interval π into two intervals of width π/. a) b) LEFT) = π f ) + f π )) = π RIGHT) = π f π ) + f π)) = π c) d) TRAP) = π + π = π MID) = π f π 4 ) + f 3π 4 )) = π 7.5. Let f ) = +. For part a), divide the interval into eight intervals of width 8. a) i) ii) LEFT8) = 8 RIGHT8) = 8 f ) + f 8 ) + f 8 ) f 7 8 ) ).86. f 8 ) + f 8 ) f 8 8 ) ) iii) TRAP8) = LEFT8) + RIGHT8) b) f ) is a decreasing function for >. Therefore RIGHT8) f ) d LEFT8). In other words, part i) is an overestimate and part ii) is an underestimate. 4

5 7.5. a) TRAP4) probably gives the best estimate. We cannot calculate MID4) because we d need information that s not in the table. LEFT4) = 3 f ) + f 3) + f 6) + f 9)) = 95 RIGHT4) = 3 f 3) + f 6) + f 9) + f )) = 96 TRAP4) = = 7.5 b) If we assume there are no points of inflection, the graph is either concave up or concave down. By plotting points we see that it is concave down. Therefore TRAP4) is an underestimate a) The graph of y = is the upper half of a circle of radius centered at the origin. By drawing a picture of the region bounded by this circle, the -ais, and the lines y =, y =, you can see that d = A + B, where A is the area of a certain triangle and B is the area of a certain sector of a circle. Recall that the area of a triangle is bh, and the area of a sector of a circle is r θ, where r is the radius of the circle and θ is the angle subtended by the sector in radians). If you draw the diagram, you ll see that b = h =, r =, θ = π/4. Therefore A = / and B = π/4. b) Using the methods of this section, LEFT5).335, RIGHT5).466, TRAP5).88, MID5).875. The eact value is + π So, the LEFT error is.38, the RIGHT error is.4474, the TRAP error is.33, and the MID error is.66. So, for this function, MID is the most accurate, followed by TRAP. LEFT and RIGHT are much less accurate as usual) a) This area looks essentially the same as Figure 7.7 a) in your book. It etends infinitely far along the positive -ais. However, the area is infinite whereas the area in Figure 7.7 a) is actually finite.) b) This area looks essentially the same as figure 7.9 a) in your book. It etends infinitely far along the positive y-ais. 5

6 7.7.4 a) e is a bell-shaped curve; see figure 5.67 in your book. The integral represents the entire area between the -ais and the curve, infinitely far in both directions. b) Using a calculator, e d e d.7646 e d.774 e d.7745 e d.7745 c) It appears that b d = lim d = lim 4 + b 4 + b ln 4 + Since lim b ln 4 + b =, the integral diverges. b = lim b ln 4 + b ln This integral is improper because v is undefined at v =. To evaluate it, we must split it into two integrals, v dv and v dv. But, notice that Therefore the integral diverges. dv = lim dv = lim ln a) =. v a + a v a + 6

7 7.7. We find the antiderivative with the help of the substitution u = ln, du = d. Namely, ln d = u du = u + C = ln ) + C. This integral is improper because the integrand is undefined at =. So, we have to take the limit from the right: ln d = lim Therefore the integral diverges a + ln ) a = lim a + ln a) )) =. a) The graph starts at the origin, increases to a maimum at =, and then decreases back towards the -ais as. Your graph should show these three features clearly, although it s okay if your sketch is not labeled with specific values.) b) People are getting sick fastest when the rate of infection is highest. Therefore we should set dr dt = to find the maimum value: dr dt = e.5t.5)te.5t = 5e.5t t), which is zero only at t =. Therefore t = is the moment at which the rate of infection is highest. c) The total number of sick people is te.5t dt. We ll use integration by parts to find an antiderivative: let u = t, dv = e.5t dt, so du = dt, v = e.5t. te.5t dt = uv v du = te.5t e.5t ) dt) = e.5t t 4) + C Now, we can calculate the original improper integral: te.5t dt = lim e.5t t 4) b [ ] = lim b e.5b b 4) ) 4) However, we know that the limit above is equal to zero because e b times any polynomial goes to zero as b. Therefore the total number of infected people is 4. b 7