Find the midpoint Riemann Sum of cos(x^2) with n = 4, from [0, 2] - CORRECT ANSWERS-Mid S4 = (1)(1/2)[cos(.25^2) + cos(.75^2) + cos(1.25^2) + cos(1.75^2)] Mid S4 = (1)(1/2)[cos(.625) + cos(.5625) + cos(1.5625) cos(3.0625)] Mid S4 = .824 If the function f

Find the midpoint Riemann Sum of cos(x^2) with n = 4, from [0, 2] - CORRECT ANSWERS-Mid S4 = (1)(1/2)[cos(.25^2) + cos(.75^2) + cos(1.25^2) + cos(1.75^2)] Mid S4 = (1)(1/2)[cos(.625) + cos(.5625) + cos(1.5625) cos(3.0625)] Mid S4 = .824 If the function f is continuous for all real numbers and if f(x) = (x^2-7x +12)/(x -4) when x ≠ 4 then f(4) = - CORRECT ANSWERS-Factor numerator so f(x) = (x-3)(x-4)/(x-4) = x-3 f(4)=4-3 f(4) = 1 If f(x) = (x^2+5) if x 2, & f(x) = (7x -5) if x ≥ 2 for all real numbers x, which of the following must be true? I. f(x) is continuous everywhere. II. f(x) is differentiable everywhere. III. f(x) has a local minimum at x = 2. - CORRECT ANSWERS-At f(2) both the upper and lower piece of the discontinuity is 9 so the function is continuous everywhere. At f'(2) the upper piece is 4 and lower piece is 7 so f(x) is not differentiable everywhere. Since the slopes of the function on the left and right are both positive the function cannot have a local minimum or maximum at x= 2. Only I is true. For the function f(x) = (ax^3-6x), if x ≤ 1, & f(x) = (bx^2+4), x 1 to be continuous and differentiable, a = ..... - CORRECT ANSWERS-for the function to be continuous f(1) has to equal f(1): a(1^3) -6(1) = b(1^2) +4 a -6 = b +4 b=a-10 for the functions to be differentiable f'(1) has to equal f'(1): 3a(1^2) -6 = 2b(1) 3a -6 = 2b plug b from the first equation in to find a: 3a -6 = 2(a -10) a = -14 Find k if f(x) = (k) at x = 4 and f(x) = ((x^2 -16)/(x-4)) - CORRECT ANSWERS-1. f(4) exists and is equal to 8 2. lim from the left and right are both 8 3. lim f(x) as x approaches 4 is 8 which equals f(4) k must equal 8 If f(x) is continuous and differentiable and f(x) = (ax^4 +5x) for x ≤ 2, & f(x)= (bx^2 -3) for x 2 , then b =... - CORRECT ANSWERS-Plug x = 2 into both pieces. f(x) = (16a +10) for x ≤ 2, & (4b -6) for x 2 They must be equal to be continuous 16a +10 = 4b -6 a=.25b-1 Take the derivative of both pieces of this function and plug in x = 2 f(x) = (32a +5) for x ≤ 2, & f(x) = (4b -3) for x 2 They must be equal to be differentiable 32a +5 = 4b -3 plug in the first equation to find b 32(.25b-1)+5= 4b-3 b = 6 If f is continuous for a ≤ x ≤ b, then at any point x = c, where a c b, which of the following must be true? a. f(c) = (f(b) - f(a))/(b-a) b. f(a) = f(b) c. f(c) = 0 d. lim f(x) as x approaches c = f(c) - CORRECT ANSWERS-In order for f(x) to be continuous at point c, there are three conditions that need to be fulfilled: 1. f(c) exists 2. lim f(x) as x approaches c exists 3. lim f(x) as x approaches c = f(c) Answers a ,b , and c are not necessarily true. If the function f(x) = (3ax^2+2bx +1), x ≤ 1, & f(x) = (ax^4-4bx^2-3x), x 1 , is differentiable for all real values of x, then b = ... - CORRECT ANSWERS-For the function to be continuous both pieces must be equal at 1. Plug in x = 1 into both pieces and set them equal to each other. f(x) = (3a +2b +1), x ≤ 1, & f(x) = (a -4b -3), x 1 3a +2b +1 = a -4b -3 2a +6b = -4. For the function to be differentiable the derivative of both pieces must be equal at 1. Plug x = 1 into the derivative of both pieces f'(x) = (6a +2b), x 1, & f'(x) = (4a -8b -3), x 1 6a +2b = 4a -8b -3 2a +10b = -3 Solving the simultaneous equations, we get b = 1/4 Let F(x) = the integral of (cos(t/2) + (3/2))dt on the closed interval from [0, 4pi]. Approximate F(2pi) using four inscribed rectangles. - CORRECT ANSWERS-Area = (pi/2)[(cos(pi/4) + 3/2) + (cos(pi/2) + 3/2) + (cos(3pi/4) + 3/2) + (cos(pi) + 3/2)] Area = (pi/2)[(3/2 + 1/sqrt(2)) + (3/2) + ((3/2) - 1/sqrt(2)) + (1/2)] = 5pi/2 Find the left-sided Riemann Sum of sin(x^2) with n = 4, from [0, 4] - CORRECT ANSWERS-Left S4 = (1)[sin(0^2) + sin(1^2) + sin(2^2) + sin(3^2)] Left S4 = (1)[sin(0) + sin(1) + sin(4) + sin(9)] Left S4 = .497 Find the right-sided Riemann Sum of sin(x^2) with n = 4, from [0, 4] - CORRECT ANSWERS-Right S4 = (1)[sin(1^2) + sin(2^2) + sin(3^2) + sin(4^2)] Right S4 = (1)[sin(1) + sin(4) + sin(9) + sin(16)] Right S4 = .201 Find the midpoint Riemann Sum of sin(x^2) with n = 4, from [0, 4] - CORRECT ANSWERS-Mid S4 = (1)[sin(.5^2) + sin(1.5^2) + sin(2.5^2) + sin(3.5^2)] Mid S4 = (1)[sin(.25) + sin(2.25) + sin(6.25) + sin(12.25)] Mid S4 = .681 Find the left-sided Riemann Sum of ln(x^2) with n=2, from [1, 3] - CORRECT ANSWERS-Left S2 = (1)[ln(1^2) + ln(2^2)] Left S2 = (1)[ln(1) + ln(4)] Left S2 = 1.386 Find the right-sided Riemann Sum of cos(x^2) with n = 3, from [2, 5] - CORRECT ANSWERS-Right S3 = (1)[cos(3^2) + cos(4^2) + cos(5^2)] Right S3 = (1)[cos(9) + cos(16) + cos(25)] Right S3 = -.878 Approximate the area between the x-axis and h(x) = 1/(7-x) from x = 2 to x = 5 using a left Riemann sum with 3 equal subdivisions. - CORRECT ANSWERS-Area = (1)[h(2) +h(3) +h(4)] Area = (1)[1/5 +1/4 +1/3] Area = 47/60 Use the trapezoidal approximation for the integral of (sinx)^2dx from [0, 1] with n = 4 to three decimal places. - CORRECT ANSWERS-Trapezoid = (1/2)(1/4)[(sin(0))^2 +2(sin(1/4))^2 +2(sin(1/2))^2 +2(sin(3/4))^2 +(sin(1))^2] Trapezoid = .277 What is the trapezoidal approximation for the integral of (e^x)dx from [0, 1] with n = 4 sub intervals to three decimal places. - CORRECT ANSWERS-Trapezoid = (1/2)(3/4)[(e^(0)) +2(e^(3/4)) +2(e^(6/4)) +2(e^(9/4)) +(e^(3))] Trapezoid = 10.972 Use the trapezoid rule with n = 4 to approximate the area between the curve f(x) = x^3 -x^2 and the x-axis from x = 3 to x =4 - CORRECT ANSWERS-Trapezoid = (1/2)(1/4)[(3^3 -3^2) +2(3.25^3 -3.25^2) +2(3.5^3 -3.5^2) +2(3.75^3 -3.75^2) +(4^3 -4^2)] Trapezoid = 31.516 Use the trapezoid rule with n = 4 to approximate the area between the curve f(x) = x^3 -x and the x-axis from x = 3 to x =7 - CORRECT ANSWERS-Trapezoid = (1/2)(1)[(3^3 -3) +2(4^3 -4) +2(5^3 -5) +2(6^3 -6) +(7^3 -7)] Trapezoid = 570 Use the trapezoid rule with n = 4 to approximate the area between the curve f(x) = x^2 + 1 and the x-axis from x = 3 to x =7 - CORRECT ANSWERS-Trapezoid = (1/2)(1)[(3^2 +1) +2(4^2 +1) +2(5^2 +1) +2(6^2 +1) +(7^2 +1)] Trapezoid = 110 Use the trapezoid rule with n = 6 to approximate the area between the curve f(x) = 3x^3 - 4 and the x-axis from x = 0 to x =6 - CORRECT ANSWERS-Trapezoid = (1/2)(1)[(3(0^3) - 4) + 2(3(1^3) - 4) + 2(3(2^3) - 4) + 2(3(3^3) - 4) + 2(3(4^3) - 4) + 2(3(5^3) - 4) + (3(6^3) - 4)] Trapezoid = 975 Use the trapezoid rule with n = 4 to approximate the area between the curve f(x) = 2x^3 - 1 and the x-axis from x = 2 to x =6 - CORRECT ANSWERS-Trapezoid = (1/2)(1)[(2(2^3)-1) +2(2(3^3)-1) +2(2(4^3)-1) +(2(5^3)-1) +(2(6^3)-1)] Trapezoid = 527.5 Find the area of the polar equation r = 4cos θ - CORRECT ANSWERS-A = (1/2) ∫(4cos θ)^2dθ from [0, 2pi] plug into calculator A = 8pi + 8sin(pi) Find the area inside the first curve R = 2 + sin θ and outside the second curve r = 3sin θ - CORRECT ANSWERS-Find the positions of intersection by setting the equations equal to each other and solving for θ. θ=-pi/2, pi/2 (These are your bounds) A = (1/2)∫(2+sinθ)^2 - (3sinθ)^2 dθ from [-pi/2, pi/2] plug into calculator A = 9pi/4 Find the area bounded by the two curves. y = x, y = ∜x - CORRECT ANSWERS-set functions equal to find the points of intersection; x = 0,1 (These are your bounds) A = ∫(∜x) - (x)dx from [0,1] A = (4x^(5/4))/5 -(x^2)/2 from [0,1] A = 3/10 Find the area bounded by the two curves. y =x^2, y = x - CORRECT ANSWERS-set functions equal to find the points of intersection; x = 0,1 (These are your bounds) A = ∫(x)-(x^2)dx from [0,1] A = (x^2)/2 - (x^3)/3 from [0,1] A = 1/6 Find the area bounded by the two curves. y =x^2-6x+10, y = -x^2+6x-6 - CORRECT ANSWERS-set functions equal to find the points of intersection; x = 2,4 (These are your bounds) A = ∫(-x^2+6x-6) - (x^2-6x+10)dx from [2,4] plug into calculator A = 8/3 Find the area bounded by the two curves. y = 3 - x, y = 4 -(x-1)^2 - CORRECT ANSWERS-set functions equal to find the points of intersection; x = 0,3 (These are your bounds) A = ∫[4 -(x-1)^2 - (3 -x)]dx from [0,3] plug into calculator A = 9/2 Find the area bounded by the two curves. y = 4x - x^2, y = 8x -2x^2 - CORRECT ANSWERS-set functions equal to find the points of intersection; x = 0,4 (These are your bounds) A = ∫[8x -2x^2 -(4x - x^2)]dx from [0,4] plug into calculator A = 32/3 Find the area bounded by the two curves. y = 4x, y = 8x -2x^2 - CORRECT ANSWERS-set functions equal to find the points of intersection; x = 0,2 (These are your bounds) A = ∫[8x -2x^2 -4x]dx from [0,2] A = ∫[(4x - 2x^2)]dx from [0,2] A = 2x^2 - (2x^3)/3 from [0,2] A = 8/3 Is the function continuous at the point x = 2? f(x) = x+1, x 2 f(x) = 2x-1, x≥2 - CORRECT ANSWERS-Condition 1: Does f(2) exist? Yes 2(2) -1 = 3 Condition 2: Does lim f(x) as x approaches 2 exist? Left limit is 3, Right limit is 3 Because the two limits are the same the limit exists. Condition 3: Does lim f(x) as x approaches 2 = f(2)? The two equal each other so the function is continuous at x = 2. Is the function continuous at the point x = 2? f(x) = x+1, x 2 f(x) = 2x-1, x2 - CORRECT ANSWERS-Condition 1: Does f(2) exist? No. The function of x is defined if x is greater than or less than 2, but not if x is equal to 2. The function is not continuous at x =2. Is the function continuous at the point x = 2? f(x) = x+1, x 2 f(x) = 2x+1, x≥2 - CORRECT ANSWERS-Condition 1: Does f(2) exist? Yes 2(2) +1 = 5 Condition 2: Does lim f(x) as x approaches 2 exist? Left limit is 3, Right limit is 5 Because the two limits don't match the limit does not exist so the function is not continuous at x = 2. Integrate (x^2 +2x -1)/(x^3 -x) - CORRECT ANSWERS-This is a partial fractions integral. Rewrite integrand as (x^2 +2x -1)/(x^3 -x) = (A/x) +(B/(x+1)) +(C/(x-1)) multiply both sides by x^3-x x^2 + 2x - 1 = A(x+1)(x-1) +Bx(x-1) + Cx(x+1) let x = -1 and solve for B B = -1 let x = 0 and solve for A A = 1 let x = 1 and solve for C C = 1 Plug numbers into integral and integrate ∫(1/x) -1/(x+1) + 1/(x-1) dx = ln|x| - ln|x+1| + ln|x-1| + C = ln|(x(x-1))/(x+1)| + C ∫(18x-17)/(2x-3)(x+1)dx = - CORRECT ANSWERS-This is a partial fractions integral. Write integrand as: 18x-17/(2x-3)(x+1)= (A/(2x-3))+ (B/(x+1)) Multiply both sides by (2x-3)(x+1) Now we have: 18x-17=A(x+1) + B(2x-3) 18x-17=Ax+A+2Bx-3B 18x-17=(Ax+2Bx) + (A-3B) 18x-17=x(A+2B) + (A-3B) Thus (A+2B) = 18 and (A-3B)= -17 Solve these to get: A=4 and B=7 We can now rewrite the original integral as: ∫(4/(2x-3))+(7/(x+1)) dx Integrate to get: 2ln|2x-3|+7ln|x+1|+C ∫(x-18)/(x+3)(x-4) dx = - CORRECT ANSWERS-Use partial fractions to evaluate the integral. First, we write the intragrand as: (x-18)/(x+3)(x-4)= A/(x+3) + B/(x-4) Multiply both sides by: (x+3)(x-4) Now we have: x-18 = A(x-4)+B(x+3) x-18= Ax-4A+Bx+3B x-18= (Ax+Bx) + (3B-4A) x-18= x(A+B) + (3B-4A) Thus (A+B)=1 and (3-4A)= -18 Solve this using simultaneous equations to get A=3 and B=-2. We can now rewrite the original integral as: ∫(3/(x+3)) + (-2/(x-4))dx = ∫3dx/(x+3) - ∫2dx/(x-4) Find the area of the polar equation r = 2cos 3θ - CORRECT ANSWERS-A = (1/2) ∫(2cos3θ)^2dθ from [0, 2pi] plug into calculator A = (1/6)(pi) Find the area of the polar equation r = sin2θ + 2 - CORRECT ANSWERS-A = (1/2) ∫ (sin2θ + 2)^2 dθ from [0, 2pi] plug into calculator A = 9pi/4 Find the midpoint Riemann Sum of cos(x^2) with n = 2, from [0, 2] - CORRECT ANSWERS-Mid S2 = (1)[cos(.5^2) + cos(1.5^2)] Mid S2 = (1)[cos(.25) + cos(2.25)] Mid S2 = .341 Is the function continuous at the point x = 4? f(x) = (x/(x-4)), x 4 f(x) = ((2x-1)/(x-4)), x4 - CORRECT ANSWERS-No. The function has an infinite discontinuity at x = 4