Calculus BC Project
Find k so that f(x) = ((x^2 -16)/(x-4)), x ≠ 4, & (k), x = 4.
1. f(4) exists, it's equal to k 2. lim from the left and right are both 8 3. lim f(x) as x approaches 4 is 8 = f(4) is 8 k must equal 8
Find the area of the polar equation r = sin2θ + 2
A = (1/2) F (sin2θ + 2)^2 dθ from [0, 2pi] A = (1/2)(1/2) F(8sin(2θ) - cos(4θ) + 9) dθ from [0, 2pi] A = (1/4))(9θ - (1/4)sin(4θ) - 4cos(2θ) [0, 2pi] A = 9pi/4
Find the area of the polar equation r = 2cos 3θ
A = (1/2) F(2cos3θ)^2dθ from [0, 2pi] A = (1/2)(4) Fcos3θ^2dθ from [0, 2pi] A = (2)(1/12)(6θ + sin(6θ)) [0, 2pi] A = (1/6)(pi)
Find the area of the polar equation r = 4cos θ
A = (1/2) F(4cos θ)^2dθ from [0, 2pi] A = (1/2)(16) Fcosθ^2dθ from [0, 2pi] A = (8)(x/2 +sin2x/4) [0, 2pi] A = (8)(pi + sin(pi) - 0) A = 8pi + 8sin(pi)
Find the area inside the first curve and outside the second curve. R = 2 + sin θ, r = 3sin θ
A = (1/2)F(2+sinθ)^2 -(1/2)F(3sinθ)^2 dθ [-pi/2, pi/2] A = 2(1/2)F4 + 4sinθ + sinθ^2dθ - 9Fsinθ^2 dθ [0, pi/2] A = F4+4sinθ+(1/2)(1-cos2θ) dθ - (9/2)F(1-cos2θ) dθ [0, pi/2] A = F(9/2)+4sinθ -(1/2)cos2θ dθ - (9/2)F(1-cos2θ) dθ [0, pi/2] A = (9θ/2)-4cosθ-sin2θ/4 - 9θ/2 - 9sin2θ/4 [0, pi/2] A = 9pi/4
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.
Area = (1)[h(2) +h(3) +h(4)] Area = (1)[1/5 +1/4 +1/3] Area = 47/60
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.
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
If f(x) = (x^2 +5) if x < 2, & (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.
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.
Is the function continuous at the point x = 2? f(x) = x+1, x < 2 f(x) = 2x-1, x>2
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
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.
Is the function continuous at the point x = 2? f(x) = x+1, x < 2 f(x) = 2x-1, x≥2
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.
For the function f(x) = (ax^3 -6x), if x ≤ 1, & (bx^2 +4), x > 1 to be continuous and differentiable, a = .....
Continuous a(1^3) -6(1) = b(1^2) +4 a -6 = b +4 Differentiable 3a(1^2) -6 = 2b(1) 3a -6 = 2b 3a -6 = 2(a -10) a = -14
Integrate (x^2)/[(x-3)(x+2)^2]
F(A/(x-3)) + (B/(x+2)) + (C/(x+2)^2) x^2 = A(x+2)^2 + B(x+2)(x-3) + C(x-3) x^2 = Ax^2 +4Ax +4A +Bx^2-Bx -6B +Cx+3C SYSTEMS of EQUATIONS x^2 eq: 1 = A +B x eq: 0 = 4A -B +C # eq: 0 = 4A -6B -3C 0 = 4A -B +C 0 = 4A -6B -3C 0 = 12A -3B +3C 0 = 4A -6B -3C 0 = 16A -9B 9 = 9A+9B 9 = 25A A = 9/25 1 = 9/25 +B B = 16/25 0 = 36/25 - 16/25 + C C = -20/25 = (1/25)F(9/(x-3)) + (16/(x+2)) - (20/(x+2)^2) = (1/25) (9ln[x-3] + (16ln[x+2]) +(20/(x+2)) + C
Integrate (x^2 -x +6)/(x^3 +3x)
F(A/x) + (Bx+C)/(x^2+3)dx x^2 -x +6 = Ax^2 +3A +Bx^2 +Cx x^2: 1 = A +B, B = -1 x: -1 = C, C = -1 #: 6 = 3A, A = 2 F(2/x) - (x+1)/(x^2 + 3) dx = 2ln[x] - (1/2)ln[x^2 +3] - (1/sqrt(3))arctan(x/sqrt(3))+C
Integrate (x^2 +2x -1)/(x^3 -x)
F(A/x) +(B/(x+1)) +C/(x-1)dx x^2 + 2x - 1 = A(x+1)(x-1) +Bx(x-1) + Cx(x+1) let x = -1 B = -1 let x = 0 A = 1 let x = 1 C = 1 F(1/x) -1/(x+1) + 1/(x-1) + C ln[x] - ln[x+1] + ln[x-1] ln[(x(x-1))/(x+1)] + C
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. Ff(x)dx = f(c) e. lim f(x) as x approaches c = f(c)
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 ,c, and d are not necessarily true.
Find the left-sided Riemann Sum with n = 2, ∑ln(x^2 ) from [1, 3]
Left S2 = (1)[ln(1^2) + ln(2^2)] Left S2 = (1)[ln(1) + ln(4)] Left S2 = 1.386
Find the left-sided Riemann Sum with n = 4, ∑sin(x^2 ) from [0, 4]
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 midpoint Riemann Sum with n = 2, ∑cos(x^2 ) from [0, 2]
Mid S2 = (1)[cos(.5^2) + cos(1.5^2)] Mid S2 = (1)[cos(.25) + cos(2.25)] Mid S2 = .341
Find the midpoint Riemann Sum with n = 4, cos(x^2 ) from [0, 2]
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
Find the midpoint Riemann Sum with n = 4, ∑sin(x^2 ) from [0, 4]
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
Is the function continuous at the point x = 4? f(x) = (x/(x-4)), x < 4 f(x) = ((2x-1)/(x-4)), x>4
No. The function has an infinite discontinuity at x = 4
If the function f(x) = (3ax^2 +2bx +1), x ≤ 1, & (ax^4 -4bx^2 -3x), x > 1 , is differentiable for all real values of x, then b = ...
Plug in x = 1 into both pieces. f(x) = (3a +2b +1), x ≤ 1, & (a -4b -3), x > 1 , so 3a +2b +1 = a -4b -3, which can be simplified to 2a +6b = -4. Plug x = 1 into the derivative of both pieces f'(x) = (6a +2b), x < 1, & (4a -8b -3), x > 1 , so we need 6a +2b = 4a -8b -3 which can be simplified to 2a +10b = -3 Solving the simultaneous equations, we get a = -11/4 & b = 1/4
If f(x) is continuous and differentiable and f(x) = (ax^4 +5x), x ≤ 2, & (bx^2 -3), x > 2 , then b =...
Plug x = 2 into both pieces. f(x) = (16a +10), x ≤ 2, & (4b -6), x > 2 16a +10 = 4b -6 Derivative of both pieces of this function and plug x = 2 f(x) = (32a +5), x ≤ 2, & (4b -3), x > 2 , so we need 32a +5 = 4b -3 a = 1/2 & b = 6
Find the area bounded by the two curves. y =x^2, y = x
Pts of intersection; x = 0,1 A = F(x)dx - F(x^2)dx [0,1] A = (x^2)/2 - (x^3)/3 [0,1] A = 1/6
Find the area bounded by the two curves. y = x, y = ∜x
Pts of intersection; x = 0,1 A = F(∜x)dx - F(x)dx [0,1] A = (4x^(5/4))/5 -(x^2)/2 [0,1] A = 3/10
Find the area bounded by the two curves. y = 4x, y = 8x -2x^2
Pts of intersection; x = 0,2 A = F[8x -2x^2 -4x]dx [0,2] A = F[(4x - 2x^2)][0,2] A = 2x^2 - (2x^3)/3 [0,2] A = 8/3
Find the area bounded by the two curves. y = 3 - x, y = 4 -(x-1)^2
Pts of intersection; x = 0,3 A = F[4 -(x-1)^2 - (3 -x)]dx [0,3] A = (3x^2)/2 + (x^3)/3 [0,3] A = 9/2
Find the area bounded by the two curves. y = 4x - x^2, y = 8x -2x^2
Pts of intersection; x = 0,4 A = F[8x -2x^2 -(4x - x^2)]dx [0,4] A = F[(4x - x^2)][0,4] A = 2x^2 - (x^3)/3 [0,4] A = 32/3
Find the area bounded by the two curves. y =x^2-6x+10, y = -x^2+6x-6
Pts of intersection; x = 2,4 A = F(-x^2+6x-6)dx - F(x^2-6x+10)dx [2,4] A = (-2x^3)/3 + 6x^2 -16x [2,4] A = 8/3
Find the right-sided Riemann Sum with n = 3, ∑cos(x^2) from [2, 5]
Right S3 = (1)[cos(3^2) + cos(4^2) + cos(5^2)] Right S3 = (1)[cos(9) + cos(16) + cos(25)] Right S3 = -.878
Find the right-sided Riemann Sum with n = 4, ∑sin(x^2 ) from [0, 4]
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
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
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
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
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) = x^2 + 1 and the x-axis from x = 3 to x =7
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 = 4 to approximate the area between the curve f(x) = x^3 -x and the x-axis from x = 3 to x =7
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^3 -x^2 and the x-axis from x = 3 to x =4
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 trapezoidal approximation for the integral of (sinx)^2dx from [0, 1] with n = 4 to three decimal places.
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.
Trapezoid = (1/2)(3/4)[(e^(0)) +2(e^(3/4)) +2(e^(6/4)) +2(e^(9/4)) +(e^(3))] Trapezoid = 10.972
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) =
f(x) = (x -3)(x -4)/(x-4) so f(4) = 1
