HW Section 5.3 - 1 11/11points|.3.002 x Let g(x)= f(t)dt...

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Unformatted text preview: 1. 11/11 points | Previous AnswersSCalcET8 5.3.002. x Let g(x) = f(t) dt, where f is the function whose graph is shown. 0 (a) Evaluate g(x) for x = 0, 5, 10, 15, 20, 25, and 30. g(0) = 0 g(5) = 12.5 g(10) = 0 0 g(25) = 37.5 ­12.5 0 g(30) = 100 12.5 g(15) = ­12.5 g(20) = 0 0 37.5 100 (b) Estimate g(35). (Use the midpoint to get the most precise estimate.) g(35) = 150 155 (c) Where does g have a maximum and a minimum value? minimum x = 15 15 maximum x = 35 35 (d) Sketch a rough graph of g. Solution or Explanation Click to View Solution 2. 8/8 points | Previous AnswersSCalcET8 5.3.003. x Let g(x) = f(t) dt, where f is the function whose graph is shown. 0 (a) Evaluate g(0), g(4), g(8), g(12), and g(24). g(0) = 0 g(4) = 32 g(8) = 80 g(12) = 112 g(24) = 48 0 32 80 112 48 (b) On what interval is g increasing? (Enter your answer using interval notation.) $$(0,12) (c) Where does g have a maximum value? x = 12 12 (d) Sketch a rough graph of g. Solution or Explanation Click to View Solution 3. 1/1 points | Previous AnswersSCalcET8 5.3.009. Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. s g(s) = 4 (t − t8)3dt g'(s) = $$(s−s8)3 Solution or Explanation s f(t) = (t − t8)3 and g(s) = 4 8 3 (t − t8)3dt, so by FTC1, g'(s) = f(s) = (s − s ) . 4. 1/1 points | Previous AnswersSCalcET8 5.3.007. Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. x g(x) = t4 + t6 dt 0 g'(x) = $$√x4+x6 Solution or Explanation f(t) = x t4 + t6 and g(x) = 0 t4 + t6 dt , so by FTC1, g'(x) = f(x) = x4 + x6 . 5. 1/1 points | Previous AnswersSCalcET8 5.3.013. Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. ex h(x) = 4 ln(t) dt 1 h'(x) = $$4exx Solution or Explanation du dh dh du Let u = ex . Then = ex . Also, = , so dx dx du dx x h'(x) = u e d d du du 4 ln(t) dt = 4 ln(t) dt · = 4 ln(u) = (4 ln(ex )) · ex = 4xex . dx 1 du 1 dx dx 6. 1/1 points | Previous AnswersSCalcET8 5.3.019. Evaluate the integral. 3 1 (x2 + 2x − 3) dx $$323 Solution or Explanation 3 1 3 1 3 1 5 32 x + x2 − 3x = (9 + 9 − 9) − + 1 − 3 = 9 − − = 3 3 3 3 1 (x2 + 2x − 3) dx = 7. 1/1 points | Previous AnswersSCalcET8 5.3.025. Evaluate the integral. π π/6 sin θ dθ $$1+√32 Solution or Explanation π π/6 sin θ dθ = −cos θ π π/6 = −cos π − −cos π = −(−1) − (− 6 3 /2) = 1 + 3 /2 8. 1/1 points | Previous AnswersSCalcET8 5.3.027. Evaluate the integral. 1 0 (u + 3)(u − 4) du $$−736 Solution or Explanation 1 0 (u + 3)(u − 4) du = 1 0 (u2 − u − 12) du = 1 1 3 1 1 1 73 u − u2 − 12u = − − 12 − 0 = − 3 2 3 2 6 0 9. 1/1 points | Previous AnswersSCalcET8 5.3.029. Evaluate the integral. 4 6 + x2 x 1 dx $$1225 Solution or Explanation 4 6 + x2 4 dx = x 1 1 6 x + = 12x1/2 + x2 x 4 dx = 1 (6x−1/2 + x3/2) dx 2 5/2 4 2 2 64 2 122 x = 12(2) + (32) − 12(1) + (1) = 24 + − 12 − = 5 5 5 5 5 5 1 10.1/1 points | Previous AnswersSCalcET8 5.3.031. Evaluate the integral. π/3 π/4 csc(t) cot(t) dt $$−23+3√23 Solution or Explanation π/3 π/4 csc(t) cot(t) dt = −csc(t) π/3 π/4 = −csc π 3 − −csc π 4 = − 2 3 3 − (− 2 ) = − 2 3 3 + 2 11.1/1 points | Previous AnswersSCalcET8 5.3.035. Evaluate the integral. 2 v3 + 4v5 v2 1 dv $$332 Solution or Explanation 2 v3 + 4v5 v2 1 2 dv = v + 4v3 dv = 1 2 1 2 33 v + v4 = 2 2 1 12.1/1 points | Previous AnswersSCalcET8 5.3.037. Evaluate the integral. 1 (7xe + 2ex ) dx 0 $$−2+2e+71+e Solution or Explanation 1 1 e + 1 7 7 (7xe + 2ex ) dx = 7x + 2ex = + 2e − (0 + 2) = + 2e − 2 e + 1 e + 1 e + 1 0 0 13.1/1 points | Previous AnswersSCalcET8 5.3.039. Evaluate the integral. √3 4 1/√3 1 + x 2 dx $$2π3 Solution or Explanation √3 4 2 1/√3 1 + x dx = 4 arctan(x) √3 1/√3 2 = 4 π − π = 4 π = π 3 3 6 6 14.1/1 points | Previous AnswersSCalcET8 5.3.043. Evaluate the integral. π f(x) dx where f(x) = 0 ­1 6 sin x if 0 ≤ x < π/2 7 cos x if π/2 ≤ x ≤ π ­1 Solution or Explanation Click to View Solution 15.1/1 points | Previous AnswersSCalcET8 5.3.044. Evaluate the integral. 4 f(x) dx where f(x) = −4 32/3 3 if −4 ≤ x ≤ 0 5 − x2 if 0 < x ≤ 4 32/3 Solution or Explanation Click to View Solution 16.1/1 points | Previous AnswersSCalcET8 5.3.050. Use a graph to give a rough estimate of the area of the region that lies beneath the given curve. Then find the exact area. y = x−3, 3 ≤ x ≤ 6 1/24 1/24 Solution or Explanation Click to View Solution 17.2/2 points | Previous AnswersSCalcET8 5.3.053. Evaluate the integral and interpret it as a difference of areas. 2 x3 dx −1 $$154 Illustrate with a sketch. Solution or Explanation 2 x3 dx = −1 1 4 2 1 15 x = 4 − = = 3.75 4 4 4 −1 18.1/1 points | Previous AnswersSCalcET8 5.3.055. What is wrong with the equation? 4 4 −2 7 x−3 dx = x = 288 −2 −3 −3 f(x) = x−3 is continuous on the interval [−3, 4] so FTC2 cannot be applied. f(x) = x−3 is not continuous on the interval [−3, 4] so FTC2 cannot be applied. f(x) = x−3 is not continuous at x = −3, so FTC2 cannot be applied. The lower limit is less than 0, so FTC2 cannot be applied. There is nothing wrong with the equation. Solution or Explanation f(x) = x−3 is not continuous on the interval [−3, 4], so FTC2 cannot be applied. In fact, f has an infinite discontinuity at x = 0, so 4 −3 x−3 dx does not exist. 19.1/1 points | Previous AnswersSCalcET8 5.3.057. What is wrong with the equation? π π/3 3 sec θ tan θ dθ = 3 sec θ π = −9 π/3 f(θ) = 3 sec θ is not continuous at θ = π/3 so FTC2 cannot be applied. f(θ) = 3 sec θ tan θ is not continuous on the interval [π/3, π] so FTC2 cannot be applied. There is nothing wrong with the equation. The lower limit is not equal to 0, so FTC2 cannot be applied. f(θ) = 3 tan θ is not continuous on the interval [π/3, π] so FTC2 cannot be applied. Solution or Explanation f(θ) = 3 sec θ tan θ is not continuous on the interval [π/3, π], so FTC2 cannot be applied. In fact, f has an infinite discontinuity at x = π/2, so π π/3 3 sec θ tan θ dθ does not exist. 20.1/1 points | Previous AnswersSCalcET8 5.3.059. Find the derivative of the function. 7x g(x) = 2x u2 − 1 u2 + 1 7x du Hint: f(u) du = 2x 0 7x f(u) du + 2x f(u) du 0 g'(x) = $$−2·4x2−14x2+1+7·49x2−149x2+1 Solution or Explanation Click to View Solution 21.1/1 points | Previous AnswersSCalcET8 5.3.061. Find the derivative of the function. x2 F(x) = x F '(x) = $$−ex4+2xex8 4 et dt Solution or Explanation x2 F(x) = x 4 et dt = 0 x 4 et dt + x2 0 4 et dt = − x 0 4 et dt + x2 0 4 et dt 4 4 2 4 8 d F'(x) = −ex + e(x ) · (x2) = −ex + 2xex dx 22.1/1 points | Previous AnswersSCalcET8 5.3.063. Find the derivative of the function. sin x y = ln(8 + 4v)dv cos x y'(x) = $$sin(x)ln(8+4cos(x))+cos(x)ln(8+4sin(x)) Solution or Explanation y = sin x 0 ln(8 + 4v)dv = cos x cos x y' = −ln(8 + 4 cos x) · sin x ln(8 + 4v)dv + 0 cos x ln(8 + 4v)dv = − 0 23.1/1 points | Previous AnswersSCalcET8 5.3.069. 7 If f(5) = 15, f ' is continuous, and 5 f '(x) dx = 18, what is the value of f(7)? 33 Solution or Explanation Click to View Solution 0 ln(8 + 4v)dv d d cos x + ln(8 + 4 sin x) · sin x = sin x ln(8 + 4 cos x) + cos x ln(8 + 4 sin x) dx dx f(7) = 33 sin x ln(8 + 4v)dv + 24.7/7 points | Previous AnswersSCalcET8 5.3.073. x Let g(x) = f(t) dt, where f is the function whose graph is shown. 0 (a) At what values of x do the local maximum and minimum values of g occur? xmin = 12 12 (smaller x­value) xmin = 28 28 (larger x­value) xmax = 3 xmax = 20 4 (smaller x­value) 20 (larger x­value) (b) Where does g attain its absolute maximum value? x = 36 36 (c) On what interval is g concave downward? (Enter your answer using interval notation.) $$(2,8)∪(16,24)∪(32,36) (d) Sketch the graph of g. Solution or Explanation Click to View Solution 25.7/7 points | Previous AnswersSCalcET8 5.3.074. x Let g(x) = f(t) dt, where f is the function whose graph is shown. 0 (a) At what values of x do the local maximum and minimum values of g occur? xmin = 4 4 (smaller x­value) xmin = 8 8 (larger x­value) xmax = 2 2 (smaller x­value) xmax = 6 6 (larger x­value) (b) Where does g attain its absolute maximum value? x = 2 2 (c) On what interval is g concave downward? (Enter your answer using interval notation.) $$(1,3)∪(5,7)∪(9,10) (d) Sketch the graph of g. Solution or Explanation Click to View Solution 26.1/1 points | Previous AnswersSCalcET8 5.3.075. Evaluate the limit by first recognizing the sum as a Riemann sum for a function defined on [0, 1]. n i 5 lim n → ∞ 2/3 i = 1 n6 i + 2 n 2/3 Solution or Explanation n i 5 i + 2 n n6 lim n → ∞ i = 1 n = lim n → ∞ i = 1 i 5 n5 + i n 1 1 − 0 = lim n n n → ∞ 1 1 6 1 1 1 2 x + x2 = + − 0 = = 6 2 6 2 3 0 n i = 1 1 i 5 i + = (x5 + x) dx n n 0 27.1/1 points | Previous AnswersSCalcET8 5.3.076. Evaluate the limit by first recognizing the sum as a Riemann sum for a function defined on [0, 16]. lim n → ∞ 128/3 16 + n 16 n 32 + n 128/3 Solution or Explanation Click to View Solution 48 + n + 16n n 28.7/8 points | Previous AnswersSCalcET8 5.3.082. Let 0 if x < 0 x if 0 ≤ x ≤ 1 f(x) = 2 − x if 1 < x ≤ 2 0 if x > 2 and x g(x) = 0 f(t) dt. (a) Find an expression for g(x) similar to the one for f(x). $$0 if x < 0 $$x22 if 0 ≤ x ≤ 1 g(x) = $$−x22+2x−1 if 1 < x ≤ 2 $$1 if x > 2 (b) Sketch the graph of f. Sketch the graph of g. (c) Where is f differentiable? (Enter your answer using interval notation.) $$(−∞,∞) Where is g differentiable? (Enter your answer using interval notation.) $$(−∞,∞) Solution or Explanation Click to View Solution 29.1/1 points | Previous AnswersSCalcET8 5.3.505.XP. Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. tan x y = t dt 5t + 9 y' = $$√5tan(x)+√tan(x)sec2(x) Solution or Explanation Click to View Solution 30.1/1 points | Previous AnswersSCalcET8 5.3.506.XP. Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. y = 2 2 − 3x u3 1 + u2 du y' = $$3(2−3x)31+(2−3x)2 Solution or Explanation Click to View Solution ...
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