# Answer solution solution the average value of a

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Answer: Solution: SOLUTION The average value of a function f on the interval [ a , b ] is f ave = 1 b - a Z b a f ( x ) dx . So the average value of f ( x ) = cos ( 16 x ) on the interval [ 0 , π / 2 ] is f ave = 1 ( π / 2 - 0 ) Z π / 2 0 cos ( 16 x ) dx . And we have f ave = 1 ( π / 2 - 0 ) Z π / 2 0 cos ( 16 x ) dx = 2 π h sin ( 16 x ) 16 i π / 2 0 = 2 π h sin ( 8 π ) - sin ( 0 ) 16 i = 2 π 0 - 0 16 = 0 Answer(s) submitted: 0 (correct) Correct Answers: 0 20. (1 point) Find the average value of the function f ( t ) = - 2 te - t 2 for the interval [ 0 , 8 ] . Answer: Solution: SOLUTION The average value of a function f on the interval [ a , b ] is f ave = 1 b - a Z b a f ( t ) dt . So the average value of f ( t ) = - 2 te - t 2 on the interval [ 0 , 8 ] is f ave = 1 8 - 0 Z 8 0 - 2 te - t 2 dt . We use a substitution to evaluate the integral Z 8 0 - 2 te - t 2 dt . We let u = - t 2 , so that du = - 2 t dt . Changing the limits to match the new variable, the lower limit of x = 0 corresponds to u = 0, and the upper limit of x = 8 cor- responds to u = - 64. Therefore Z 8 0 - 2 te - t 2 dt = Z - 64 0 e u du = e u - 64 0 = e - 64 - 1 = - ( 1 - e - 64 ) And the average value of f is f ave = 1 8 - 0 Z 8 0 - 2 te - t 2 dt = - 1 8 1 - e - 64 Answer(s) submitted: -(1/8)(1-eˆ-64) (correct) Correct Answers: [eˆ(-8ˆ2)-1]/8 21. (1 point) In a certain city the temperature (in degrees Fahrenheit) t hours after 9am was approximated by the function T ( t ) = 30 + 5sin π t 12 Determine the temperature at 9 am. Determine the temperature at 3 pm. Find the average temperature during the period from 9 am to 9 pm. Solution: SOLUTION The temperature at 9 am is T ( 0 ) = 30. The temperature at 3 pm is T ( 6 ) = 30 + 5sin ( π 2 ) = 30 + 5 = 35. The average temperature during the period from 9 am to 9 pm is 1 12 Z 12 0 30 + 5sin π t 12 dt To evaluate this integral we perform the substitution u = π t 12 . Then du = π 12 dt , so 1 12 Z 12 0 30 + 5sin π t 12 dt = 1 12 Z π 0 ( 30 + 5sin u ) 12 π du = 1 π Z π 0 ( 30 + 5sin u ) du = 1 π 30 u - 5cos u π 0 = 1 π [ 30 π - 5cos ( π ) - ( - 5cos0 )] = 1 π [ 30 π + 2 ( 5 )] = 30 + 10 π = 33 . 1831 F Answer(s) submitted: 30 35 (score 0.666666686534882) Correct Answers: 30 35 33.1830988618379 6
22. (1 point) Using the method of u -substitution, Z 4 3 ( 5 x - 5 ) 5 dx = Z b a f ( u ) du where u = (enter a function of x ) du = dx (enter a function of x ) a = (enter a number) b = (enter a number) f ( u ) = (enter a function of u ). The value of the original integral is . Note: You can earn full credit if the last answer box is correct and all other answer boxes are either blank or correct. Solution: SOLUTION u = 5 x - 5. du = 5 dx a = 5 · 3 - 5 = 10 b = 5 · 4 - 5 = 15 f ( u ) = 1 5 u 5 . The value of the original integral is: Z 4 3 ( 5 x - 5 ) 5 dx = Z 15 10 1 5 u 5 du = 1 5 u 6 6 15 10 = 1 30 h ( 15 ) 6 - ( 10 ) 6 i = 2078125 6 Answer(s) submitted: 5x-5 5 10 15 uˆ5/5 (score 0.75) Correct Answers: 5 * x - 5 5 10 15 u**5 / 5 346354.166666667 23. (1 point) If f is continuous and Z 4 0 f ( x ) dx = 18 , evalu- ate Z 2 0 f ( 2 x ) dx . Answer: Solution: Let u = 2 x . Then du = 2 dx so Z 2 0 f ( 2 x ) dx = Z 4 0 f ( u ) 1 2 du = 1 2 Z 4 0 f ( u ) du = 18 2 Answer(s) submitted: 18/2 (correct) Correct Answers: 18/2 24. (1 point) If f is continuous and Z 9 0 f ( x ) dx = 3 , find Z 3 0 x f ( x 2 ) dx . Solution: Let u = x 2 . Then du = 2 xdx so Z 3 0 x f ( x 2 ) dx = Z 9 0 f ( u ) 1 2 du = 1 2 Z 9 0 f ( u ) du = 3 2 Answer(s) submitted: 3/2 (correct) Correct Answers: 3/2 Generated by c WeBWorK, , Mathematical Association of America 7
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