# 6.5_MAT_266_ONLINE_B_Spring_2020..Section_6.5.pdf - Larena...

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Larena Whitney-krug Boerner MAT 266 ONLINE B Spring 2020 Assignment Section 6.5 due 03/29/2020 at 11:59pm MST 1. (1 point) Given the following integral and value of n, approximate the following integral using the methods indicated (round your answers to six decimal places): Z 1 0 e - 2 x 2 dx , n = 4 (a) Trapezoidal Rule (b) Midpoint Rule (c) Simpson’s Rule Solution: SOLUTION For each part below, a = 0 , b = 1 and n = 4. So Δ x = b - a n = 1 4 = 0 . 25, and x k = a + k Δ x so x 0 = 0, x 1 = 1 4 , x 2 = 1 2 , x 3 = 3 4 , and x 4 = 1. (a) The trapezoidal Rule with n = 4 is T 4 = Δ x 2 h f ( x 0 )+ 2 f ( x 1 )+ 2 f ( x 2 )+ 2 f ( x 3 )+ f ( x 4 ) i . So with f ( x ) = e - 2 x 2 T 4 = 1 / 4 2 h f ( 0 )+ 2 f ( 1 4 ) + 2 f ( 1 2 ) + 2 f ( 3 4 ) + f ( 1 ) i = 1 8 h 1 + 2 e - 2 / 16 + 2 e - 2 / 4 + 2 e - 2 ( 9 ) / 16 + e - 2 i 1 8 h 1 + 1 . 76499 + 1 . 21306 + 0 . 649305 + 0 . 135335 i 0 . 595336 (b) The midpoint Rule with n = 4 is M 4 = Δ x h f ( x 1 )+ f ( x 2 )+ f ( x 3 )+ f ( x 4 ) i , where x k = 1 2 ( x k - 1 + x k ) . So x 1 = 1 8 , x 2 = 3 8 , x 3 = 5 8 , and x 4 = 7 8 . So with f ( x ) = e - 2 x 2 M 4 = 1 4 h f ( 1 8 ) + f ( 3 8 ) + f ( 5 8 ) + f ( 7 8 ) i = 1 4 h e - 2 / 64 + e - 2 ( 9 ) / 64 + e - 2 ( 25 ) / 64 + e - 2 ( 49 ) / 64 i 1 4 h 0 . 969233 + 0 . 75484 + 0 . 457833 + 0 . 216265 i 0 . 599543 (c) Simpson’s Rule with n = 4 is S 4 = Δ x 3 h f ( x 0 )+ 4 f ( x 1 )+ 2 f ( x 2 )+ 4 f ( x 3 )+ f ( x 4 ) i . So with f ( x ) = e - 2 x 2 S 4 = 1 / 4 3 h f ( 0 )+ 4 f ( 1 4 ) + 2 f ( 1 2 ) + 4 f ( 3 4 ) + f ( 1 ) i = 1 12 h 1 + 4 e - 2 / 16 + 2 e - 2 / 4 + 4 e - 2 ( 9 ) / 16 + e - 2 i 1 12 h 1 + 3 . 52999 + 1 . 21306 + 1 . 29861 + 0 . 135335 i 0 . 598083 Correct Answers: 0.595336917818471 0.599542841266713 0.598082840202805 2. (1 point) A radar gun was used to record the speed of a runner during the first 5 seconds of a race (see table). Use Simpson’s rule to estimate the distance the runner covered dur- ing those 5 seconds. t (s) 0 0 . 5 1 1 . 5 2 2 . 5 3 3 . 5 4 4 . 5 v (ft/s) 0 2 . 55 3 . 15 4 . 15 8 . 4 8 . 55 9 . 6 9 . 8 10 . 45 10 . 75 Estimated distance: Solution: SOLUTION We have Δ t = 0 . 5 and n = 10 sub-intervals in the given chart. Simpson’s rule with n = 10 is S 10 = Δ t 3 h f ( t 0 ) + 4 f ( t 1 ) + 2 f ( t 2 )+ 4 f ( t 3 )+ 2 f ( t 4 )+ 4 f ( t 5 )+ 2 f ( t 6 )+ 4 f ( t 7 )+ 2 f ( t 8 )+ 4 f ( t 9 )+ f ( t 10 ) i . So for the above S 10 = 0 . 5 3 h f ( 0 )+ 4 f ( 0 . 5 )+ 2 f ( 1 )+ 4 f ( 1 . 5 )+ 2 f ( 2 )+ 4 f ( 2 . 5 )+ 2 f ( 3 )+ 4 f ( 3 . 5 )+ 2 f ( 4 )+ 4 f ( 4 . 5 )+ f ( 5 ) i . S 10 = 1 6 h 0 + 4 ( 2 . 55 ) + 2 ( 3 . 15 ) + 4 ( 4 . 15 ) + 2 ( 8 . 4 ) + 4 ( 8 . 55 ) + 2 ( 9 . 6 )+ 4 ( 9 . 8 )+ 2 ( 10 . 45 )+ 4 ( 10 . 75 )+ 10 . 75 i . S 10 36 . 1917. Correct Answers: 36.1916666666667 3. (1 point) Consider the integral approximation T 20 of R 4 0 4 e - x 4 dx . Does T 20 overestimate or underestimate the exact value? A. overestimates B. underestimates 1
Find the error bound for T 20 without calculating T N using the result that Error ( T N ) M ( b - a ) 3 12 N 2 , where M is the least upper bound for all absolute values of the second derivatives of the function 4 e - x 4 on the interval [ a , b ] . Error ( T 20 ) Solution: Solution: Let f ( x ) = 4 e - x 4 . Then f 00 ( x ) = 0 . 25 e - x 4 . f 00 ( x ) > 0 on [0, 4], so f(x) is concave up, and T 20 overes- timates the integral (drawing a picture may help to understand why this is). Since | f 00 ( x ) | = 0 . 25 e - x 4 has its maximum value on [0, 4] at x = 0, we can take K 2 = f 00 ( 0 ) = 0 . 25, and Error ( T 20 ) K 2 ( 4 - 0 ) 3 12 N 2 = 0 . 25 ( 4 ) 3 12 ( 20 ) 2 = 0 . 00333333. Correct Answers: A 0.00333333 4. (1 point) Use six rectangles to find an estimate of each type for the area under the given graph of f from x = 0 to x = 12 . 1. Take the sample points from the left-endpoints. Answer: L 6 = 2. Is your estimate L 6 an underestimate or overestimate of the true area?