Inozemtseva_MAT_266_Online_B_Fall_2023.bairwin.Section_6.5

Bryan Irwin Inozemtseva MAT 266 Online B Fall 2023 Assignment Section 6.5 due 10/26/2023 at 11:59pm MST Problem 1. (1 point) Given the following integral and value of n, approximate the fol- lowing integral using the methods indicated (round your answers to six decimal places): Z 1 0 e − 5 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 ∆ x = b − a n = 1 4 = 0 . 25, and x k = a + k ∆ 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 = ∆ 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 − 5 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 − 5 / 16 + 2 e − 5 / 4 + 2 e − 5 ( 9 ) / 16 + e − 5 i ≈ 1 8 h 1 + 1 . 46323 + 0 . 57301 + 0 . 120109 + 0 . 00673795 i ≈ 0 . 395386 (b) The midpoint Rule with n = 4 is M 4 = ∆ 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 − 5 x 2 M 4 = 1 4 h f ( 1 8 ) + f ( 3 8 ) + f ( 5 8 ) + f ( 7 8 ) i = 1 4 h e − 5 / 64 + e − 5 ( 9 ) / 64 + e − 5 ( 25 ) / 64 + e − 5 ( 49 ) / 64 i ≈ 1 4 h 0 . 924849 + 0 . 495036 + 0 . 14183 + 0 . 0217504 i ≈ 0 . 395866 (c) Simpson’s Rule with n = 4 is S 4 = ∆ 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 − 5 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 − 5 / 16 + 2 e − 5 / 4 + 4 e − 5 ( 9 ) / 16 + e − 5 i ≈ 1 12 h 1 + 2 . 92646 + 0 . 57301 + 0 . 240219 + 0 . 00673795 i ≈ 0 . 395536 Answer(s) submitted: • 0.395386 • 0.395866 • 0.395535 (correct) Correct Answers: • 0.395386016800421 • 0.395866307143549 • 0.395535727340605 1

Problem 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 during those 5 seconds. t (s) 0 0 . 5 1 1 . 5 2 2 . 5 3 3 . 5 4 4 . 5 5 v (ft/s) 0 2 . 55 6 . 35 6 . 9 7 . 8 9 . 25 9 . 8 10 . 5 10 . 6 10 . 7 10 . 7 Estimated distance: Solution: SOLUTION We have ∆ t = 0 . 5 and n = 10 sub-intervals in the given chart. Simpson’s rule with n = 10 is S 10 = ∆ 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 ( 6 . 35 ) + 4 ( 6 . 9 ) + 2 ( 7 . 8 ) + 4 ( 9 . 25 ) + 2 ( 9 . 8 )+ 4 ( 10 . 5 )+ 2 ( 10 . 6 )+ 4 ( 10 . 7 )+ 10 . 7 i . S 10 ≈ 39 . 9. Answer(s) submitted: • 39.9 (correct) Correct Answers: • 39.9 Problem 3. (1 point) Consider the integral approximation T 20 of R 10 0 2 e − x 4 dx . Does T 20 overestimate or underestimate the exact value? • A. underestimates • B. overestimates Find the error bound for T 20 without calculating T N using the re- sult 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 2 e − x 4 on the interval [ a , b ] . Error ( T 20 ) ≤ Solution: Solution: Let f ( x ) = 2 e − x 4 . Then f ′′ ( x ) = 0 . 125 e − x 4 . f ′′ ( x ) > 0 on [0, 10], so f(x) is concave up, and T 20 overestimates the integral (drawing a picture may help to understand why this is). Since | f ′′ ( x ) | = 0 . 125 e − x 4 has its maximum value on [0, 10] at x = 0, we can take K 2 = f ′′ ( 0 ) = 0 . 125, and Error ( T 20 ) ≤ K 2 ( 10 − 0 ) 3 12 N 2 = 0 . 125 ( 10 ) 3 12 ( 20 ) 2 = 0 . 0260417. Answer(s) submitted: • B • 0.026041 (correct) Correct Answers: • B • 0.0260417 2

The midpoints are ¯ x k = x k − 1 + x k 2 , k = 1 , 2 ,..., 6. The Midpoint Sum with n = 6 is M 6 = ∆ x h f ( ¯ x 1 )+ f ( ¯ x 2 )+ f ( ¯ x 3 )+ f ( ¯ x 4 )+ f ( ¯ x 5 )+ f ( ¯ x 6 ) i = 2 h f ( 1 )+ f ( 3 )+ f ( 5 )+ f ( 7 )+ f ( 9 )+ f ( 11 ) i ≈ 2 h 7 . 0 + 6 . 8 + 6 . 3 + 5 . 6 + 4 . 8 + 3 . 6 i ≈ 68 . 1 Answer(s) submitted: • 71.8 • Overestimate • 63.6 • Underestimate • 67.6 (correct) Correct Answers: • 71.7778 • Overestimate • 63.7778 • Underestimate • 68.1111 4

Problem 5. (1 point) Use four rectangles to find an estimate of each type for the area under the given graph of f from x = 1 to x = 9 . 1. Take the sample points from the left-endpoints. Answer: L 4 = 2. Is your estimate L 4 an underestimate or overestimate of the true area? • Choose one • Underestimate • Overestimate 3. Take the sample points from the right-endpoints. Answer: R 4 = 4. Is your estimate R 4 an underestimate or overestimate of the true area? • Choose one • Underestimate • Overestimate 5. Use the Trapezoid Rule with n = 4 . Answer: T 4 = 6. Is your estimate T 4 an underestimate or overestimate of the true area? • Choose one • Underestimate • Overestimate Note: You can click on the graph to enlarge the image. Solution: SOLUTION We have a = 1 , b = 9 and n = 4. So ∆ x = b − a n = 8 4 = 2, and x k = a + k ∆ x so x 0 = 1, x 1 = 3 , x 2 = 5, x 3 = 7, and x 4 = 9. 1. The function together with the rectangles for the Left Rule is plotted below. We use the picture to estimate the values of the function at the required values of x . The left Rule with n = 4 is L 4 = ∆ x h f ( x 0 )+ f ( x 1 )+ f ( x 2 )+ f ( x 3 ) i = 2 h f ( 1 )+ f ( 3 )+ f ( 5 )+ f ( 7 ) i ≈ 2 h 4 + 4 . 3 + 4 . 7 + 5 . 4 i ≈ 36 . 8 2. Because the function is concave up, the Left Sum is an underestimate. (This is also apparent from the picture). 3. The function together with the rectangles for the Right Rule is plotted below. We use the picture to estimate the values of the function at the required values of x . The Right Sum with n = 6 is R 4 = ∆ x h f ( x 1 )+ f ( x 2 )+ f ( x 3 )+ f ( x 4 ) i = 2 h f ( 3 )+ f ( 5 )+ f ( 7 )+ f ( 9 ) i ≈ 2 h 4 . 3 + 4 . 7 + 5 . 4 + 6 . 3 i ≈ 41 . 4 4. Because the function is concave up, the Right Sum is an overestimate. (This is also apparent from the picture). 5. The function together with the trapezoids is plotted below. We use the picture to estimate the values of the function at the required values of x . 5

Problem 6. (1 point) The graph of a function f is given below. Estimate R 8 0 f ( x ) dx us- ing four subintervals with (a) right endpoints, (b) left endpoints, and (c) midpoints. (a) R 8 0 f ( x ) dx ≈ (b) R 8 0 f ( x ) dx ≈ (c) R 8 0 f ( x ) dx ≈ Solution: SOLUTION We have a = 0 , b = 8 and n = 4. So ∆ x = b − a n = 8 4 = 2, and x k = a + k ∆ x so x 0 = 0, x 1 = 2 , x 2 = 4, x 3 = 6, and x 4 = 8. (a) The function together with the rectangles for the Right Rule is plotted below. We use the picture to estimate the values of the function at the required values of x . R 8 0 f ( x ) dx ≈ ∆ x h f ( x 1 )+ f ( x 2 )+ f ( x 3 )+ f ( x 4 ) i = 2 h f ( 2 )+ f ( 4 )+ f ( 6 )+ f ( 8 ) i = 2 h 1 + 2 +( − 2 )+ 1 i = 4 (b) The function together with the rectangles for the Left Rule is plotted below. We use the picture to estimate the values of the function at the required values of x . R 8 0 f ( x ) dx ≈ ∆ x h f ( x 0 )+ f ( x 1 )+ f ( x 2 )+ f ( x 3 ) i = 2 h f ( 0 )+ f ( 2 )+ f ( 4 )+ f ( 6 ) i = 2 h 2 + 1 + 2 +( − 2 ) i = 6 (c) The function together with the rectangles for the midpoint Rule is plotted below. We use the picture to estimate the values of the function at the required values of x . The midpoints are ¯ x 1 = 1 , ¯ x 2 = 3 , ¯ x 3 = 5 , ¯ x 4 = 7. R 8 0 f ( x ) dx ≈ ∆ x h f ( ¯ x 1 )+ f ( ¯ x 2 )+ f ( ¯ x 3 )+ f ( ¯ x 4 ) i = 2 h f ( 1 )+ f ( 3 )+ f ( 5 )+ f ( 7 ) i = 2 h 3 + 2 + 1 +( − 1 ) i = 10 Answer(s) submitted: • 4 • 6 • 10 (correct) Correct Answers: • 2*(1+2+(-2)+1) • 2*(2+1+2+(-2)) • 2*(3+2+1+(-1)) 7

Problem 7. (1 point) Given the following graph of the function y = f ( x ) and n = 6 , an- swer the following questions about the area under the curve from x = 0 to x = 6. 1. Use the Trapezoidal Rule to estimate the area. Answer: T 6 = 2. Use Simpson’s Rule to estimate the area. Answer: S 6 = Note: You can click on the graph to enlarge the image. Solution: SOLUTION We have a = 0 , b = 6 and n = 6. So ∆ x = b − a n = 6 6 = 1, and x k = a + k ∆ x so x 0 = 0, x 1 = 1 , x 2 = 2, x 3 = 3, x 4 = 4, x 5 = 5, and x 6 = 6. We use the graph of the function to estimate the values of the func- tion at the required values of x : 1. T 6 = ∆ x 2 h f ( x 0 )+ 2 f ( x 1 )+ 2 f ( x 2 )+ 2 f ( x 3 )+ 2 f ( x 4 )+ 2 f ( x 5 )+ f ( x 6 ) i = 1 2 h f ( 0 )+ 2 f ( 1 )+ 2 f ( 2 )+ 2 f ( 3 )+ 2 f ( 4 )+ 2 f ( 5 )+ f ( 6 ) i ≈ 1 2 h 5 + 2 · 6 + 2 · 8 + 2 · 6 + 2 · 5 . 5 + 2 · 5 + 7 i = 36 . 5 2. S 6 = ∆ x 3 h f ( x 0 )+ 4 f ( x 1 )+ 3 f ( x 2 )+ 4 f ( x 3 )+ 3 f ( x 4 )+ 4 f ( x 5 )+ f ( x 6 ) i = 1 3 h f ( 0 )+ 4 f ( 1 )+ 2 f ( 2 )+ 4 f ( 3 )+ 2 f ( 4 )+ 4 f ( 5 )+ f ( 6 ) i ≈ 1 3 h 5 + 4 · 6 + 2 · 8 + 4 · 6 + 2 · 5 . 5 + 4 · 5 + 7 i = 35 . 6667 Answer(s) submitted: • 36.5 • 35.6667 (correct) Correct Answers: • 1/2*(2*1+2*3+2*1+2*0.5+2)+5*6 • 1/3*(4*1+2*3+4*1+2*0.5+2)+5*6 8

Problem 9. (1 point) The left, right, Trapezoidal, and Midpoint Rule approximations were used to estimate R 2 0 f ( x ) dx , where f is the function whose graph is shown below. The estimates were 0.7811, 0.8675, 0.8632, and 0.9540, and the same number of subintervals were used in each case. (a) Which rule produced which estimate? ? 1. Right-hand estimate ? 2. Left-hand estimate ? 3. Trapezoidal Rule estimate ? 4. Midpoint Rule estimate (b) Between which two approximations does the true value of R 2 0 f ( x ) dx lie? • A. 0 . 7811 < R 2 0 f ( x ) dx < 0 . 8632 • B. 0 . 8675 < R 2 0 f ( x ) dx < 0 . 9540 • C. No conclusions can be drawn. • D. 0 . 8632 < R 2 0 f ( x ) dx < 0 . 8675 Solution: SOLUTION (a) Let L , R , T and M be the Left, Right, Trapezoidal, and Midpoint Rule approximations, respectively. Since the function is concave up, we have M ≤ Z 2 0 f ( x ) dx ≤ T Since the function is decreasing, we have R ≤ Z 2 0 f ( x ) dx ≤ L The above inequalities also imply that T is more accurate than L and M is more accurate than R . Therefore: R ≤ M ≤ Z 2 0 f ( x ) dx ≤ T ≤ L Ordering the given values, gives: 0 . 7811 < 0 . 8632 ≤ Z 2 0 f ( x ) dx ≤ 0 . 8675 < 0 . 9540 Thus 1. Left-hand estimate = 0.9540 2. Right-hand estimate = 0.7811 3. Trapezoidal Rule estimate = 0.8675 4. Midpoint Rule estimate = 0.8632 (b) Since M ≤ R 2 0 f ( x ) dx ≤ T , we have that 0 . 8632 < Z 2 0 f ( x ) dx < 0 . 8975 Answer(s) submitted: • 7811 • 9540 • 8675 • 8632 • D (correct) Correct Answers: • 7811 • 9540 • 8675 • 8632 • D 10

Problem 10. (1 point) Consider the four functions shown below. On the first two, an approximation for R b a f ( x ) dx is shown. 1. 2. 3. 4. (Click on any graph to get a larger version.) 1. For graph number 1, Which integration method is shown? • A. left rule • B. midpoint rule • C. right rule • D. trapezoid rule Is this method an over- or underestimate? • A. under • B. over 2. For graph number 2, Which integration method is shown? • A. trapezoid rule • B. midpoint rule • C. left rule • D. right rule Is this method an over- or underestimate? • A. under • B. over 3. On a copy of graph number 3, sketch an estimate with n = 2 subdivisions using the midpoint rule. Is this method an over- or underestimate? • A. over • B. under 4. On a copy of graph number 4, sketch an estimate with n = 2 subdivisions using the trapezoid rule. Is this method an over- or underestimate? • A. over • B. under Solution: SOLUTION We can see by inspection or by looking at how the functions in- crease/decrease or their concavity that graph 1 shows the left rule and gives an under estimate. Similarly, graph 2 shows the left rule and gives an under estimate. The graphs showing the numerical estimates for graphs 3 and 4 are shown below. 3. 4. (Click on any graph to get a larger version.) Then graph 3, showing the midpoint rule, gives an under estimate. Then graph 4, showing the trapezoid rule, gives an over estimate. Answer(s) submitted: • A • A • C • A • B • A (correct) Correct Answers: • A • A • C • A • B • A 11