# Let t n be the n th trapezoidal approximation to j n

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Chapter 6 / Exercise 72
Applied Calculus
Berresford/Rockett
Expert Verified
Let T N be the N th trapezoidal approximation to J N . Calculate T 4 and show that T 4 approximates J to three decimal places.
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Chapter 6 / Exercise 72
Applied Calculus
Berresford/Rockett
Expert Verified
948 C H A P T E R 7 TECHNIQUES OF INTEGRATION SOLUTION T 4 is the 4 th trapezoidal approximation to J 4 D R 4 0 e x 2 dx . We divide the interval OE0; 4 into four subintervals, with endpoints 0 , 1 , 2 , 3 , and 4 . Then T 4 D 1 2 1 h e 0 2 C 2e 1 2 C 2e 2 2 C 2e 3 2 C e 4 2 i 0:8863185 We have T 4 J 0:8863185 p 2 0:8863185 0:8862269 0:0000916 53. Let f .x/ D sin .x 2 / and I D Z 1 0 f .x/ dx . (a) Check that f 00 .x/ D 2 cos .x 2 / 4x 2 sin .x 2 / . Then show that j f 00 .x/ j 6 for x 2 OE0; 1 . Hint: Note that j 2 cos .x 2 / j 2 and j 4x 2 sin .x 2 / j 4 for x 2 OE0; 1 . (b) Show that Error( M N / is at most 1 4N 2 . (c) Find an N such that j I M N j 10 3 . SOLUTION (a) Taking derivatives, we get f 0 .x/ D 2x cos .x 2 / f 00 .x/ D 2x. sin .x 2 / 2x/ C 2 cos .x 2 / D 2 cos .x 2 / 4x 2 sin .x 2 /: On the interval OE0; 1 ; j f 00 .x/ j D j 2 cos .x 2 / 4x 2 sin .x 2 / j j 2 cos .x 2 / j C j 4x 2 sin .x 2 / j 2 C 4 D 6: (b) Using K 2 D 6 , we get Error .M N / K 2 .1 0/ 3 24N 2 D 6 24N 2 D 1 4N 2 : (c) To ensure that M N has error at most 10 3 , we must find N such that 1 4N 2 1 10 3 : This gives us N 2 10 3 4 D 250 ) N p 250 15:81: Thus let N D 16: 54. The error bound for M N is proportional to 1=N 2 , so the error bound decreases by 1 4 if N is increased to 2N . Compute the actual error in M N for R 0 sin x dx for N D 4 , 8, 16, 32, and 64. Does the actual error seem to decrease by 1 4 as N is doubled? SOLUTION The exact value of the integral is Z 0 sin x dx D cos x ˇ ˇ ˇ ˇ 0 D . 1/ .1/ D 2: To compute M 4 , we have x D . 0/=4 D =4 , and midpoints =8; 3 =8; 5 =8; 7 =8: With this data, we get M 4 D 4 sin 8 C sin 3 8 C sin 5 8 C sin 7 8 2:052344: The values for M 8 ; M 16 ; M 32 ; and M 64 are computed similarly: M 8 D 8 sin 16 C sin 3 16 C C sin 15 16 2:012909 I M 16 D 16 sin 32 C sin 3 32 C C sin 31 32 2:0032164 I M 32 D 32 sin 64 C sin 3 64 C C sin 63 64 2:00080342 I
S E C T I O N 7.8 Numerical Integration 949 M 64 D 64 sin 128 C sin 3 128 C C sin 127 128 2:00020081: Now we can compute the actual errors for each N : Error .M 4 / D j 2 2:052344 j D 0:052344 Error .M 8 / D j 2 2:012909 j D 0:012909 Error .M 16 / D j 2 2:0032164 j D 0:0032164 Error .M 32 / D j 2 2:00080342 j D 0:00080342 Error .M 64 / D j 2 2:00020081 j D 0:00020081 The actual error does in fact decrease by about 1=4 each time N is doubled. 55. Observe that the error bound for T N (which has 12 in the denominator) is twice as large as the error bound for M N (which has 24 in the denominator). Compute the actual error in T N for R 0 sin x dx for N D 4 , 8, 16, 32, and 64 and compare with the calculations of Exercise 54. Does the actual error in T N seem to be roughly twice as large as the error in M N in this case? SOLUTION The exact value of the integral is Z 0 sin x dx D cos x ˇ ˇ ˇ ˇ 0 D . 1/ .1/ D 2: To compute T 4 , we have x D . 0/=4 D =4 , and endpoints 0; =4; 2 =4; 3 =4; : With this data, we get T 4 D 1 2 4 sin .0/ C 2 sin 4 C 2 sin 2 4 C 2 sin 3 4 C sin . / 1:896119: The values for T 8 ; T 16 ; T 32 ; and T 64 are computed similarly: T 8 D 1 2 8 sin .0/ C 2 sin 8 C 2 sin 2 8 C C 2 sin 7 8 C sin . / 1:974232 I T 16 D 1 2 16 sin .0/ C 2 sin 16 C 2 sin 2 16 C C 2 sin 15 16 C sin . / 1:993570 I T 32 D 1 2 32 sin .0/ C 2 sin 32 C 2 sin 2 32 C C 2 sin 31 32 C sin . / 1:998393 I T 64 D 1 2 64 sin .0/ C 2 sin 64 C 2 sin 2 64 C C 2 sin 63 64 C sin . / 1:999598: