MATH 180 FALL 1995 10:00 SECTION FONG

ASSIGNMENT #17

3.1, #9. We get upper and lower estimates for the speed at t = 5 by summing upper and lower estimates for the speed on the intervals between consecutive readings of the acceleration. The speed at t = 0 is 0, so these estimates are the left and the right sums

9.81 + 8.03 + 6.53 + 5.30 + 4.41 = 34.16 m/sec
8.03 + 6.53 + 5.30 + 4.41 + 3.61 = 27.96 m/sec

Their average is 31.06 m/sec. Since consecutive changes in acceleration are decreasing, the graph of acceleration against time is concave-up. The average of the left and right sums approximates the area under the graph by trapezoids. So the new estimate is greater than the true speed.

3.2, #1, 2, 5. Left and right sums are given in the following table:

∫_{0}^{3} x^{3} dx  ∫_{0}^{π/2} cos x dx  ∫_{2}^{3} sin(t^{2}) dt
n  Left Right  Left Right  Left Right
2  .0625 .5625  1.3408 .5554  −.3950 .1895
10  .2025 .3025  1.0765 .9194  −.0921 .0248
50  .2401 .2601  1.0156 .9842  −.0430 −.0196
250  .2480 .2520  1.0031 .9969  −.0336 −.0289

x^{3} and cos x are monotonic on the given intervals, so their integrals are between the left and right sums. We take the averages of the left and right sums for n = 250, namely .25 and 1.00, as estimates for the values of the integrals. In #5 sin(t^{2}) is not monotonic on [2, 3], so its integral may not be between left and right sums. Thus the average −.0312 of the left and right sums for n = 250 may not be a reaonable estimate for the integral.

3.2, #13. Since 2^{x} is increasing on [1, 2], its integral is between left and right sums. For one decimal accurancy, (right sum) − (left sum) should be less than 0.1. Thus the number n of subintervals must satisfy the inequality (2^{2} − 2^{1})(2 − 1)/n < 0.1 and n > 20. For n = 21 the left and right sums are respectively 2.8380 and 2.9333. Their average 2.8857 is accurate to one decimal place.

3.2, #18. t^{2} is increasing on [1, 3]. We reason as in #13. The number n of subintervals must satisfy the inequality (3^{2} − 1^{2})(3 − 1)/n < 0.1. Thus n > 160. An approximation for the integral was not asked for.