Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 85. A major complaint of professors teaching calculus is that students don't have the appropriate background to work through the calculus course successfully. This text is targeted directly at this underprepared audience. This is a single-variable (2-semester) calculus text that incorporates a conceptual re-introduction to key precalculus ideas throughout the exposition as appropriate. This is the ideal resource for those schools dealing with poorly prepared students or for schools introducing a slower paced, integrated precalculus/calculus course | Simpson s Rule and Error Estimates 821 We call it 2 because we need n data points for M and n 1 different data points for r . Sometimes it will be convenient to write 2Mn3 rn where 2n. Simpson s rule generally gives a substantially better numerical estimate than either the midpoint or the trapezoidal sums. To get some appreciation for this try it out on some integrals that you can compute exactly. Below we do this for the integral 5 1 x dx. EXAMPLE SOLUTION Compare M T and Simpson s rule the weighted average for 5 X dx where n 4 8 50 and 100. Note that jf 1 dx ln x 1 ln 5 - ln 1 ln 5 . a n 4 M4 . T4 . . . Simpson s rule 2M43 r4 . . . b n 8 M8 . T8 . . . Simpson s rule 2M83 r8 . . . c n 50 M50 . r50 . . Simpson s rule LL . . . d n 100 M100 . T100 . . Simpson s rule 2M100 r100 . Like the trapezoidal sum Simpson s rule is a convenient tool for approximating fa x dx even when a formula for is not available. For instance the values of x referred to in the discussion below could be measurements taken by a surveyor estimating the area of a plot of land or body of water. Suppose our aim is to approximate f x dx. We can collect values of x at equally spaced intervals and use Simpson s rule to approximate the deflnite integral. Partition the interval a into n equal subintervals each of length Ax . We need to use a weighted average of the midpoint and trapezoidal sums therefore on each subinterval we need not only the value of at each endpoint for T but also the value of in the middle for M . Although we are chopping a into n equal subintervals we will use 2n 1 values of n 1 values for Tn and n values for M . Let k xk for k 0 . 2n where x0 x1 x2 . x2n are as indicated below. Xk a k I A x Ax 2 Ax for k . Ax 2 7 x0 x1 x2 À a for Mn x3 for Mn x4 . . . x2n _2 x2n-1 Î for Mn x2n II b Figure In the following exercise you will show that .
