Just as the trapezoidal rule is the average of the left-hand and right-hand rules for estimating definite integrals, Simpson\u2019s rule may be obtained from the midpoint and trapezoidal rules by using a weighted average. It can be shown that:<\/p>\r\n
[latex]{S}_{2n}=\\left(\\frac{2}{3}\\right){M}_{n}+\\left(\\frac{1}{3}\\right){T}_{n}[\/latex].<\/p>\r\n
It is also possible to put a bound on the error when using Simpson\u2019s rule to approximate a definite integral. The bound in the error is given by the following rule:<\/p>\r\n\r\n [latex]\\text{Error in}{S}_{n}\\le \\frac{M{\\left(b-a\\right)}^{5}}{180{n}^{4}}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/section> Use [latex]{S}_{2}[\/latex] to approximate [latex]{\\displaystyle\\int }_{0}^{1}{x}^{3}dx[\/latex]. Estimate a bound for the error in [latex]{S}_{2}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558799\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558799\"]\r\n Since [latex]\\left[0,1\\right][\/latex] is divided into two intervals, each subinterval has length [latex]\\Delta x=\\frac{1 - 0}{2}=\\frac{1}{2}[\/latex]. The endpoints of these subintervals are [latex]\\left\\{0,\\frac{1}{2},1\\right\\}[\/latex]. If we set [latex]f\\left(x\\right)={x}^{3}[\/latex], then<\/p>\r\n [latex]{S}_{4}=\\frac{1}{3}\\cdot \\frac{1}{2}\\left(f\\left(0\\right)+4f\\left(\\frac{1}{2}\\right)+f\\left(1\\right)\\right)=\\frac{1}{6}\\left(0+4\\cdot \\frac{1}{8}+1\\right)=\\frac{1}{4}[\/latex]. Since [latex]{f}^{\\left(4\\right)}\\left(x\\right)=0[\/latex] and consequently [latex]M=0[\/latex], we see that<\/p>\r\n\r\n This bound indicates that the value obtained through Simpson\u2019s rule is exact. A quick check will verify that, in fact, [latex]{\\displaystyle\\int }_{0}^{1}{x}^{3}dx=\\frac{1}{4}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section> Use [latex]{S}_{6}[\/latex] to estimate the length of the curve [latex]y=\\frac{1}{2}{x}^{2}[\/latex] over [latex]\\left[1,4\\right][\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558699\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558699\"]\r\n The length of [latex]y=\\frac{1}{2}{x}^{2}[\/latex] over [latex]\\left[1,4\\right][\/latex] is [latex]{\\displaystyle\\int }_{1}^{4}\\sqrt{1+{x}^{2}}dx[\/latex]. If we divide [latex]\\left[1,4\\right][\/latex] into six subintervals, then each subinterval has length [latex]\\Delta x=\\frac{4 - 1}{6}=\\frac{1}{2}[\/latex], and the endpoints of the subintervals are [latex]\\left\\{1,\\frac{3}{2},2,\\frac{5}{2},3,\\frac{7}{2},4\\right\\}[\/latex]. Setting [latex]f\\left(x\\right)=\\sqrt{1+{x}^{2}}[\/latex],<\/p>\r\n\r\n After substituting, we have<\/p>\r\n\r\nerror bound for Simpson\u2019s Rule<\/h3>\r\nLet [latex]f\\left(x\\right)[\/latex] be a continuous function over [latex]\\left[a,b\\right][\/latex] having a fourth derivative, [latex]{f}^{\\left(4\\right)}\\left(x\\right)[\/latex], over this interval. If [latex]M[\/latex] is the maximum value of [latex]|{f}^{\\left(4\\right)}\\left(x\\right)|[\/latex] over [latex]\\left[a,b\\right][\/latex], then the upper bound for the error in using [latex]{S}_{n}[\/latex] to estimate [latex]{\\displaystyle\\int }_{a}^{b}f\\left(x\\right)dx[\/latex] is given by\r\n