Just as the trapezoidal rule is the average of the left-hand and right-hand rules for estimating definite integrals, Simpson’s rule may be obtained from the midpoint and trapezoidal rules by using a weighted average. It can be shown that:
It is also possible to put a bound on the error when using Simpson’s rule to approximate a definite integral. The bound in the error is given by the following rule:
error bound for Simpson’s Rule
Let [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
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].
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
[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
[latex]\text{Error in }{S}_{2}\le \frac{0{\left(1\right)}^{5}}{180\cdot {2}^{4}}=0[/latex].
This bound indicates that the value obtained through Simpson’s rule is exact. A quick check will verify that, in fact, [latex]{\displaystyle\int }_{0}^{1}{x}^{3}dx=\frac{1}{4}[/latex].
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].
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],
Watch the following video to see the worked solution to the example above.
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