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Section 1.23 Numerical integration: error bounds
Theorem 1.23.1 . Error bounds.
Let \(f\) be an integrable function on \([a,b]\text{,}\) let \(n\) be a positive integer, and let
\begin{equation*}
a=x_0\lt x_1\lt x_2\lt \cdots \lt x_n=b\text{.}
\end{equation*}
be partition of \([a,b]\) into \(n\) subintervals of equal length \(\Delta x=\frac{b-a}{n}\text{.}\)
Right/left Riemann sums.
Let \(R_n\) be either the right or left Riemann sum for this partition. If \(f'\) is continuous and \(\vert f'(x)\vert \leq M\) for all \(x\) in \([a,b]\text{,}\) then
\begin{equation*}
\left\vert\int_a^b f(x)\, dx - R_n\right\vert\leq \frac{M(b-a)^2}{2n}\text{.}
\end{equation*}
Midpoint Riemann sum.
Let \(M_n\) be the midpoint Riemann sum for this partition. If \(f''\) is continuous and \(\vert f''(x)\vert \leq M\) for all \(x\) in \([a,b]\text{,}\) then
\begin{equation*}
\left\vert\int_a^b f(x)\, dx - M_n\right\vert\leq \frac{M(b-a)^3}{24n^2}\text{.}
\end{equation*}
Trapezoidal estimate.
Let \(T_n\) be the \(n\) -th trapezoidal estimate of \(\int_a^b f(x)\, dx\text{.}\) If \(f''\) is continuous and \(\vert f''(x)\vert\leq M\) for all \(x\) in \([a,b]\text{,}\) then
\begin{equation*}
\left\vert\int_a^b f(x)\, dx - T_n\right\vert\leq\frac{M(b-a)^3}{12n^2}\text{.}
\end{equation*}
Simpson’s rule.
Suppose \(n\) is even, and let \(S_n\) be the \(n\) -th Simpson’s rule estimate of \(\int_a^bf(x)\, dx\text{.}\) If \(f^{(4)}\) is continuous and \(\vert f^{(4)}(x)\vert\leq M\) for all \(x\) in \([a,b]\text{,}\) then
\begin{equation*}
\left\vert\int_a^b f(x)\, dx - S_n\right\vert\leq\frac{M(b-a)^5}{180n^4}\text{.}
\end{equation*}
Example 1.23.2 . Estimating \(\ln 4\text{:}\) error bounds.
Let \(f(x)=\frac{1}{x}\text{.}\) Recall that \(\ln 4=\int_1^4 f(x)\, dx\text{.}\)
Compute bounds for the errors in (a) the \(n=10\) trapezoidal estimate of \(\ln 4\) and (b) the \(n=10\) Simpson’s rule estimate of \(\ln 4\text{.}\)
Example 1.23.3 . Estimating \(\pi\text{:}\) error bounds.
Let \(f(x)=\frac{4}{x^2+1}\text{.}\) We have
\begin{align*}
f''(x)\amp =\frac{8(3x^2-1)}{(x^2+1)^3}\\
f^{(4)}(x)\amp =\frac{96(5x^4-10x^2+1)}{(x^2+1)^5}\text{.}
\end{align*}
Recall that \(\pi=\int_0^1 f(x)\, dx\text{.}\) Compute bounds for the errors in estimating \(\pi\) using (a) the \(n=10\) trapezoidal rule, and (b) the \(n=10\) Simpson’s rule.
Find (a) an \(n\) such that the \(n\) -th trapezoidal estimate of \(\pi\) is within \(10^{-9}\) of the actual value, and (b) an \(n\) such that the \(n\) -th Simpson’s rule estimate of \(\pi\) is within \(10^{-9}\) of the actual value.
