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The antiderivatives of many functions either cannot be expressed or cannot be expressed easily in closed form (that is, in terms of known functions). Consequently, rather than evaluate definite integrals of these functions directly, we resort to various techniques of numerical integration to approximate their values. In this section we explore several of these techniques. In addition, we examine the process of estimating the error in using these techniques.
Earlier in this text we defined the definite integral of a function over an interval as the limit of Riemann sums . In general, any Riemann sum of a function $f(x)$ over an interval $[a,b]$ may be viewed as an estimate of ${\int}_{a}^{b}f(x)dx}.$ Recall that a Riemann sum of a function $f(x)$ over an interval $[a,b]$ is obtained by selecting a partition
and a set
The Riemann sum corresponding to the partition $P$ and the set $S$ is given by $\sum _{i=1}^{n}f({x}_{i}^{*})\text{\Delta}{x}_{i}},$ where $\text{\Delta}{x}_{i}={x}_{i}-{x}_{i-1},$ the length of the i th subinterval.
The midpoint rule for estimating a definite integral uses a Riemann sum with subintervals of equal width and the midpoints, ${m}_{i},$ of each subinterval in place of ${x}_{i}^{*}.$ Formally, we state a theorem regarding the convergence of the midpoint rule as follows.
Assume that $f(x)$ is continuous on $\left[a,b\right].$ Let n be a positive integer and $\text{\Delta}x=\frac{b-a}{n}.$ If $\left[a,b\right]$ is divided into $n$ subintervals, each of length $\text{\Delta}x,$ and ${m}_{i}$ is the midpoint of the i th subinterval, set
Then $\underset{n\to \infty}{\text{lim}}{M}_{n}={\displaystyle {\int}_{a}^{b}f\left(x\right)dx}.$
As we can see in [link] , if $f(x)\ge 0$ over $[a,b],$ then $\sum}_{i=1}^{n}f({m}_{i})\text{\Delta}x$ corresponds to the sum of the areas of rectangles approximating the area between the graph of $f(x)$ and the x -axis over $\left[a,b\right].$ The graph shows the rectangles corresponding to ${M}_{4}$ for a nonnegative function over a closed interval $[a,b].$
Use the midpoint rule to estimate ${\int}_{0}^{1}{x}^{2}dx$ using four subintervals. Compare the result with the actual value of this integral.
Each subinterval has length $\text{\Delta}x=\frac{1-0}{4}=\frac{1}{4}.$ Therefore, the subintervals consist of
The midpoints of these subintervals are $\left\{\frac{1}{8},\frac{3}{8},\frac{5}{8},\frac{7}{8}\right\}.$ Thus,
Since
we see that the midpoint rule produces an estimate that is somewhat close to the actual value of the definite integral.
Use ${M}_{6}$ to estimate the length of the curve $y=\frac{1}{2}{x}^{2}$ on $[1,4].$
The length of $y=\frac{1}{2}{x}^{2}$ on $[1,4]$ is
Since $\frac{dy}{dx}=x,$ this integral becomes ${\int}_{1}^{4}\sqrt{1+{x}^{2}}\phantom{\rule{0.1em}{0ex}}dx}.$
If $[1,4]$ is divided into six subintervals, then each subinterval has length $\text{\Delta}x=\frac{4-1}{6}=\frac{1}{2}$ and the midpoints of the subintervals are $\left\{\frac{5}{4},\frac{7}{4},\frac{9}{4},\frac{11}{4},\frac{13}{4},\frac{15}{4}\right\}.$ If we set $f\left(x\right)=\sqrt{1+{x}^{2}},$
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