PlanetMath: example of estimating a Riemann integral

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example of estimating a Riemann integral

(Example)

The left hand rule, right hand rule, midpoint rule, and composite trapezoidal rule, all with , will be used in turn to estimate the Riemann integral $\displaystyle \int\limits_{-1}^2 x^2 \, dx$ .

Since , $\displaystyle \frac{b-a}{n}=\frac{2-(-1)}{6}=\frac{3}{6}=\frac{1}{2}$ .

Left hand rule:

$\begin{pspicture}(-3,-1)(3,5) \psset{unit=0.8cm} \pspolygon[fillstyle=solid,fill... ....41)(0,0) \psdots(-1,1)(-0.5,0.25)(0,0)(0.5,0.25)(1,1)(1.5,2.25) \end{pspicture}$

$\begin{displaymath}\begin{array}{rl} \displaystyle \sum_{j=1}^n \left(a+(j-1) \l... ...frac{19}{4} \ & \ & =\displaystyle \frac{19}{8} \end{array}\end{displaymath}$
Right hand rule:

$\begin{pspicture}(-3,-1)(3,5) \psset{unit=0.8cm} \pspolygon[fillstyle=solid,fill... ...4.41)(0,0) \psdots(-0.5,0.25)(0,0)(0.5,0.25)(1,1)(1.5,2.25)(2,4) \end{pspicture}$

$\begin{displaymath}\begin{array}{rl} \displaystyle \sum_{j=1}^n \left(a+j \left(... ...frac{27}{4} \ & \ & =\displaystyle \frac{27}{8} \end{array}\end{displaymath}$
Midpoint rule:

$\begin{pspicture}(-3,-1)(3,5) \psset{unit=0.8cm} \pspolygon[fillstyle=solid,fill... ...0.25,0.0625)(0.25,0.0625)(0.75,0.5625)(1.25,1.5625)(1.75,3.0625) \end{pspicture}$

$\begin{displaymath}\begin{array}{rl} \displaystyle \sum_{j=1}^n \left( \! a \!+\... ...ac{94}{16} \ & \ & =\displaystyle \frac{47}{16} \end{array}\end{displaymath}$
Composite trapezoidal rule:

$\begin{pspicture}(-3,-1)(3,5) \psset{unit=0.8cm} \pspolygon[fillstyle=solid,fill... ...0,0) \psdots(-1,1)(-0.5,0.25)(0,0)(0.5,0.25)(1,1)(1.5,2.25)(2,4) \end{pspicture}$

$\begin{displaymath}\begin{array}{rl} \displaystyle \frac{1}{2} \! \left( \! \fra... ...c{25}{2} \ & \ & \displaystyle \!= \frac{25}{8} \end{array}\end{displaymath}$

For comparison purposes, the Riemann integral will also be computed:

$\begin{pspicture}(-3,-1)(3,5) \psset{unit=0.8cm} \psline(-1,0)(-1,1) \psline(2,0... ...5,-0.5){$x$} \rput[l](-0.4,5){$y$} \parabola{<->}(2.1,4.41)(0,0) \end{pspicture}$

$\begin{displaymath}\begin{array}{rl} \displaystyle \int\limits_{-1}^2 x^2 \, dx ... ... =\displaystyle \frac{1}{3} (8-(-1)) \ & \ & =3 \end{array}\end{displaymath}$

As expected, of the three estimates, the ones obtained from the midpoint rule and the composite trapezoidal rule are closest to the actual value of the Riemann integral. Their errors are $\displaystyle \frac{1}{16}$ and $\displaystyle \frac{1}{8}$ , respectively. It may seem odd that the midpoint rule should be closer to the actual value of the Riemann integral even though, in the graph for the composite trapezoidal rule, the approximating trapezoids are barely distinguishable from the parabola. Actually though, this is to be expected, as the maximum error for the midpoint rule is half of the maximum error for the composite trapezoidal rule.

"example of estimating a Riemann integral" is owned by Wkbj79.

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See Also: left hand rule, right hand rule, midpoint rule

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Cross-references: parabola, trapezoids, graph, Riemann integral, composite trapezoidal rule, midpoint rule, right hand rule, left hand rule
There are 2 references to this entry.

This is version 28 of example of estimating a Riemann integral, born on 2007-04-21, modified 2008-03-12.
Object id is 9236, canonical name is ExampleOfEstimatingARiemannIntegral.
Accessed 2056 times total.

Classification:

AMS MSC:	26A42 (Real functions :: Functions of one variable :: Integrals of Riemann, Stieltjes and Lebesgue type)
	41-01 (Approximations and expansions :: Instructional exposition )
	28-00 (Measure and integration :: General reference works )

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