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Riemann Sum Approximation

The definite integral of a continuous function ‘f’ over an interval [a, b] is computed as ab f(x)dx = lim Σf(xk*)∆x, where the sum that appears on the right side is called Riemann sum.  In this formula, the interval [a, b] is divided into n subintervals of width ∆x = (b-a)/n, and xk* denotes an arbitrary point in the kth sub-interval  It follows that as n increases the Riemann sum will eventually be a good approximation to the integral, which we denote by writing

                  ab f(x)dx ≈ Σf(xk*)∆x
                 ab f(x)dx ≈ ∆x[f(x1*) + f(x2*) + …….+f(xn*)]

Here we denote the values of ‘f’ at the endpoints of the subintervals by

y0=f(a), y1=f(x1), y2=f(x2), ………, yn-1=f(xn-1), yn=f(b) and we will denote the values of f at the midpoints of the subintervals by ym1, ym2, ……ymn

Trapezoidal Approximation

The left-hand and right hand endpoint approximations are rarely used in applications; however, if we take the average of the left-hand and right hand endpoint approximations, we obtain a result, called the trapezoidal approximation, which is commonly used as,
                 ab f(x)∆x ≈  (b-a)/2n [ y0 + 2y1 + …..+ 2yn-1 + yn]
The name trapezoidal approximation can be explained by considering the case in which f(x)≥0 on [a, b], so that ab f(x)dx represents the area under f(x) over [a, b].  Geometrically, the trapezoidal approximation formula results if we approximate this area by the sum of the trapezoidal areas as shown in the figure

Left end point Approximation:  The formula for evaluating left end point approximation is given by
                ab f(x)dx =  (b-a)/n [ y0 + y1 + ……….. + yn-1]
Right Endpoint Approximation:  The formula for evaluating right end point approximation is given by
                       ab f(x)dx = (b-a)/n [y1 + y2 + …………. + yn]
Mid-Point Approximation:  The formula for evaluating midpoint approximation is given by
              ab f(x)dx = (b-a)/n [ym1 + ym2 + ………..+ ymn] where m1, m2 ……mn represents the mid values.
Example: Use Trapezoidal rule to approximate 0πsinx dx using n=10 sub intervals
Solution: a=0, b=π n=10 and f(x)=sinx,   (b-a)/n = π/10

0π sinx dx = (π/20)[y0 + 2y1 + 2y2 + …………+2yn-1 + yn]
                = (π/20) [ sin(0) + 2sin(π/10) + 2sin(2π/10) + ……..sin(10π/10)
                = 1.983523538

Comparison of the Midpoint and Trapezoidal Approximations

The table below shows the comparison between midpoint and trapezoidal approximations for the function ln 2 = 12(1/x)dx with n=10 subdivisions

                    Midpoint Approximation


12 (1/x)dx = (0.1)(6.928353603) = 0.692835360

                          Trapezoidal Approximation


12 (1/x)dx = (0.05)(13.875428063) = 0.693771403

The value of ln 2 is rounded to nine decimal places and we have seen that midpoint approximation produces a more accurate result than the trapezoidal approximation.  Hence we can conclude that,
If f be a continuous on [a, b] and let |EM| and |ET| be the absolute errors that result from the midpoint and trapezoidal approximations of abf(x)dx using n subintervals.
a)    If the graph of f is either concave up or concave down on (a, b), then |EM|<|ET|, which means that the error from the midpoint approximation is less than from the trapezoidal approximation.
b)    If the graph of ‘f’ is concave down on (a, b) then Tn <ab f(x)dx < Mn
c)    If the graph of ‘f’ is concave up on (a,b), then Mn < ab f(x)dx < Tn

Simpson’s Rule

Simpson’s Rule is given by


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