Riemann Sum Approximation
The definite integral of a continuous function ‘f’ over an interval [a, b] is computed as a∫b 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
a∫b f(x)dx ≈ Σf(xk*)∆x
a∫b 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
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,
a∫b 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 a∫b 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
a∫b f(x)dx = (b-a)/n [ y0 + y1 + ……….. + yn-1]
Right Endpoint Approximation: The formula for evaluating right end point approximation is given by
a∫b f(x)dx = (b-a)/n [y1 + y2 + …………. + yn]
Mid-Point Approximation: The formula for evaluating midpoint approximation is given by
a∫b f(x)dx = (b-a)/n [ym1 + ym2 + ………..+ ymn] where m1, m2 ……mn represents the mid values.
Example: Use Trapezoidal rule to approximate 0∫πsinx dx using n=10 sub intervals
Solution: a=0, b=π n=10 and f(x)=sinx, (b-a)/n = π/10
0∫π sinx dx = (π/20)[y0 + 2y1 + 2y2 + …………+2yn-1 + yn]
= (π/20) [ sin(0) + 2sin(π/10) + 2sin(2π/10) + ……..sin(10π/10)
Comparison of the Midpoint and Trapezoidal Approximations
The table below shows the comparison between midpoint and trapezoidal approximations for the function ln 2 = 1∫2(1/x)dx with n=10 subdivisions
1∫2 (1/x)dx = (0.1)(6.928353603) = 0.692835360
1∫2 (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 a∫bf(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 <a∫b f(x)dx < Mn
c) If the graph of ‘f’ is concave up on (a,b), then Mn < a∫b f(x)dx < Tn
Simpson’s Rule is given by
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