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9.6: Reimann Sums

  • Page ID
    1255
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    Approximating the area under a function curve by summing a finite number of rectangles in a Riemann Sum can yield very accurate results. Intuitively, we know, however, that the more sub-intervals we have the better the result. Taking the limit of the Riemann Sum as the subintervals get smaller (number of rectangles gets larger) should asymptotically give the true area. For some function curves, the Riemann limit can be evaluated algebraically; for complex curves, the area can only be determined by brute force numerical computations of Riemann Sums.


    Limits and Reimann Sums

    Earlier, the area under a curve was defined in terms of a limit of sums:

    \[ A = \lim_{n \to +∞} S(P) = \lim_{n \to +∞} T(P) \nonumber\]

    where

    \[ S(P) = \sum_1^n m_i(x_i − x_i − 1) = m_1 (x_1−x_0) + m_2(x_2−x_1)+…+ mn(xn−xn−1), \nonumber\]

    \[ T(P) = \sum_1^n M_i (x_i−x_i−1)= M_1(x_1−x-0)+M_2(x_2−x_1)+…+M_n(x_n−x_n−1), \nonumber\]

    S(P) and T(P) are examples of Riemann Sums.

    In general, Riemann Sums are of form \( \sum_{i=1}^n f ( x_i^∗) △x  \nonumber\) where each \( x_i^∗ \nonumber\) is the value we use to find the length of the rectangle in the ith sub-interval. For example, the maximum function value in each sub-interval to find the upper sums and the minimum function in each sub-interval to find the lower sums. But since the function is continuous, we could have used any points within the sub-intervals to find the limit.

    To make use of the concept of limit, we make the width of each rectangle approach 0, which is equivalent to making the number of rectangles, n, approach infinity. By doing so, we find the exact area under the curve,

    \[ limn→∞An=limn→∞∑i=1nf(xi)△x. \nonumber\]

    We now define the most general situation as follows:

    If f is continuous on [a,b], and:

    1. The interval [a,b] is divided into n sub-intervals of equal width △x, with △x=b−an, and
    2. The endpoints of these sub-intervals are x0=a,x1,x2,...,xn=b, and
    3. x∗1,x∗2,…,x∗n are any sample points in these sub-intervals, then the definite integral of f from x=a to x=b is

    ∫abf(x)dx=limn→∞∑i=1nf(x∗i)△x.

    provided the limit exists.

    If the above limit exists, f is said to be integrable on the closed interval [a,b] and the definite integral exists.

    Note that the sample point x∗i can be any sample point in the i-th sub-interval, with common choices being right, or mid-point, or left.

    For example, evaluate the Riemann Sum for f(x)=x3 from x=0 to x=3 using n=6 sub-intervals, and take the sample points to be the midpoints of the sub-intervals.

    If we partition the interval [0, 3] into n=6 equal sub-intervals, then each sub-interval will have length 3−06=12. So we have △x=12 and

     

     

     

     

     


    Additional Resources

    Video - Riemann Sums and Integrals

    Practice - Riemann Sums


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