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Riemann Sums and Integrals Definition: Let f (x) be a continuous function on the interval [a, b]. Then the definite integral of f from a to b is defined as b Z f (x) dx = lim a n X n→∞ f (x∗i )∆x i=1 b−a where ∆x = , xi = a + i∆x, and xi−1 ≤ x∗i ≤ xi . n ∗ But what is this xi ? Once we have chopped our interval [a, b] into n rectangles of equal width (∆x = (b − a)/n), our ith rectangle lies on top of the sub-interval [xi−1 ,xi ]. x∗i represents any point in the interval [xi−1 ,xi ]. By using this x∗i notation, we include left-hand sums, right-hand sums, midpoint sums, and even the possibility that we choose a random x-value within each subinterval. Riemann Sums Sums of the form n X f (x∗i )∆x are called Riemann sums. i=1 • When x∗i = xi−1 , we have a Left Hand Sum. • When x∗i = xi , we have a Right Hand Sum. • When x∗i = 12 (xi−1 + xi ), we have a Midpoint Sum. Based off of our definition, we expect these Riemann sums to approach the same value, as we send the number of rectangles to infinity. Monotonic Functions 1. When f (x) is strictly increasing on [a, b], we always have b Z LHS(n) ≤ f (x)dx ≤ RHS(n) a Z Example: Suppose we want to approximate xdx using a Riemann sum with 4 rect0 angles. 1 4 (a) LHS(4) (b) RHS(4) Figure 1: Riemann Sums 2. When f (x) is strictly decreasing on [a, b], we always have b Z RHS(n) ≤ f (x)dx ≤ LHS(n) a (a) LHS(4) (b) RHS(4) Figure 2: Riemann Sums So, in the case that f (x) is either always increasing, or always decreasing, we can use LHS(n) and RHS(n) as the best bounds on the definite integral of f (x) over [a, b]. In particular, if we want to guarantee that our estimate is within some given level of accuracy, p, of the true answer, it suffices to ensure that |LHS(n) − RHS(n)| < p ⇔ |f (b) − f (a)| 2 b−a <p n But what if f (x) is neither always increasing, nor always decreasing? Example: From the Riemann Sums Lab, Part 4. We want to use Riemann sums with 4 Z 4 4 x+ rectangles to estimate dx. x 1 4−1 3 Here, ∆x = = , and xi = 1 + i × ∆x. 4 4 • Largest sum: in each interval, choose the value of x that maximizes f (x). Note that this is not one of the named Riemann sums, as the choice of x∗i changes in each interval. 4 Z 1 4 x+ dx ≈ (f (1) + f (10/4) + f (13/4) + f (4))(3/4) = 13.9356 x • Smallest sum: in each interval, choose the value of x that minimizes f (x). Note that this is not one of the named Riemann sums, as the choice of x∗i changes in each interval. Z 1 4 4 x+ dx ≈ (f (7/4) + f (2) + f (10/4) + f (13/4))(3/4) = 12.4624 x (a) Largest possible Riemann sum (b) Smallest possible Riemann sum Figure 3: Riemann Sums Practice Problems Z 1. Suppose you want to estimate appropriate sum in Σ notation. 5 ln(x)dx. For each of the following, write down an 2 (a) LHS(n) (b) RHS(n) 3 (c) MPS(n) 2. Compute the following by recognizing them as definite integrals, and using the Fundamental Theorem of Calculus. 2 n X 10i 10 (a) lim n→∞ n n i=1 2 n X 10i 10 (b) lim 2+ n→∞ n n i=1 2 ! n X 10 10i (c) lim 2+ n→∞ n n i=1 n X 1 1 (d) lim n→∞ 1 + (k/n)2 n i=1 n X πk πk π (e) lim cos + n→∞ 2n 2n 2n i=1 4