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Riemann Sums and the Definite Integral Lesson 5.3

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Why? Why is the area of the yellow rectangle at the end = a b

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Review We partition the interval into n sub-intervals Evaluate f(x) at right endpoints of k th sub-interval for k = 1, 2, 3, … n a b f(x)

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Review Sum We expect S n to improve thus we define A, the area under the curve, to equal the above limit. a b f(x) Look at Goegebra demo Look at Goegebra demo

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Riemann Sum 1.Partition the interval [a,b] into n subintervals a = x 0 < x 1 … < x n-1 < x n = b Call this partition P The k th subinterval is x k = x k-1 – x k Largest x k is called the norm, called ||P|| 2.Choose an arbitrary value from each subinterval, call it

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Riemann Sum 3.Form the sum This is the Riemann sum associated with the function f the given partition P the chosen subinterval representatives We will express a variety of quantities in terms of the Riemann sum

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The Riemann Sum Calculated Consider the function 2x 2 – 7x + 5 Use x = 0.1 Let the = left edge of each subinterval Note the sum

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The Riemann Sum We have summed a series of boxes If the x were smaller, we would have gotten a better approximation f(x) = 2x 2 – 7x + 5

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The Definite Integral The definite integral is the limit of the Riemann sum We say that f is integrable when the number I can be approximated as accurate as needed by making ||P|| sufficiently small f must exist on [a,b] and the Riemann sum must exist

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Example Try Use summation on calculator.

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Example Note increased accuracy with smaller x

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Limit of the Riemann Sum The definite integral is the limit of the Riemann sum.

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Properties of Definite Integral Integral of a sum = sum of integrals Factor out a constant Dominance

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Properties of Definite Integral Subdivision rule a b c f(x)

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Area As An Integral The area under the curve on the interval [a,b] a c f(x) A

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Distance As An Integral Given that v(t) = the velocity function with respect to time: Then Distance traveled can be determined by a definite integral Think of a summation for many small time slices of distance

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Assignment Section 5.3 Page 314 Problems: 3 – 47 odd

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