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mlerma_courses_math214-2-02f_notes_c2-all

# mlerma_courses_math214-2-02f_notes_c2-all - CHAPTER 1...

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CHAPTER 1 Integrals 1.1. Areas and Distances. The Definite Integral 1.1.1. The Area Problem. The Definite Integral. Here we try to find the area of the region S under the curve y = f ( x ) from a to b , where f is some continuous function. b a O Y X y=f(x) S In order to estimate that area we begin by dividing the interval [ a,b ] into n subintervals [ x 0 ,x 1 ], [ x 1 ,x 2 ], [ x 2 ,x 3 ], ... , [ x n - 1 ,x n ], each of length Δ x = ( b - a ) /n (so x i = a + i Δ x ). b a O Y X 6

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1.1. AREAS AND DISTANCES. THE DEFINITE INTEGRAL 7 The area S i of the strip between x i - 1 and x i can be approximated as the area of the rectangle of width Δ x and height f ( x * i ), where x * i is a sample point in the interval [ x i ,x i +1 ]. So the total area under the curve is approximately the sum n i =1 f ( x * i x = f ( x * 1 x + f ( x * 2 x + · · · + f ( x * n x. This expression is called a Riemann Sum . The estimation is better the thiner the strips are, and we can iden- tify the exact area under the graph of f with the limit: A = lim n →∞ n i =1 f ( x * i x As long as f is continuous the value of the limit is independent of the sample points x * i used. That limit is represented b a f ( x ) dx , and is called definite integral of f from a to b : b a f ( x ) dx = lim n →∞ n i =1 f ( x * i x The symbols at the left historically were intended to mean an infinite sum, represented by a long “S” (the integral symbol ), of infinitely small amounts f ( x ) dx . The symbol dx was interpreted as the length of an “infinitesimal” interval, sort of what Δ x becomes for infinite n . This interpretation was later abandoned due to the difficulty of reasoning with infinitesimals, but we keep the notation. Remark : Note that in intervals where f ( x ) is negative the graph of y = f ( x ) lies below the x -axis and the definite integral takes a negative value. In general a definite integral gives the net area between the graph of y = f ( x ) and the x -axis, i.e., the sum of the areas of the regions where y = f ( x ) is above the x -axis minus the sum of the areas of the regions where y = f ( x ) is below the x -axis. 1.1.2. Evaluating Integrals. We will soon study simple and ef- ficient methods to evaluate integrals, but here we will look at how to evaluate integrals directly from the definition. Example : Find the value of the definite integral 1 0 x 2 dx from its definition in terms of Riemann sums.
1.1. AREAS AND DISTANCES. THE DEFINITE INTEGRAL 8 Answer : We divide the interval [0 , 1] into n equal parts, so x i = i/n and Δ x = 1 /n . Next we must choose some point x * i in each subinterval [ x i - 1 ,x i ]. Here we will use the right endpoint of the interval x * i = i/n . Hence the Riemann sum associated to this partition is: n i =1 i n 2 1 /n = 1 n 3 n i =1 i 2 = 1 n 3 2 n 3 + 3 n 2 + n 6 = 2 + 3 /n + 2 /n 2 6 . So: 1 0 x 2 dx = lim n →∞ 2 + 3 /n + 2 /n 2 6 = 1 3 . In order to check that the result does not depend on the sample points used, let’s redo the computation using now the left endpoint of each subinterval: n i =1 i - 1 n 2 1 /n = 1 n 3 n i =1 ( i - 1) 2 = 1 n 3 2 n 3 - 3 n 2 + n 6 = 2 - 3 /n + 2 /n 2 6 .

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mlerma_courses_math214-2-02f_notes_c2-all - CHAPTER 1...

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