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Unformatted text preview: 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 , x 1 ], [ x 1 , x 2 ], [ x 2 , x 3 ], . . . , [ x n 1 , x n ], each of length x = ( b a ) /n (so x i = a + i x ). b a O Y X 6 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 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 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 xaxis 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 xaxis, i.e., the sum of the areas of the regions where y = f ( x ) is above the xaxis minus the sum of the areas of the regions where y = f ( x ) is below the xaxis. 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 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 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 ....
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This note was uploaded on 02/05/2012 for the course MATH 214 taught by Professor Lerna during the Fall '10 term at FIU.
 Fall '10
 Lerna
 Math, Integrals

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