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Chapter 5 Integral In chapter 3 we used the tangent and velocity problems to introduce the derivative, which is the basic idea in differential calculus. In this chapter, we begin with the area and the distance problems, because they contains a basic idea in integral calculus. 1 Section 1 Areas and Distances 1. The Area Problem Find the area of the region of S that lies under the curve y=f(x) from a to b. y f x xa S xb a b A natural idea is that we first approximate the reign S by so many small rectangles and then we take the Limit of the sum of these rectangle areas as we increase the number of rectangles. 2 Example 1 Use rectangle to estimate the area under the parabola y x 2 from 0 to 1. Approximate each strip by a rectangle whose base is the same as the strip, whose height is the same as the right edge of the strip. n Ln Rn 10 0.2850000 0.3850000 20 0.3087500 0.3587500 o 1 R4=0.46875 30 0.3168519 0.3501852 40 0.3234000 0.3434000 100 0.3283500 0.3383500 1000 0.3328335 0.3338335 1 lim Ln lim Rn L4=0.2187 n n 3 3 Example 2 For the reigon S in Example 1 ,show that the sum of the areas of the uper approximating rectangles appracches 1/3. 4 Apply the idea to the more general region S 1. Divide the interval [a,b] into n subintervals, and each subintervals has same length. x0 , x1 , x1 , x2 xn1 , xn ba x n 2. Approximate the strip Si by small rectangles Si f xi x 3. Make a sum Rn f x1 x f x2 x f xn x Ln f x0 x f x1 x f x2 x f xn1 x 4.Find the limit lim Rn lim Ln n n If the two limits exist and they are the same, then we have S lim Ln lim Rn n n 5 It can be proved that, if f is continuous on [a, b], then lim Rn and lim Ln exist, lim Ln lim Rn n n n n In fact, instead of using the left endpoint and right endpoint, we could take the height of the ith rectangle to be the value of f at any number x i in the subinterval [xi-1, xi]. x1 , x 2 , , x n are called the sample points. n A lim f x1 x f x 2 x f x n x n lim f x i x n i 1 6 Example 3 Let A be the area of the reigon that lies under the graph of f(x)=cosx between x=0 and x=b, where 0 b / 2 (a) Using right endpoints, find an expression for A as a limit. Do not evaluate the limit. (b) Estimate the area for the case b / 2 by taking the sample points to be midpoints and using four subintervals. M 4 1.006 7 1. The Distance Problem distance velocity time Example 4 Suppose the odometer on our car is broken and we want to estimate the distance driven over a 30-second time interval. We take speedometer readings every five second and record them in the following table: Time(s) 0 5 10 15 20 25 30 Velocity(km/h) 27 34 38 46 51 50 45 1km/h=1000/3600m/s Time(s) 0 5 10 15 20 25 30 Velocity(m/s) 7.5 9.4 10.6 12.8 14.2 13.9 12.5 8 Time(s) 0 5 10 15 20 25 30 Velocity(km/h) 7.5 9.4 10.6 12.8 14.2 13.9 12.5 During the first seconds the velocity doesn’t change very much, so we can estimate the distance traveled by assuming that the velocity is constant. 7.5m/s 5s 37.5m Similarly, during the second time interval the velocity is approximately constant and we take it to be the velocity when t=5s. 9.4m/s 5s 47m If we add similar estimates for the other time intervals, we obtain an estimate for the total distance traveled: 7.5 5 9.4 5 10 .6 5 12 .8 5 14 .2 5 13 .9 5 342 m 9.4 5 10.6 5 12.8 5 14.2 5 13.9 5 12.5 5 342m If we had wanted a more accurate estimate, we should partition the time Interval more brief. 9 v y 15 yx 2 10 5 o x 1 10 20 30 t L4=0.2187 L6 342m In general, suppose an object moves with velocity v=f(t), where ba 1. Divide t0 , t1 , t1, t2 tn1, tn t n 2. Approximate di f ti t di f ti 1 t n n 3. Make a sum Rn f ti t Ln f ti 1 t i 1 i 1 n n 4.Find the limit d lim f ti t lim f ti 1 t n i 1 n i 1 10 Section 2 The Definite Integral lim f x1 x f x2 x f xn x n lim f x x n i n i 1 Except that the area and distance problems can be expressed in the form of this type of limit of a sum, many practical problem can be considered as this type of limit of a sum, even when f is not necessarily a positive function. Definition of A Definite Integral If f is a function defined for a x b , we divided the interval [a, b] into n subintervals of equal width x b a / n . We let x0 a , x1 , x2 , , xn b be the endpoints of these subintervals and we let x1 , x 2 , , x n be any sample points in these subintervals, so x i lies in the ith subinterval x i 1 , x i . Then the definite integral of f from a to b is n f x dx lim f x i x b a n i 1 11 If this type of limit exits, we say that f is integrable on [a, b]. Note 1: Symbol The function is the integrand Upper limit of integration x is the Variable of Integration b Integral sign a f (x)dx Integral f from a to b Lower limit of integration When you find the value of integral, You have evaluated the integral. f x dx f t dt f u du b b b Note 2 : a a a f x dx b Note 3 : is an algebra sum of areas. 12 a Not all functions are integrable. The following theorem shows that the most commonly occurring functions are in fact integrable. Theorem If f is continuous on [a, b], or if f has only a finite number of jump discontinuities, then f is integrable on [a, b]; that is, the definite integral a f x dx exists. b According to this theorem, if f is integrable, that means the following type of limit exists and give the same value no matter how we choose the sample points. f x dx lim f x x b n i a n i 1 To simplify the calculation of integral, we often take the sample points to be right endpoints, left endpoints and even midpoints. f x dx b a n n n x xi lim f xi 1 x lim f xi x lim f i 1 x 13 n i 1 n i 1 n i 1 2 14 (1) (2) (3) (4) (5) (6) 15 (7) 4 3 x dx. 1 2 Example 6 Use the properties of integrals to evaluate 0 f x dx 17 and f x dx 12 . 10 8 Example 7 If it is known that 0 0 f x dx. 10 find 8 4 Example 8 Use the last property to estimate x dx . 1 Example 9 Evaluate the area of the reigon bounded by y x 3 and y x. 16 Section 3 The Fundamental Theorem of Calculus(TFTC) Differential Integral Calculus Calculus Newton, Leibniz (Tangent problem) (Area problem) They discovered the closed relation between the two seeming unrelated problems. TFTC gives the precise inverse relationship between the derivative and the integral. 17 x f t dt f x a Vary upper limit function ——变上限函数 18 Find the derivative of the following function 19 Section 4 Indefinite Integral We saw that the second part of TFTC provides a very powerful method for evaluating the denifite integral of a function. That is we just need to find the antiderivative of the function then calculate the difference between the upper limit and lower limit of integration. Now we try to explore the methods to find antiderivatives of functions. In mathematic, we called all the antiderivatives of a function indefinite integral. Denote f x dx F x C , where [F x C ] f x Note: 20 the difference between the definite integral and the indefinite integral. Function Particular Function Particular Antiderivative Antiderivative cos x sin x cf x CF x sin x cos x sec2 x tan x f x g x F x G x csc2 x cot x x n 1 n x n 1 sec x tan x sec x n 1 csc xcot x csc x The process of finding the antiderivatives is opposite to the process of finding the derivatives. We can obtain the formulas for indefinite integral by inversing those formulas. 21 Table of Indefinite Integrals Cf x dx C f x dx [ f x g x ]dx f x dx g x dx x n 1 kdx kx C x n dx n 1 C sin xdx cos x C cos xdx sin x C sec2 xdx tan x C csc2 xdx cot x C sec x tan xdx sec x C csc x cot xdx csc x C Any formula can be verified by differentiating the function on the right side and obtain the integrand. 22 cos Example 3 Evaluate sin2 d 1 tan d 2 Example 4 Evaluate x 2 x dx 2 Example 5 Evaluate 1 23 Section 5 The Substitution Rule 24 25 x 2 Example 10 Find the value 6 1 dx 2 1 tan x Example 11 Find the value dx 1 1 x x 2 4 26

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