Chapter 5 Integrals 5.1 Areas and Distances We start with a problem how can we calculate the area under a given function ie, the area between the function and the x-axis? If the curve happens to be something easy like a horizontal line - this isn t too hard, but if it s a curve, the problem becomes more difficult. Let s start by drawing a rectangle that s f(a) tall by (b a) wide as an estimate: Area = f(a) (b a) How can we get a better estimate? Let s try 2 rectangles: Area = f(a) x 1 + f(c) x 2 How could we get an even better estimate?
Ex: Estimate the area under the curve f(x) = x 2 between x = 0 and x = 1 by constructing 8 rectangles (the book calls them strips ) of equal width. This seems like a relatively simple situation, until you realize that you get different answers depending on whether you draw the rectangles from the left or from the right: Using left endpoints, we get: f(0) " + f " " + f # " " + f $ " " + f % " " + f & " " + f ' " " + f ( " " We can see this will be a lower bound, because all the rectangles fit completely under the curve. Using right endpoints, we get: f " " + f # " " + f $ " " + f % " " + f & " " + f ' " " + f ( " " + f 1 " We can see this will be an upper bound, because all the rectangles extend to above the curve. By computing both, we get: 0.2734375 A 0.3984375 Hmmm if only we had a technique for looking at a progression of smaller and smaller rectangles.
Definition: The Area A of the region S that lies under the graph of the continuous function f is the limit of the sum of the areas of approximating rectangles (using right endpoints): A = lim - / R - = lim - / f x x + f x # x + + f(x - ) x 1) It can be proven that this limit always exists. 2) It can also be shown that you get the same limit value A from using left endpoints. 3) In fact, you can choose any x-value in each interval to calculate the height. The values that are chosen - x 1 *, x 2 *,, x n * - are called the sample points. Using sigma notation we get: A = lim - / - :; f(x : ) x (Riemann sum) The Distance Problem My odometer is broken, but I d still like to calculate how far I ve driven using my speedometer and a stopwatch (held by my passenger, for safety.) I know D = R*T and I have the following data: Time (s) 0 5 10 15 Speed (ft/s) 0 20 25 10 How can I estimate the distance I ve travelled?
Using Left Endpoints: 0 * 5 + 20 * 5 + 25 * 5 = 225 ft Using Right Endpoints: 20 * 5 + 25 * 5 + 10 * 5 = 275 ft 1) How can we improve our estimate? 2) Do we know whether the true distance lies between the two estimates?
5.2 The Definite Integral The limit of the sum from 5.1: lim f x x + f x # x + + f(x - ) x = lim - / - / - :; f(x : ) x where x = = -, can be written as: f x dx and is called the definite integral of f from a to b, as long as the limit exists and is the same for all choices of x i. If the limit exists, then f is said to be integrable over [a,b] Properties of the Integral 1) c dx = c(b a) where c is any constant 2) f x + g x dx = f x dx + g x dx 3) c f x dx = c f x dx where c is any constant 4) [ f x g x ]dx E 5) f x dx + f x dx E = f x dx = f x dx 6) If f(x) 0 for a x b, then f x dx 7) If f(x) g(x) for a x b, then f x dx g x dx 8) If m f(x) M for a x b, then m (b a) f x dx M(b a) - g x dx 0
Ex: (#33) The graph of f is shown. Evaluate each integral by interpreting it in terms of areas. # a) f x dx d) f x dx K
5.3 The Fundamental Theorem of Calculus (Cue exciting music) The Fundamental Theorem of Calculus, Part I If f is continuous on [a,b], then the function g defined by g(x) = M f t dt a x b is continuous on [a,b] and differentiable on (a,b), and g (x) = f(x). Ex: Find N NM M R sec t dt (Hint: You ll need the Chain Rule) The Fundamental Theorem of Calculus, Part 2 If f is continuous on [a, b], then f x dx = F(b) F(a) where F is any antiderivative of f. That is, a function s. t. F = f. Why can it be *any* antiderivative of f?
So, differentiation and integration are inverse processes. Ex: Evaluate these integrals 1) 1 + x + 3x # dx % 2) 4 t t dt Note: Be careful when rushing in to evaluate integrals. Look at this example: $ = 1 x # dx It seems like it would be easy to find an antiderivative and evaluate it, but first is this function continuous over [-1, 3]?
5.4 Indefinite Integrals Definite vs. Indefinite Integrals: $ 2x dx number: is a definite integral. It s a string of symbols that represent a $ 2x dx = x 2 ] 0 3 = 3 2 0 2 = 9 2x dx is an indefinite integral. It s a string of symbols that represent a function, or family of functions: 2x dx = x 2 + C Table of Indefinite Integrals a f x dx = a f(x) dx [f(x) + g(x)] dx = f x dx + g x dx kdx = kx + C x n dx = MXYZ -[ + C (n -1) e x dx = e x + C M dx = ln x + C cos x dx = sinx + C sin x dx = - cos x + C sec 2 x dx = tan x + C sec x tan x dx = sec x + C
Net Change Theorem The integral of a rate of change is the net change F ] x dx = F b F(a)
5.5 The Substitution Rule Ex: a) Find N NM (x# + 3x) $ b) Find x # + 3x # 6x + 9 dx The Substitution Rule If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then f g x g ] x dx = f u du Ex: Evaluate 2x + 1 dx using the Substitution Rule
Ex: Evaluate 1 + x # x & dx using the Substitution Rule Integrals of Symmetric Functions If f is even [ f(-x) = f(x) ], then If f is odd [ f(-x) = -f(x) ], then = = f x dx f x dx = 0 = 2 f x dx