Chapter 5 Integrals 51 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 xaxis? 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: HT µ flat bit Area f(a) (b a) How can Txtk 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: % n 8 ~ n 't Using left endpoints, we get: f(0) " + f " I ax I ax " + f # " " + f $ " " + f % " " + f & " We can see this will be a lower bound, because all the rectangles fit completely under the curve " + f ' " " + f ( " " 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: 02734375 A 03984375 Left Right 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 / 0 lim / 3 4 4 + 3 4 # 4 + + 3(4 ) 4 " # " " " * 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 xvalue 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:,* F) ax A lim :; 3(4 : ) 4 (Riemann sum) / hig fcxda If 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? More data points smaller ax 2) Do we know whether the true distance lies between the two estimates? ## non "
52 The Definite Integral The limit of the sum from 51: lim 3 4 4 + 3 4 # 4 + + 3(4 ) 4 lim :; 3(4 : ) 4 / / where 4, can be written as: 3 4?4 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] fsum as i 1,33, n Properties of the Integral 1) @?4 c(b a) where c is any constant 2) 3 4 + A 4?4 3 4?4 + A 4?4 3) @ 3 4?4 c 3 4?4 where c is any constant 4) [ 3 4 A 4 ]?4 E 5) 3 4?4 + 3 4?4 E 3 4?4 3 4?4 6) If f(x) 0 for a x b, then 3 4?4 7) If f(x) g(x) for a x b, then 3 4?4 A 4?4 8) If m f(x) M for a x b, then m (b a) 3 4?4 G(H I) A 4?4 0
Ex: (#33) The graph of f is shown Evaluate each integral by interpreting it in terms of areas 5 # a) 3 4?4 J in, P8f It 'z 4 + 3 K d) 3 4?4 J I pain n a y (10) + ( 8 ) 2
53 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 M g(x) 3 L?L 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 L?L ( Ex 4) (Hint: You ll need the Chain Rule), f * sect dt see ( x 4) 4 3 The Fundamental Theorem of Calculus, Part 2 If f is continuous on [a, b], then 3 4?4 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 + 4 + 34 #?4 J x+iz+ II ( it ' +P ) (0+0*03) ( z 's ) 02 'EorE % 2) 4 L L?L J ',ij f I I l f fo4@rttfddteoett5olgezeydtxeiyiieeieiheiaii Z+l3ztEE 8 32 6 6513,250*2 Note: Be careful when rushing in to evaluate integrals Look at this example: 12815 $ 1 4 #?4 [ xtdx It seems like it would be easy to find an antiderivative and evaluate it, but first is this function continuous over [1, 3]? ifengto
54 Indefinite Integrals Definite vs Indefinite Integrals: $ 24?4 J number: $ 24?4 J is a definite integral It s a string of symbols that represent a x 2 ] 0 3 3 2 0 2 9 24?4 is an indefinite integral It s a string of symbols that represent a function, or family of functions: 24?4 x 2 + C t ta o Table of Indefinite Integrals I 3 4?4 a 3(4)?4 [f(x) + g(x)] dx 3 4?4 + A 4?4 W?4 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 \ ] 4?4 \ H \(I) gab ( Position ) # Speed
55 The Substitution Rule Ex: a) Find N NM (4# + 34) $ x2t3x5 (2 +3) ( x2t3 ) tci#6et9 ) Sf ), b) Find 4 # + 34 # 64 + 9?4 (#D3tC v What if we didn't just ] (12+3 56 +9) do porta? f( '+3 56 +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 3 A 4 A ] 4?4 3 `?` Let U 43X du( 2 +31? f(u)23du ' 033 i(x43d'tc Itc Ex: Evaluate 24 + 1?4 using the Substitution Rule Let u 2 +1 ( 2 ) duzdx d dx fo ±Su due ±+C u c±u +c (zx+de+c
Ex : ) ( 2+5x ) 2 ( zxt 5) dx Let u 2t5X du (2 +5) dx f u2du Is + C ( 45x)3 3 + C ex : SEIF St Let ux2t4 du Zxdx fjtdy d xdx
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Ex: Evaluate 1 + # # J # & )# using the Substitution Rule Let u1+ 2 du Zxdx U 1 2 df d Fttx5dx FxdxYxdx fhf(ud' {F0?zu+D Integrals of Symmetric Functions If f is even [ f(x) f(x) ], then * # )# If f is odd [ f(x) f(x) ], then * # )# 2 * # )# J 0
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