Integrals, areas, Riemann sums

Transcription

1 Itegrals, areas, Riema sums October 5, 2017 Itegrals, areas, Riema sums

2 We had the ith breakfast yesterday morig: There are still lots of slots available Breakfast #10, ext Moday (October 9) at 9AM. Itegrals, areas, Riema sums

3 Siged area Fid the area eclosed betwee the cubic y = x 3 x 2 2x ad the x-axis from x = 1 to x = y = x 3 x 2 2x Pro-tip: the aswer is ot 2 1 (x 3 x 2 2x) dx! Itegrals, areas, Riema sums

4 The itegral 2 1 (x 3 x 2 2x) dx adds the positive area betwee 1 ad 0 to the egative area betwee 0 ad 2, thereby gettig the icorrect aswer 9/4. The correct aswer may be writte as 2 1 x 3 x 2 2x dx, but that s ot especially helpful because we ca t itegrate absolute values very well. The best move is to divide the regio of itegratio ito the two segmets [ 1, 0] ad [0, 2]. Itegrals, areas, Riema sums

5 The positive area the becomes 0 1 (x 3 x 2 2x) dx 2 0 (x 3 x 2 2x) dx = 2F(0) F( 1) F(2), where F is a atiderivative of x 3 x 2 2x, say F(x) = x 4 /4 x 3 /3 x 2. With this choice, The total area is F (0) = 0, F( 1) = 5/12, F(2) = 8/3. 2F(0) F( 1) F(2) = 5/12 + 8/3 = 37/12. Itegrals, areas, Riema sums

6 How is area actually defied? Area is a limit of Riema sums. To defie b a [a, b] ito equal pieces [a, a + b a f (x) dx: Choose a iteger 1 ad divide up ], [a + b a, a + 2 b a ],... [a + ( 1) b a, b]. Each iterval has legth x = (b a)/. The last edpoit b is a + b a. Itegrals, areas, Riema sums

7 There are itervals. Choose x 1 i the first iterval, x 2 i the secod iterval, etc. The Riema sum attached to these choices is b a (f (x 1 ) + f (x 2 ) + + f (x )). It s the sum of the areas of rectagles, each havig base b a. The heights of the rectagles are f (x 1), f (x 2 ),..., f (x ). The Riema sum is a approximatio to the true area. As ad the rectagles get thier, the approximatio gets better ad better. Itegrals, areas, Riema sums

8 b The itegral. a f (x) dx is the limit of the Riema sums as The choices of the poits x i i the itervals is irrelevat. It is most commo to take the x i to be the left- or the right-edpoits of the itervals. Oe could take them to be i the middle of the itervals. Itegrals, areas, Riema sums

9 A simple example To see 0 x dx as a limit of Riema sums, divide the iterval [0, 1] ito equal pieces ad let the x i be the right edpoits of the resultig small itervals: The Riema sum is 1 x 1 = 1, x 2 = 2,..., x =. ( ) = 1 ( ) 2 = 1 2 ( + 1) Itegrals, areas, Riema sums

10 We used that the arithmetic progressio has ( + 1) sum, a fact that ca be explaied easily o the 2 documet camera. This is a completely silly way to fid the area of a right triagle with base ad height both equal to 1. Itegrals, areas, Riema sums

11 We used that the arithmetic progressio has ( + 1) sum, a fact that ca be explaied easily o the 2 documet camera. This is a completely silly way to fid the area of a right triagle with base ad height both equal to 1. Itegrals, areas, Riema sums

12 To fid 0 x 2 dx, we d eed to kow the formula = ( + 1)(2 + 1). 6 There are similar formulas for the sum of the kth powers of the first itegers, though kowig the full formulas is ot ecessary for computig the limits of the Riema sums. The Fudametal Theorem of Calculus just tells us that x k dx = 1 for k 1, so we do t eed explicit formulas 0 k + 1 to compute itegrals. Itegrals, areas, Riema sums

13 To fid 0 x 2 dx, we d eed to kow the formula = ( + 1)(2 + 1). 6 There are similar formulas for the sum of the kth powers of the first itegers, though kowig the full formulas is ot ecessary for computig the limits of the Riema sums. The Fudametal Theorem of Calculus just tells us that x k dx = 1 for k 1, so we do t eed explicit formulas 0 k + 1 to compute itegrals. Itegrals, areas, Riema sums

14 A 2016 midterm problem Express 1 cos x dx as a limit of Riema sums. This problem came from the textbook: it s #12 of 5.3 with the absolute value sigs removed (to make the problem easier). There is o sigle correct aswer because the user (you) gets to choose the poits x i. Divide the iterval [ 1, 1] ito equal segmets ad use left edpoits for the x i. The itervals have legth 2, so the Riema sum with pieces is 2 1 ( cos 1 + 2i ). i=0 The itegral is the limit of this sum as approaches. Itegrals, areas, Riema sums

15 A 2016 midterm problem Express 1 cos x dx as a limit of Riema sums. This problem came from the textbook: it s #12 of 5.3 with the absolute value sigs removed (to make the problem easier). There is o sigle correct aswer because the user (you) gets to choose the poits x i. Divide the iterval [ 1, 1] ito equal segmets ad use left edpoits for the x i. The itervals have legth 2, so the Riema sum with pieces is 2 1 ( cos 1 + 2i ). i=0 The itegral is the limit of this sum as approaches. Itegrals, areas, Riema sums

16 A 2016 midterm problem Express 1 cos x dx as a limit of Riema sums. This problem came from the textbook: it s #12 of 5.3 with the absolute value sigs removed (to make the problem easier). There is o sigle correct aswer because the user (you) gets to choose the poits x i. Divide the iterval [ 1, 1] ito equal segmets ad use left edpoits for the x i. The itervals have legth 2, so the Riema sum with pieces is 2 1 ( cos 1 + 2i ). i=0 The itegral is the limit of this sum as approaches. Itegrals, areas, Riema sums

17 A 2016 midterm problem Express lim i=1 ( 1 2i ) ( ) 2 i the form This is problem 4 of 5.3 of the textbook. 0 f (x) dx. We ca write the expressio before takig the limit as 2 i=1 ( 1 2i ). This looks like a Riema sum for a iterval of itegratio of legth 2. Because you re asked to shoehor the problem ito a itegral 0 dx, the problem is challegig. Itegrals, areas, Riema sums

18 A 2016 midterm problem Express lim i=1 ( 1 2i ) ( ) 2 i the form This is problem 4 of 5.3 of the textbook. 0 f (x) dx. We ca write the expressio before takig the limit as 2 i=1 ( 1 2i ). This looks like a Riema sum for a iterval of itegratio of legth 2. Because you re asked to shoehor the problem ito a itegral 0 dx, the problem is challegig. Itegrals, areas, Riema sums

19 It helps to write the sum as 1 i=1 ( 2 1 2i ), to make the x term ito the expected 1. To have 1 i=1 ( 2 1 2i ) = 1 we take f (x) = 2(1 2x) = 2 4x. f ( i ), i=1 Itegrals, areas, Riema sums

20 Yet more challegig Evaluate the limit lim Oce we write the limit as i=1 ( 1 2i 0 ) ( ) 2. itegral ad be fiished. The itegral is (2 4x) dx, we ca evaluate the ] 1 (2x 2x 2 ) = 0. 0 ( This is plausible because the terms 1 2i ) are positive for i small ad egative for i ear. The first term i the paretheses is 1 2 > 0 (for > 2) ad the last term is 1. Apparetly there s cacellatio! Itegrals, areas, Riema sums

21 Substitutio The chai rule states: d dx (F(u)) = F (u) du dx. Thus, i the world of atiderivatives: F (u) du dx = F(u) + C. dx It is atural to cacel the two factors dx ad write this as F (u) du = F(u) + C. Further, if F is give as a fuctio f ad F is itroduced as a atiderivative of f, the we have the formula f (u) du = F(u) + C, where F is a atiderivative of f. Itegrals, areas, Riema sums

22 This makes sese after we do examples: Evaluate cos(x 2 )2x dx. It s up to us to itroduce u, so we set u = x 2, du dx = 2x, du = 2x dx. I terms of u, the itegral to be evaluated is cos u du = si(u) + C = si(x 2 ) + C. I other words, we computed si(x 2 ) as a atiderivative of 2x cos(x 2 ). Coclusio: if you eed to evaluate a idefiite itegral ad ca t see the atiderivative immediately, try to make the itegrate simpler by a judicious substituio u = (some fuctio of x). Itegrals, areas, Riema sums

23 This makes sese after we do examples: Evaluate cos(x 2 )2x dx. It s up to us to itroduce u, so we set u = x 2, du dx = 2x, du = 2x dx. I terms of u, the itegral to be evaluated is cos u du = si(u) + C = si(x 2 ) + C. I other words, we computed si(x 2 ) as a atiderivative of 2x cos(x 2 ). Coclusio: if you eed to evaluate a idefiite itegral ad ca t see the atiderivative immediately, try to make the itegrate simpler by a judicious substituio u = (some fuctio of x). Itegrals, areas, Riema sums

24 This makes sese after we do examples: Evaluate cos(x 2 )2x dx. It s up to us to itroduce u, so we set u = x 2, du dx = 2x, du = 2x dx. I terms of u, the itegral to be evaluated is cos u du = si(u) + C = si(x 2 ) + C. I other words, we computed si(x 2 ) as a atiderivative of 2x cos(x 2 ). Coclusio: if you eed to evaluate a idefiite itegral ad ca t see the atiderivative immediately, try to make the itegrate simpler by a judicious substituio u = (some fuctio of x). Itegrals, areas, Riema sums

25 It would be more commo to ecouter the idefiite itegral cos(x 2 )x dx; the factor 2 has disappeared. Agai, we set u = x 2 ad write du = 2x dx, x dx = 1 du. I terms of u, the itegral becomes 2 cos u 1 si(u) du = + C = si(x 2 ) + C Itegrals, areas, Riema sums