Areas ad Distaces We ca easily fid areas of certai geometric figures usig well-kow formulas: However, it is t easy to fid the area of a regio with curved sides: METHOD: To evaluate the area of the regio S we first approximate it by rectagles ad the we take the limit of the area of these rectagles as we icrease the umber of rectagles. EXAMPLE: Use rectagles to estimate the area uder the parabola f(x) = x from 0 to. Solutio: We first draw pictures: We have R = ( ) ( ad L = 0 + ( ) Thus ) ( ( 3 ) ) ( ) = ( ( ) 3 = 0 ( ) 0.875 < A < 0.6875 ( ) ) ( ) ( ) 3 ( 3 = 5 3 = 0.6875 ) = 7 3 = 0.875
If we repeat this procedure with 8 strips, we get a better approximatio: ad 0.73375 < A < 0.398375 More strips we use, better approximatio we get: We summarize this i the followig table: EXAMPLE: Use four rectagles with (a) right edpoits, (b) left edpoits, (c) middle poits to estimate the area uder the parabola y = x + from 0 to.
EXAMPLE: Use four rectagles with (a) right edpoits, (b) left edpoits, (c) middle poits to estimate the area uder the parabola y = x + from 0 to. Solutio: We first draw pictures: We have R = f ( ) f ()+ f ( ) 3 f () = 5 L = f (0) ( ) f f ()+ f ( ) 3 = 5 M = ( ) f ( ) 3 f ( ) 5 f ( ) 7 f + + 3 + 5 = 5.75 + + 3 = 3.75 = 7 6 5 6 6 65 6 =.65 For compariso s sake the exact area is A = 3 we get a better approximatio: =.6. If we repeat this procedure with 8 strips, ad R 8 = 5.875, L 8 =.875, M 8 =.6565. EXAMPLE: Use five rectagles with (a) right ed poits, (b) left edpoits, (c) middle poits to estimate the area uder the parabola y = x 3 5x +6x+5 from 0 to. Solutio: We first draw pictures: We have R 5 = ( ) 5 f + ( ) 8 5 5 f + ( ) 5 5 f + ( ) 6 5 5 f + ( ) 0 5 5 f = 8.96 5 L 5 = 5 f (0)+ ( ) 5 f + ( ) 8 5 5 f + ( ) 5 5 f + ( ) 6 5 5 f =.56 5 M 5 = ( ) 5 f + ( 5 5 f 5 + ) + ( 5 5 f 5 + 8 ) + ( 5 5 f 5 ) + ( 5 5 f 5 6 ) 5 = ( ) 5 f + ( ) 6 5 5 f + ( ) 0 5 5 f + ( ) 5 5 f + ( ) 8 5 5 f = 5. 5 For compariso s sake the exact area is A = 76 3 = 5.3. 3
EXAMPLE: For the regio S i Example, show that the sum of the areas of the upper approximatig rectagles approaches, that is 3 Solutio: We have R = ( ) lim R = 3 ( ) = 3 +... = + 3 + 3 +...+ 3 3 3 ( ) 3 +... ( ) = 3 3 3 3 +... 3 It is kow that therefore = 3( + +3 +...+ ) ++3+...+ = (+) + +3 +...+ = (+)(+) 6 3 + 3 +3 3 +...+ 3 = (+) R = + +3 +...+ ) = 3( (+)(+) 3 6 Thus we have lim R (+)(+) 6 +3+ 6 ( 6 + 3 6 ) 6 ( 3 ) 6 or = (+)(+) 6 lim R (+)(+) 6 6 ( )( ) + + ( 6 )( ) ( )( ) 6 = 3 +0+0 = 3 = 6 (+0) (+0) = 3
DEFINITION: The area A of the regio S that lies uder the graph of the cotiuous oegative fuctio f is the limit of the sum of the areas of approximatig rectagles: A R [f(x ) x+f(x ) x+...+f(x ) x] () REMARK : It ca be proved that the above limit always exists, sice we are assumig that f is cotiuous. REMARK : It ca also be show that we get the same value if we use left edpoits A L [f(x 0 ) x+f(x ) x+...+f(x ) x] () Moreover, istead of usig left edpoits or right edpoits, we could take the height of the ith rectagle to be the value of f at ay umber x i i the ith subiterval [x i,x i ]. We call the umbers x,x,...,x the sample poits. So a more geeral expressio for the area of S is A [f(x ) x+f(x ) x+...+f(x ) x] (3) NOTATION: f(x i ) x = f(x ) x+f(x ) x+...+f(x ) x Usig this otatio we ca rewrite (), (), ad (3) as A f(x i ) x A f(x i ) x i=0 A f(x i) x 5
EXAMPLE: Let A be the area of the regio that lies uder the graph of f(x) = x 3 x+ betwee x = 0 ad x =. Fid A usig the right edpoits. Solutio: We partitio [0, ] ito subitervals. We have Thus Therefore A x = b a = 0 = x =, x = 8,...,x i = i,...,x = = f(x i ) x f [ (i ) ] 3 i + [ 6i 3 3 ] 8i + ( ) i [ 6 (+) 8 3 (+) + [ 6 [ (i ) ] 3 i + [ 6i 3 8i ] 3 + [ 6 3 ] i 3 8 ] i+ [ 6(+) ( + ) ( 6 ) ] +8 = 6 6+8 = 56 6(+) ] +8 6
EXAMPLE: Let A be the area of the regio that lies uder the graph of f(x) = cosx betwee x = 0 ad x = π/. (a) Usig right edpoits, fid a expressio for A as a limit. Do ot evaluate the limit. (b) Estimate the area by takig the sample poits to be midpoits ad usig four subitervals. Solutio: (a) Sice a = 0 ad b = π/, the width of a subiterval is x = b a = π/ 0 = π So x = π, x = π, x 3 = 3π,...,x i = iπ,...,x = π = π The sum of the areas of the approximatig rectagles is ( ) iπ ( π ) R = f(x i ) x = cos From this ad () it follows that the area is A R ( ) iπ ( π ) ( π cos ) ( ) iπ cos (b) With = the subitervals of equal width x = π [ 8 are 0, π ] [ π, 8 8, π [ π, ], 3π 8 The midpoits of these subitervals are x = π 6, x = 3π 6, x 3 = 5π 6, x = 7π 6 Therefore the sum of the areas of the four approximatig rectagles is M = f(x ) x+f(x ) x+f(x 3) x+f(x ) x = [f(x )+f(x )+f(x 3)+f(x )] x = [ ( π cos +cos 6) ( ) 3π +cos 6 ( ) 5π +cos 6 ( )] 7π π 6 8.006 ] [ 3π, 8, π ]. 7
The Distace Problem THE DISTANCE PROBLEM: Fid the distace traveled by a object durig a certai time period if the velocity of the object is kow at all times. If the velocity remais costat, the the distace problem is easy to solve by meas of the formula distace = velocity time But if the velocity varies, it s ot easy to fid the distace traveled. We ivestigate the problem i the followig example. EXAMPLE: Suppose the odometer o our car is broke ad we wat to estimate the distace drive over a 30-secod time iterval. We take speedometer readigs every five secods ad record them i the followig table: Time(s) 0 5 0 5 0 5 30 Velocity (mi/h) 7 9 3 3 8 I order to have the time ad the velocity i cosistet uits, let s covert the velocity readigs to feet per secod ( mi/h = 580/3600 ft/s): Time(s) 0 5 0 5 0 5 30 Velocity (ft/s) 5 3 35 3 7 6 Durig the first five secods the velocity does t chage very much, so we ca estimate the distace traveled durig that time by assumig that the velocity is costat. If we take the velocity durig that time iterval to be the iitial velocity (5 ft/s), the we obtai the approximate distace traveled durig the first five secods 5 ft/s 5 s = 5 ft Similarly, durig the secod time iterval the velocity is approximately costat ad we take it to be the velocity whe t = 5 s. So, our estimate for the distace traveled from t = 5 s to t = 0 s is 3 ft/s 5 s = 55 ft If we add similar estimates for the other time itervals, we obtai a estimate for the total distace traveled: (5 5)+(3 5)+(35 5)+(3 5)+(7 5)+(6 5) = 35 ft We could just as well have used the velocity at the ed of each time period istead of the velocity at the begiig as our assumed costat velocity. The our estimate becomes (3 5)+(35 5)+(3 5)+(7 5)+(6 5)+( 5) = 5 ft 8
I geeral, suppose a object moves with velocity v = f(t), where a t b ad f(t) 0 (so the object always moves i the positive directio). We take velocity readigs at times t 0 (= a), t, t,...,t (= b) so that the velocity is approximately costat o each subiterval. If these times are equally spaced, the the time betwee cosecutive readigs is t = b a Durig the first time iterval the velocity is approximately f(t 0 ) ad so the distace traveled is approximately f(t 0 )t. Similarly, the distace traveled durig the secod time iterval is about f(t )t ad the total distace traveled durig the time iterval [a,b] is approximately f(t 0 ) t+f(t ) t+...+f(t ) t = f(t i ) t If we use the velocity at right edpoits istead of left edpoits, our estimate for the total distace becomes f(t ) t+f(t ) t+...+f(t ) t = f(t i ) t The more frequetly we measure the velocity, the more accurate our estimates become, so it seems plausible that the exact distace d traveled is the limit of such expressios: d f(t i ) t f(t i ) t 9
Notatio DEFINITION: a k = a m +a m+ +...+a k=m EXAMPLES: (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) a k = a +a +a 3 +a 7 b k = b 3 +b +b 5 +b 6 +b 7 k=3 7 b k = b 3 +b +b +b 0 +b +b +b 3 +b +b 5 +b 6 +b 7 k= 3 k = ++3+ = 0 3 (k ) = 0 + + = 0++ = 5 5 = ++++ = 0 5 ( ) = ( )+( )+( ) = 6 k=3 7 (k +) = +5+6+7+8 = 30 k=3 7 ( ) k (k +) = +5 6+7 8 = 6 k=3 k= = k 3 = 8 6 = 7 6 0
usig summatio otatio. 3 + 3 + 5 + 5 6 Solutio: We have 3 + 3 + 5 + 5 6 = k + 5 k + = k= k 3 k + = k=0 k + k +3 usig summatio otatio. 3 + +5 +6 +7 +8 Solutio: We have 3 + +5 +6 +7 +8 = 8 k = 6 (k +) k=3 usig summatio otatio. 3 5 5 +5 5 6 5 +7 5 8 5 Solutio: We have 3 5 5 +5 5 6 5 +7 5 8 5 = 8 ( ) k+ k 5 = 6 ( ) k+ (k +) 5 = 7 ( ) k (k +) 5 k=3 k= usig summatio otatio. a+a +a 3 +a +a 5 +a 6 Solutio: We have a+a +a 3 +a +a 5 +a 6 = 6 a k = 5 3 a k+ = k=0 k= a k+3 usig summatio otatio. a a +a 3 a +a 5 a 6 Solutio: We have a a +a 3 a +a 5 a 6 = 6 ( ) k+ a k = 9 ( ) k a k 3 k=
usig summatio otatio. a+a a 3 +a a 5 Solutio: We have a+a a 3 +a a 5 = 5 ( ) k a k = k=0 6 ( ) k+ a k usig summatio otatio. 7 0 + 8 3 + 9 0 3 5 Solutio: We have 7 0 + 8 3 + 9 0 3 5 = 0 k=7 k (k +3)(k +5) = k +6 (k +9)(k +)