Section 13.3 Area and the Definite Integral

Transcription

1 Sectio 3.3 Area ad the Defiite Itegral 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 ( 0.87 < A < ( ( ( 3 ( 3 = 3 = = 7 3 = 0.87

2 If we repeat this procedure with 8 strips, we get a better approximatio: ad < A < More strips we use, better approximatio we get: We summarize this i the followig table: REMARK: Oe ca show that the area uder the parabola f(x) = x from 0 to is /3 (see Appedix). 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.

3 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 () = L = f (0) ( ) f f ()+ f ( ) 3 = M = ( ) f ( ) 3 f ( ) f ( ) 7 f = = 3.7 = =.6 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 =.87, L 8 =.87, M 8 =.66. 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 x +6x+ from 0 to. Solutio: We first draw pictures: We have R = ( ) f + ( ) 8 f + ( ) f + ( ) 6 f + ( 0 f = 8.96 L = f (0)+ ( ) f + ( ) 8 f + ( ) f + ( ) 6 f =.6 M = ( f + ( f + ) + ( f + 8 ) + ( f ) + ( f 6 ) = ( f + ( ) 6 f + ( ) 0 f + ( ) f + ( ) 8 f =. For compariso s sake the exact area is A = 76 3 =.3. 3

4 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 L [f(x ) x+f(x ) x+...+f(x ) x] REMARK: It ca also be show that we get the same value if we use right edpoits A R [f(x ) x+f(x 3 ) x+...+f(x + ) x] DEFINITION OF A DEFINITE INTEGRAL: If f is a cotiuous fuctio defied o [a, b], the defiite itegral of f from a to b is a umber b a f(x)dx [f(x ) x+f(x ) x+...+f(x ) x] EXAMPLE: Approximate x dx, which is the area uder the graph of f(x) = x, above the 0 x-axis, ad betwee x = 0 ad x =, by usig four rectagles of equal width whose heights are the values of the fuctio at the left edpoit of each rectagle. Solutio: I Example it was show that x dx 0 ( 0 ( ( 3 = 7 3 = 0.87 REMARK:Notethatthedefiitioofthedefiiteitegralisvalidevewhef(x)takesegative values. A defiite itegral ca be iterpreted as a et area, that is, a differece of areas: b a f(x)dx = A A where A is the area of the regio above the x-axis ad below the graph of f, ad A is the area of the regio below the x-axis ad above the graph of f.

5 Applicatios 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. 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) Velocity (mi/h) I order to have the time ad the velocity i cosistet uits, let s covert the velocity readigs to feet per secod ( mi/h = 80/3600 ft/s): Time(s) Velocity (ft/s 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 ( ft/s), the we obtai the approximate distace traveled durig the first five secods ft/s s = ft Similarly, durig the secod time iterval the velocity is approximately costat ad we take it to be the velocity whe t = s. So, our estimate for the distace traveled from t = s to t = 0 s is 3 ft/s s = ft If we add similar estimates for the other time itervals, we obtai a estimate for the total distace traveled: ( )+(3 )+(3 )+(3 )+(7 )+(6 ) = 3 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 )+(3 )+(3 )+(7 )+(6 )+( ) = ft I geeral, the total distace traveled durig the time iterval [a, b] is approximately v(t ) t+v(t ) t+...+v(t ) t where t = b a 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 the expressio above: b d [v(t ) t+v(t ) t+...+v(t ) t] = a v(x)dx = s(b) s(a)

6 EXAMPLE: The Figure below shows the graph of the fuctio that gives the rate of chage of the aual maiteace charges for a certai machie. The rate fuctio is icreasig because maiteace teds to cost more as the machie gets older. Estimate the total maiteace charges over the 0-year life of the machie. Solutio: This is the situatio described i the precedig box, with F(x) beig the maiteace cost fuctio ad f(x), whose graph is give, the rate-of-chage fuctio. The total maiteace charges are the total chage i F(x) from x = 0 to x = 0 that is, the area betwee the graph of the rate fuctio ad the x-axis from x = 0 to x = 0. We ca approximate this area by usig the shaded rectagles i the Figure above. For istace, the rectagle marked with a arrow has base (from year to year 3) ad height 70 (the rate of chage at x = ), so its area is 70 = 70. Similarly, each of the other rectagles has base ad height determied by the rate of chage at the begiig of the year. Cosequetly, we estimate the area to be the sum = 3,60 Hece, the total maiteace charges over the 0 years are at least $3,60. (The ushaded areas uder the rate graph have ot bee accouted for i this estimate). EXAMPLE: The margial cost (i dollars per sowboard) for maufacturig x sowboards is give by MC(x) = (x 600) 0 Fid the amout added to the total cost whe productio goes from 30 to 00 sow boards. Solutio: The amout added is the total chage i cost from x = 30 to x = 00. Sice the margial cost fuctio is the derivative of the cost fuctio, Total chage = (x 600 dx 0 Numerical itegratio shows that the total chage i cost from x = 30 to x = 00 is $68.. 6

7 Appedix 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 ( = = ( ( = It is kow that therefore Thus we have = 3( ) = (+)(+) 6 R = 3( ) = 3 (+)(+) 6 lim R (+)(+) ( ) 6 ( 3 ) 6 or = (+)(+) 6 lim R (+)(+) 6 6 ( )( ) + + ( 6 )( ) ( )( ) 6 = = 3 = 6 (+0) (+0) = 3 7