Week 1 Notes: 1) Riem Sum Aim: Compute Are Uder Grph Suppose we wt to fid out the re of grph, like the oe o the right: We wt to kow the re of the red re. Here re some wys to pproximte the re: We cut the grph ito segmets ( =, 1, ) ove. Drw rectgle whose width is the size of the segmet, d height eig the vlue of f(t), where t is the highest poit (or lowest poit) of the segmet. We the dd up the res of ll rectgles. These re ll pproximtio of the re uder the grph. Oviously, if we cut the grph y more segmets, our pproximtio of the re is etter. Now we write the ides i mthemticl wy: Suppose [, ] is closed itervl cotiig suitervls of equl legth Δx = [x 0, x 1 ], [x 1, x ],, [x 1, x ] with = x 0 d = x. The edpoits re x 0, x 1, x,, x 1, x of the suitervls re clled grid poits d they form regulr prtitio of [, ]. I geerl, the k-th grid poit is x k = + kδx For exmple, i our cse whe =, x 0 = 1, x = 5, Δx = 5 1 = 1, d ll the poits x k = 1 + k( 1 ) re grid poits. Here is the defiitio of Riem Sum it is pproximtio of the re of f(x) uder curve:
Defiitio: Give fuctio f(x) with domi [, ], which is divided ito suitervls of equl legth Δx. Suppose x k is poit o the k th -suitervl, i.e. x k 1 x k x k. A Riem sum with -suitervls is give y S = f(x 1 )Δx + f(x )Δx + + f(x )Δx, Equtio () Notice tht i the ove defiitio, x k c e y poit i the suitervl. We c ctully e more specific o which poit we should tke i the suitervl [x k 1, x k ] = [ + (k 1)Δx, + kδx]. Upper sum Tke the vlue x k such tht f(x k ) is mximum o the k-th suitervl [x k 1, x k ]. Lower sum - Tke the vlue x k such tht f(x k ) is miimum o the k-th suitervl [x k 1, x k ]. Mid-poit sum - Tke x k = x k 1 + x k = + (k 1 ) Δx, the mid-poit of the suitervl. Right sum Tke x k = x k = + kδx, the rightmost poit of the suitervl. ) Summtio Nottios Notice tht whe we write dow the Riem sum we ofte eed to write log strig of umers, e.g. if we wt to compute the re o the first pge usig suitervls, we eed Δx = 1 : A = 1 [f(1) + f (1 + 1 ) + f (1 + 0 1 ) + + f (1 + ) + f (1 + )] It is rther clumsy to write dow such sum term y term, therefore we wt to use the sigm ottio: x k k= = x + x +1 + x + + + x 1 + x For exmple, let x k = k, the 7 k = + 4 + 5 + + 7 k= Let x k = f( + k ), the f( + k ) = f ( 17 Therefore, we c rewrite our Riem sum formul s: ) + f (1) + f (19) + + f (4 ) Riem sum = f (x k )Δx = f (x k )
Ad MID POINT Riem sum = f ( + (k 1 ) Δx) Δx RIGHT Riem sum = f ( + k Δx)Δx I prticulr, the mid-poit Riem sum o the top of this pge c e writte s f ( + (k 1 ) Δx) Δx = 1 f (1 + (k 1 ) 1 ) = 1 + 15 f (k 1 ) = 1 [0.5( k + 15 ) + 1] 1 We will ofte eed to compute the sum explicitly. How c we do tht? Here re the tools: Costt multiple rule: c i = c i Additio rule: ( i + i ) = i + i Sum of costt: c = c Sum of squres of first itegers: k = ( + 1)( + 1) Sum of first itegers: k = ( + 1) Sum of cues of first itegers: k = ( + 1) 4 Exmple: Fid the formul of 1 + + + ( + 1) Solutio: Method : 1 + + + ( + 1) = [1 + + + + ( + ) ] [ + 4 + + ( + ) ] + = i (i) + +1 = i 4i + +1 = i 4 i = = +1 ( + )( + )(4 + 5) 4( + 1)( + )( + ) ( + 1)( + ) ( + 1)( + )( + 1) [(4 + 5) 4( + )] =
+1 1 + + + ( + 1) = (i 1) +1 = (4i 4i + 1) +1 = 4 i 4 i + 1 = +1 +1 4( + 1)( + )( + ) 4( + 1)( + ) + ( + 1) = + 1 ( + 1)( + )( + 1) [4 + + ] = Exmple : Let f(x) = x +. Usig the right sum with equl suitervls, estimte the re etwee x = 1, x =. Solutio: = 1, =, Δx = 1 = RIGHT Riem sum = f ( + k Δx)Δx = f (1 + k ) = [ (1 + k ) + ] = [ 1 + 4k + 4k + ] = 1 + k + k = () + ( + 1) + ( + 1)( + 1) ) Defiite Itegrls = + 4( + 1) + 4( + 1)( + 1) We ow kow how to use Riem sum to estimte re uder grph, d to use summtio to mke the expressio of Riem sum look icer. I this sectio, we will fid the EXACT re uder grph.
Before we move o, we eed to e precise o wht re we re estimtig. For exmple, wht hppes whe we estimte re of f(x) with egtive vlues? Therefore, wheever we write dow the expressio of Riem sum, we ctully estimte the followig: Defiitio: Cosider the regio R ouded y the grph of cotiuous fuctio f d the x-xis etwee x = d x =. The et re of R is the sum of the res of the prts of R tht lie ove the x-xis mius the sum of the res of the prts of R tht lie elow the x-xis o [, ]. For exmple, the et re of the fuctio o the right etwee [, ] is give y the red re mius the lue re. As we hve oserved, if we tke, i.e. we mke more d more suitervls with ech itervl Δx 0, we c get the ccurte NET re uder grph: Defiitio: A fuctio f defied [, ] is itegrle o [, ] if the limit lim f (x k )Δx k exists for y prtitios Δx k of [, ] d y x k o the k-th itervl. This limit is clled the defiite itegrl of f from to, d is deoted s f(x)dx = lim f (x k )Δx k Do ot mess up f(x) dx (defiite itegrl, clcultig et re uder grph) d f(x) dx (ti-derivtive, the iverse of differetitio) Sice the defiitio sys we c use y prtitio d y x k, we will use the right sum to compute the defiite itegrl f(x)dx ( ) = lim f ( + k Δx)Δx = lim f ( + k ) ( ) Exmple: (Fidig defiite itegrl usig the mthemticl defiitio) Compute the re of f(x) = x + uder grph, etwee x = 1, x =.
Solutio: We use the rightmost poit of ech suitervl to estimte the re, d tke limits to get the exct re. 4) Properties of Defiite Itegrl Theorem: x + dx = lim f (1 + k 1 ) ( ) 4( + 1) = lim [ + + 4( + 1)( + 1) ] = lim [ + 4 + 4 + 4( + + 1) ] = + 4 + 0 + = 44 If f is cotiuous o [, ] or ouded o [, ] with fiite umer of discotiuities, the f is itegrle. For exmple, for the grph o the right, f(x) dx = red re lue re Let f d g e itegrle fuctios o itervl cotiig,, d c. We hve the followig properties f(x) dx = 0 [f(x) ± g(x)] dx = f(x) dx ± g(x) dx f(x) k f(x) dx = f(x) dx dx = k f(x) dx (k y costt) f(x) c dx = f(x) dx + f(x) dx c ( mi f(x)) ( ) f(x) dx ( mx f(x)) ( ) x [,] x [,] If f(x) g(x) o [, ], the f(x) dx g(x) dx