So there are two points of intersection, one being x = 0, y = 0 2 = 0 and the other being x = 2, y = 2 2 = 4. y = x 2 (2,4)

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1 Ares The motivtion for our definition of integrl ws the problem of finding the re between some curve nd the is for running between two specified vlues. We pproimted the region b union of thin rectngles nd found the ect re b tking the limit s the width of the pproimting rectngles went to zero. We cn use ectl the sme strteg to find res of more complicted regions in the plne. Here re some emples. Emple Find the re of the finite region bounded b nd 6. Solution. Our first tsk is to get good ide of wht the region looks like b sketching it. The curve is prbol. The point on this prbol with the smllest coordinte is (, ). As increses, increses so the prbol opens upwrd. The curve 6 ( 3) ( 3 ) + 9 is lso prbol. The point on this prbol with the lrgest vlue of is ( 3 /, 9 /). As moves w from, either to the right or to the left, decreses. So the prbol opens downwrd. The two prbols intersect when () So there re two points of intersection, one being, nd the other being, 4. This leds us to the sketch (,4) 6 (,) We re to find the re of the shded region. Ech point (,) in this shded region hs nd 6. When we were defining the integrl we used to denote the smllest llowed vlue of nd b to denote the lrgest llowed vlue of. Let s do tht here too. Let s lso use B() ( B stnds for bottom ) to denote the smllest llowed vlue of, when the coordinte is, nd use T() ( T stnds for top ) to denote the lrgest llowed vlue of, when the coordinte is. So, in this emple, b B() T() 6 c Joel Feldmn. 5. All rights reserved. Februr 4, 5

2 nd the shded region is { (,) b, B() T() } To find the re of this region, we use the sme strteg s we used when defining the integrl in 3. of the notes Definition of the Integrl. We pick nturl number n (tht we will lter send to infinit) nd subdivide the region into n nrrow slices, ech of width b. We pproimte the re of slice number i b the re of rectngle. The n rectngle is the ft drk verticl line in the figure below. On slice number i, the coordinte runs over ver nrrow rnge. (When we were defining the integrl we used i to denote the smllest on slice i nd we used i to denote the lrgest on slice i.) We pick number i, somewhere in tht rnge. We pproimte slice i b rectngle whose bottom is t B( i) nd whose top is t T( i). So we pproimte slice number i b rectngle of width nd of height T( i)b( i) nd so of re T( i )B( i )]. B() T( i) B( i ) T() 6 i b The desired re is lim n n T( i )B( i )] i 4 b T()B() ] d (6 )( ) ] d ] 6 d ] 3() 3 Emple c Joel Feldmn. 5. All rights reserved. Februr 4, 5

3 Emple Find the re of the finite region bounded b +6 nd. Solution. Agin we strt b sketching the region. The curve +6, or equivlentl is prbol. The point on this prbol with the smllest coordinte is (, ). As increses, increses so the prbol opens to the right. The curve is stright line of slope tht psses through,. The two curves intersect when + ( 6 ) 8 ( 4)( +) So there re two points of intersection, one being 4, 4+ 5 nd the other being, +. This leds us to the sketch (, ) (, ) +6 We cn find the re of this region b pproimting it b union of nrrow verticl rectngles, s we did in Emple. But tht s the hrd w. The es w is to pproimte it b union of nrrow horizontl rectngles. Just for prctice, here is the hrd solution. The es solution is fter it. Hrd Solution. We pproimte the region b union of nrrow verticl rectngles, ech of width. Two of those rectngles re illustrted in the sketch (, ) (, ) +6 c Joel Feldmn. 5. All rights reserved. 3 Februr 4, 5

4 In this region, runs from to b 5. The curve t the top of the region is T ( ) +6 The curve t the bottom of the region is more complicted. To the left of (,) the lower hlf of the prbol gives the bottom of the region while to the right of (,) the stright line gives the bottom of the region. So { +6 if B() if 5 The re is still given b the formul b ] T()B() d of Emple. But to ccomodte our B(), we hve to split up the domin of integrtion when we evlute the integrl. b ] ] 5 ] T()B() d T()B() d+ T()B() d ] 5 ] +6( +6) d+ +6() d d+ +6 d () d We cn evlute the first two integrls b substituting t + 6, dt d, t when, t 4 when nd t 6 when 5. Are 4 t 3 t dt ] 4 t + t dt 5 () d ] 6 ] 5 4 8] ] ( 5 5 ) ( + )] Es Solution. The es w to del to determine the re of our region is to pproimte b nrrow horizontl rectngles, rther thn nrrow verticl rectngles. Ech point (, ) in our region hs 4 nd ( 6) +. Let s use c to denote the smllest llowed vlue of, d to denote the lrgest llowed vlue of L() ( L stnds for left ) to denote the smllest llowed vlue of, when the coordinte is, nd R() ( R stnds for right ) to denote the lrgest llowed vlue of, when the coordinte is. c Joel Feldmn. 5. All rights reserved. 4 Februr 4, 5

5 So, in this emple, c d 4 L() ( 6) R() + nd the shded region is { (,) c d, L() R() } We pick nturl number n (tht we will lter send to infinit) nd subdivide the intervl c d into n nrrow subintervls, ech of width dc. We pproimte the re of n slice number i b the re of rectngle. The rectngle is the ft drk horizontl line in the figure below. On slice number i, the coordinte runs over ver nrrow rnge. We pick number i, somewhere in tht rnge. We pproimte slice i b rectngle whose left side is t L(i ) nd whose right side is t R( i ). So we pproimte slice number i b rectngle of height nd of width R(i )L( i ) nd so of re R(i)L( i) ]. (, ) (, ) R() + L( i ) R( i ) i L() ( 6) The desired re is n R( i )L(i )] lim n i d c 4 R()L() ] d ] d 6 (64+8)+ (64)+4(4+) ( +) ( 6 )] d ] 4 Emple c Joel Feldmn. 5. All rights reserved. 5 Februr 4, 5

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